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NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 Thermodynamics

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 Thermodynamics

Edited By Sumit Saini | Updated on Sep 10, 2022 05:16 PM IST

NCERT Exemplar Class 11 Chemistry solutions chapter 6 revolves around the importance of thermodynamics. Thermodynamics is the branch of science which deals with the relationship between heat and the other forms of energy. This chapter of NCERT Class 11 Chemistry Solutions deals with a variety of topics that explains the relationship between heat and other forms of energy. NCERT Exemplar Class 11 Chemistry chapter 6 solutions helps the students with detailed and explanative solutions to all the intext and exercise questions that are listed in the chapter.
Also, check - NCERT Solutions for Class 11

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: MCQ (Type 1)

Question:1

1) Thermodynamics is not concerned about.........
(i) energy changes involved in a chemical reaction.
(ii) the extent to which a chemical reaction proceeds.
(iii) the rate at which a reaction proceeds.
(iv) the feasibility of chemical reaction.
Answer:

The answer is option (iii) the rate at which a reaction proceeds.
Explanation: Thermodynamics is actually concerned with the initial and final states of a system and not with the rate at which the transformations take place. The laws can be applied to a system only when the system is in equilibrium.

Question:2

Which of the following statements is correct ?
(i) The presence of reacting species in a covered beaker is an example of open system.
(ii) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.
Answer:

The answer is option (iii) the presence of reactants in a closed vessel made up of copper is an example of a closed system.
Explanation: When the reactants are in a closed vessel, the matter cannot be exchanged. However, the energy may exchange through walls. So the reactants in a copper or a steel vessel are an example of a closed system.

Question:3

The state of gas can be described by quoting the relationship between _________ .
(i) pressure, volume, temperature
(ii) temperature, amount, pressure
(iii) amount, volume, temperature
(iv) pressure, volume, temperature, amount
Answer:

The answer is option (iv) pressure, volume, temperature, amount.
Explanation: The states of a gas can be described by the relationship between the variables such as p, V, T and the amount of gas. These variables are called state variables. Their values are only concerned with the current state of the system.

Question:4

The volume of gas is reduced to half from its original volume. The specific heat will ________ .
(i) reduce to half
(ii) be doubled
(iii) remain constant
(iv) increase four times
Answer:

The answer is option (iii) remain constant.
Explanation: The property of Specific heat capacity is an intensive property since it refers to the quantity of heat needed to elevate the temperature of one unit mass of a substance by one-degree Celsius. It depends on the volume of the substance.

Question:5

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
(i)
2C_{4}H_{10}(g)+13O_{2}(g)\rightarrow 8CO_{2}(g)+10H_{2}O_{(l)}\Delta _{c}H=-2658.0\; kJ\; mol^{-1}
(ii)
C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(g)\Delta _{c}H=-1329.0\; kJ\; mol^{-1}
(iii)
C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(l)\Delta _{c}H=-2658.0\; kJ\; mol^{-1}
(iv)
C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(g)\Delta _{c}H=+2658.0 kJ\; mol^{-1}
Answer:

The answer is the option (iii).
Explanation: The definition of the enthalpy of combustion is when the substance undergoes combustion and the enthalpy change that happens per mole in that substance. All other reactants are assumed to be in their standard states at that time.

Question:6

\Delta _{f}U^{o} of formation of CH_{4}(g) at a certain temperature is –393 kJ mol–1. The value of \Delta _{f}H^{\ominus } is
(i) zero
(ii) <\Delta _{f}U^{o}
(iii) >\Delta _{f}U^{o}
(iv) equal to \Delta _{f}U^{o}
Answer:

The answer is option (ii) <\Delta _{f}U^{o}
Explanation: The reaction for the formation of CH_{4} is:
CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O
\Delta n_{g}=1-3=-2
\Delta H^{o}=\Delta U^{o}+\Delta n_{g}RT
\Delta n_{g}=-2
Hence, \Delta H^{o}<\Delta _{f}U^{o}

Question:7

In an adiabatic process, no transfer of heat takes place between system and surroundings.Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.
(i) q = 0, \Delta T \neq 0, w = 0
(ii) q \neq 0, \Delta T = 0, w = 0
(iii) q = 0, \Delta T = 0, w = 0
(iv) q = 0, \Delta T < 0, w \neq 0
Answer:

The answer is option (iii) q = 0, \Delta T = 0, w = 0
Explanation: w =0 in the situation of free expansion since the volume is also constant. Als0, q=0 considering the process is adiabatic. The first law of thermodynamics state that \Delta U=q+w=0 which is a constant.

Question:8

The pressure-volume work for an ideal gas can be calculated by using the expression
w=-\int_{v_{i}}^{v_{f}}P_{ex}dV
The work can also be calculated from the pV - plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from Vi to Vf, choose the correct option.

(i) w (reversible) = w (irreversible)
(ii) w (reversible) < w (irreversible)
(iii) w (reversible) > w (irreversible)
(iv) w (reversible) = w (irreversible) + pex. \Delta V
Answer:

The answer is option (ii) w (reversible) < w (irreversible).
Explanation: In the situation of irreversible compression, the area under the curve is always larger comparativel

Question:9

The entropy change can be calculated by using the expression \Delta S = (q_{rev}/T).
When water freezes in a glass beaker, choose the correct statement amongst the following :

(i) \Delta S (system) decreases, but \Delta S (surroundings) remains the same.
(ii) \Delta S (system) increases but \Delta S (surroundings) decreases.
(iii) \Delta S (system) decreases, but \Delta S (surroundings) increases.
(iv) \Delta S (system) decreases, and \Delta S (surroundings) also decreases.
Answer:

The answer is option (iii) \Delta S (system) decreases, but \Delta S (surroundings) increases.
Explanation: As we all know in the case of an exothermic process, heat is released, which leads to a decrease in the surrounding entropy and as a result, the system entropy increases. As freezing is an exothermic process, we choose this option.

Question:10

On the basis of thermochemical equations (a), (b) and (c), find out which of the algebraic relationships given in options (i) to (iv) is correct.
(a) C (graphite)+O_{2}(g)\rightarrow CO_{2}(g);\Delta _{r}H=x\; kJ\; mol^{-1}
(b) C (graphite)+\frac{1}{2}O_{2}(g)\rightarrow CO(g);\Delta _{r}H=y\; kJ\; mol^{-1}
(c) CO _{(g)} + \frac{1}{2}O_{2(g)}\rightarrow CO_{2(g)} ; \Delta _{r}H= z\; kJ\; mol^{-1}

(i) z = x + y
(ii) x = y – z
(iii) x = y + z
(iv) y = 2z – x

Answer:

The answer is option (iii) x = y + z
x =y + z is the correct option as Algebraic sum of y and z will give x.
Explanation: When y is subtracted from x, we get z. Explained below:
C (graphite)+O_{2}(g)\rightarrow CO_{2}(g);\Delta _{r}H=x\; kJ\; mol^{-1}________(i)
C (graphite)+\frac{1}{2}O_{2}(g)\rightarrow CO(g);\Delta _{r}H=y\; kJ\; mol^{-1}________(ii)
When subtracted , we receive:
CO _{(g)} + \frac{1}{2}O_{2(g)}\rightarrow CO_{2(g)} ; \Delta _{r}H= z\; kJ\; mol^{-1}________(iii)
So, z= x-y, or, easily put, x=y+z. Hence, option c.

Question:11

Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct?
(i) C(g) + 4H(g) \rightarrow CH_{4}(g); \Delta _{r}H=x\; kJ \; mol^{-1}
(ii)
C(graphite, s) + 2H_{2}(g) \rightarrow CH_{4}(g); \Delta _{r}H = y \; kJ \; mol ^{-1}

(i) x = y
(ii) x = 2y
(iii) x >y
(iv) x< y
Answer:

The answer is option (iii).
Explanation: In the case of reaction b, the bonds are broken, but in the reaction a, the bonds are not broken. Although, same bonds are formed in both the cases.

Question:12

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(i) is always negative
(ii) is always positive
(iii) maybe positive or negative
(iv) is never negative
Answer:

The answer is option (iii) maybe positive or negative
Explanation: It can be positive or negative as the reaction could be exothermic or endothermic.

Question:13

Enthalpy of sublimation of a substance is equal to
(i) enthalpy of fusion + enthalpy of vapourisation
(ii) enthalpy of fusion
(iii) enthalpy of vapourisation
(iv) twice the enthalpy of vapourisation.
Answer:

The answer is option (i) enthalpy of fusion + enthalpy of vapourisation
Explanation: In case of enthalpy of sublimation, solid converts to liquid and then to gas. In this process, the conversion to liquid required enthalpy of fusion and the conversion to vapour requires the enthalpy of vapourisation.

Question:14

Which of the following is not correct?
(i) \Delta G is zero for a reversible reaction.
(ii) \Delta G is positive for a spontaneous reaction
(iii) \Delta G is negative for a spontaneous reaction
(iv) \Delta G is positive for a non-spontaneous reaction.
Answer:

The answer is option (ii) \Delta G is positive for a spontaneous reaction
Explanation: As we all know that for a spontaneous reaction, \Delta G is always negative.

NCERT Exemplar Class 11 Chemistry Solutions chapter 6: MCQ (Type 2)

Question:15

Thermodynamics mainly deals with
(i) interrelation of various forms of energy and their transformation front one form to another.
(ii) energy changes in the processes which depend only on initial and final states of the microscopic system containing a few molecules.
(iii) how and at what rate these energy transformations are carried out.
(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.
Answer:

The answer is option (i) and (iv)
Explanation: The prerequisite conditions required for the thermodynamics laws to be applicable is for the system to be in equilibrium. Also, they are relevant to the macroscopic systems and not to the microscopic systems which have very few numbers of molecules.

Question:16

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system
(i) q_{p} will be negative
(ii) \Delta _{r}H will be negative
(iii) q_{p} will be positive
(iv) \Delta _{r}H will be positive.
Answer:

The answer is options (i) and (ii)
Explanation: In an exothermic reaction, the heat is released to the surroundings of the system. So,q_{p} and \Delta _{r}H will be negative.

Question:17

The spontaneity means, having the potential to proceed without the assistance of an external agency. The processes which occur spontaneously are
(i) flow of heat from colder to warmer body.
(ii) gas in a container contracting into one comer.
(iii) gas expanding to fill the available volume.
(iv) burning carbon in oxygen to give carbon dioxide.
Answer:

The answer is option (iii) and (iv).
Explanation: Both the processes, namely, the filling of empty space by gas, as well as the burning of carbon which leads to a formation of carbon dioxide, are spontaneous processes.

Question:18

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression w = -nRT In V_{f} / V_{i}
A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.

(i) Work done at 600 K is 20 times the work done at 300 K.
(ii) Work done at 300 K is twice the work done at 600 K
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) \Delta U=0 in both cases.
Answer:

The answer is (iii) and (iv).
Explanation: Now, q = -w, in the case of an isothermal reversible change
Q=-w=nRT\; In\frac{vf}{vi}
\frac{w(600k)}{w(300k)}=\frac{1\times R\times 600k\; In\frac{10}{1}}{1\times R\times 300K\; In\frac{10}{1}}=2
In case of ideal gases, \Delta U = 0 for the expansion, which is isothermal. There is no change in the internal energy as the temperature is constant, and, hence \Delta U = 0.

Question:19

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
2Zn(s) + O_{2}(g) \rightarrow 2ZnO(s); \Delta H=-693.8 \; kJ \; mol^{-1}

(i) The enthalpy of two moles ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(iii) 8 kJ mol -1 energy is evolved in the reaction.
(iv) 693.8 kJ mol-1 energy is absorbed in the reaction.
Answer:

The answers is options (i) and (iii).
Explanation:
\Delta _{f}H=\sum _{i}a_{i}\Delta _{f}H-\sum _{i}b_{i}\Delta _{f}H
The reaction depicted above is exothermic in nature, and, hence, the reactant enthalpy is more than the product enthalpy.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Short Answer Type

Question:20

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?
Answer:

1 mol of water or 18g of water has an enthalpy of change for vaporisation. Now it is given that the change of enthalpy required for 1 mole of water is 40.79 kJ/mol.
For 2 moles of water, the enthalpy change will be 2 x 40.79 = 81.58 KJ . ThereforeThe standard enthalpy of vaporisation of water is \Delta _{vap} H = 40.79\; kJ / mol

Question:21

One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has a higher enthalpy of vapourisation?
Answer:

Water has a greater enthalpy of vaporisation than acetone. Acetone molecules have a weak force of attraction between them, and hence they require less heat to break the molecular bonds.

Question:22

Standard molar enthalpy of formation, \Delta _{f}H^{\ominus } is just a special case of enthalpy of reaction, \Delta _{r}H^{\ominus } . Is the \Delta _{r}H^{\ominus } for the following reaction same as \Delta _{f}H^{\ominus } ? Give a reason for your answer.
CaO(s) + CO_{2}(g) \rightarrow CaCO_{3}(s); \Delta _{f}H^{\ominus }=-178.3\; kJ mol^{-1}

Answer:

No, since CaCO3 has been formed from other compounds and not from its constituent elements.

Question:23

The value of \Delta _{f}H^{\ominus } for NH_{3} is – 91.8 kJ mol–1. Calculate the enthalpy change for the following reaction :
2NH_{3}(g) \rightarrow N_{2}(g) + 3H_{2}(g)

Answer:

The enthalpy change for the given reaction can be calculated as: -

N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3} (g)

\Delta _{f} H = -91.8\; kg/mol
The enthalpy of the reaction is +91.8 kg/mol. This happens as with the reversal of reaction, the value of\Delta _{r} H also gets reversed.

Question:24

Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is \Delta _{r}Hand \Delta _{r}H_{1}, \Delta _{r}H_{2},\Delta _{r}H_{3}….. represent enthalpies of intermediate reactions leading to product B. What will be the relation between \Delta _{r}H for overall reaction and \Delta _{r}H_{1},\Delta _{r}H_{2} etc. for intermediate reactions.
Answer:

According to Hess’s law, \Delta _{r} H = \Delta _{r} H_{1} + \Delta _{r} H_{2} +\Delta _{r} H_{3}.....
This is so because during the reaction A\rightarrow B, B’s formation undergoes various intermediate reactions, with the overall value of the enthalpy being \Delta _{r}H.

Question:25

The enthalpy of atomisation for the reaction CH_{4}(g)\rightarrow C(g) + 4H (g)\; is \; 1665 \; kJ mol^{-1}. What is the bond energy of C-H bond?
Answer:

The reaction presented in the question is
CH_{4}(g)\rightarrow C(g) + 4H (g)
Now, \Delta _{a} H = 1665 \; kJ / mol^{-1}
The mean bond enthalpy of the C-H bond should be used here. For the atomisation of 4 moles of C-H bonds, the value is 1665 kJ/mol-1 So, per mole energy = 1665/4 = 416.2 kJ / mol-1

Question:27

Given that \Delta H = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:

Entropy in an isolated system increases with the increase in the disorder in the system. As diffusion is a spontaneous process, the process also will be a spontaneous one.

Question:28

Heat has randomising influence on a system, and the temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
Answer:

Temperature and the entropy change are both inversely proportional quantities. Q and T are related to the entropy in case of reversible reaction in the following manner:
\Delta S=q_{rev}/T

Question:29

Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Answer:

Yes. The temperature of the surroundings and the system will be identical when we look at the condition of thermal equilibrium. The enthalpy varies in an inversely proportionate manner in case of system and surroundings. If the enthalpy of the system increases, the enthalpy of surroundings deceases.

Question:30

At 298 K, Kp for the reaction N_{2}O_{4}(g)\rightleftharpoons 2NO_{2} (g) is 0.98. Predict whether the reaction is spontaneous or not.
Answer:

As per the information given in the question, Kp for the above reaction is 0.98
\Delta _{y}G = \Delta _{y}H- T \Delta _{y}G = -RT\; ln K
For a process to be of spontaneous nature, \Delta G should be negative
As \Delta_{y} G=-RT\; ln K_{p}
Hence \Delta_{y} G has a negative connotation, and the reaction given is of spontaneous nature

Question:32

The standard molar entropy of H_{2}O(l) is 70 J K-1 mol-1. Will the standard molar entropy of H_{2}O(s) be more, or less than 70 J K-1 mol-1?
Answer:

The standard molar entropy of H2O (s) will be less than 70 JK-1mol-1. This is because entropy decreases when ice is formed from water as molecules reduce their movement and are arranged in an orderly fashion.

Question:33

Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy.
Answer:

Out of the options mentioned, the state functions are: Temperature, Enthalpy, Entropy, Free energy and the path functions are Work, Heat, etc.

Question:34

The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Answer:

As the intermolecular forces of attraction is lower in acetone than in water, the enthalpy of vaporisation of water is higher than that of acetone.

Question:35

Which quantity out of \Delta _{r}G and \Delta _{r}G^{\circleddash } will be zero at equilibrium?
Answer:

\Delta _{r}G will always be zero.
\Delta _{r}G^{\circleddash } is zero for K = 1 because \Delta _{r}G^{\circleddash }= – RT lnK, \Delta _{r}G^{\circleddash } will be non zero for other values of K.

Question:36

Predict the change in internal energy for an isolated system at constant volume.
Answer:

There is no change in internal energy for an isolated system at a constant volume if there is no transfer of heat as work which are the conditions of an isolated system. Hence, \Delta H=0.

Question:37

Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Answer:

The conditions in which heat becomes path independent are:
a) when the volume remains the same
\Delta V=0,so\; \Delta U=q+p\; \Delta V=q
As \Delta U is a state function, q is also a state function.
b) when the pressure remains the same
\Delta U=q-p\; \Delta V
So, q=p\; \Delta V+\Delta U=\Delta H which is a state function
So, at a constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions.

Question:38

Expansion of a gas in a vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?
Answer:

Free expansion is known as the expansion of gas in vacuum. The equation can be written as
(-w)=p(V_{2}-V_{1})=0\times (5-1)=0
q =0, when we consider isothermal expansion.
Referring to the first law of thermodynamics we have
q=\Delta U+(-w)
0=\Delta U+0
Hence, \Delta U=0

Question:39

Heat capacity (C_{p}) is an extensive property but specific heat (c) is an intensive property. What will be the relation between (C_{p}) and c for 1 mol of water?
Answer:

The quantity of heat needed to raise the temperature of one mole of the substance by one degree Celsius is known as its heat capacity.
For the molar heat capacity calculation, we multiply 18 and the value of specific heat
C_{p}=18\times c
We know that, C (specific heat) = 4.18\; J\; g^{-1}K^{-1} Heat capacity,
Therefore, C_{p}=18\times 4.18\; JK^{1}=75.24\; JK^{-1}

Question:40

The difference between C_{p} \; and \; C_{v} can be derived using the empirical relation H = U + pV. Calculate the difference between C_{p} \; and \; C_{v} for 10 moles of an ideal gas.
Answer:

In case of an ideal gas, the difference between C_{p}\; and \; C_{v} is nR, where where n represents the number of moles and R is the universal gas constant.
Therefore, for 10 moles of an ideal gas-
C_{p} - C_{v}= 10\; R
C_{p} - C_{v}= 10 \times 4.18\; J
= 41.84 J

Question:41

If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Answer:

We know that, 1 mole of carbon = 12g
To calculate the molar enthalpy, change of graphite, we need to multiply enthalpy change for 1g carbon along with its molar mass. Which as per the information provided in the question can be calculated as,
- 20.7 \times 12 = -248.4 \; kJ mol^{-1} (negative sign depicts exothermic reaction as there is heath involved)

Question:43

The enthalpy of vapourisation of CCl_{4} is 30.5 kJ mol–1 . Calculate the heat required for the vapourisation of 284 g of CCl_{4} at constant pressure. (Molar mass of CCl_{4} = 154 g mol–1).
Answer:

As per the information provided in the question, for one mole of CCl_{4} (154 g), the heat of vaporisation required is 30.5 kJ/mol .
Hence for the vaporisation of 284 g of CCl_{4}, we require:
= \frac{284 \; g}{154 \; g \; mol^{-1}} \times 30.5\; kJ\; mol^{-1}
=56.2\; kJ

Question:44

The enthalpy of reaction for the reaction:
2H_{2} (g) + O_{2}(g) \rightarrow 2H_{2}O(l) is \Delta_{r}H^{\circleddash }= - 572 kJ \; mol^{-1}. What will be standard enthalpy of formation of H_{2}O(l) ?

Answer:

Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol–1
Now the enthalpy of formation for H_{2}O will be half the enthalpy of the value in the given equation. So, now we can calculate that:
\Delta _{f}H=\frac{1}{2}\times \Delta _{y}H=-\frac{52}{2}=-286\; kJ\; mol^{-1}

Question:45

What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, p_{ext} in a single step as shown in Fig. 6.2. Explain graphically.

Answer:

Assumption: Cylinder is filled with one mole gas, and the piston is frictionless. Let the pressure of gas inside be p and the volume of gas be V_{I}.
Piston is moved towards the inside to make the external pressure (P_{ext}) equal to p. Now, let us assume that this change take place in a single step, hence, V is the final volume. The work done by the piston is depicted in the graph shown below by shading the area.
P_{ext}\Delta V=AV_{I}(or\; BV_{II})\times (V_{I}-V_{II})

Question:46

How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
Answer:

When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesimal amounts as compression happens from initial volume Vi to the final volume Vf. The below graph depicts the work done with the shaded area.

Question:47

Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b)
\frac{1}{2}H_{2}(g)+\frac{1}{2}Cl_{2}(g)\rightleftharpoons HCl (g) \; \Delta _{r}H^{\ominus }=-92.32\; kJ\; mol^{-1}
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

Answer:

(a) Throwing a stone from ground to roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:

In process b), potential energy/enthalpy change is contributing factor to the spontaneity.

Question:48

Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain.

Answer:

No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

Question:50

An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case?
(Given that 1 L bar = 100 J)

Answer:

We know that the amount of work done =-p_{ext}\Delta V
On substituting the values in the formula, we get,
-2\; bar\times (50-10)L=-80\; L\; bar
According to the described problem,1L\; Bar = 100\; J

Therefore, -80\; L\; bar=(-80\times 100)=-8000\; J

=-8\; kJ, which is the amount of work done
The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Matching Type

Question:51

Match the following:

A

B

(i) Adiabatic Process

(a) Heat

(ii) Isolated system

(b) At constant volume

(iii) Isothermal change

(c) First law of thermodynamics

(iv) Path function

(d) No exchange of energy and matter

(v) State function

(e) No transfer of heat

(vi) \Delta U=q

(f) Constant temperature

(vii)Law of conservation of energy

(g) Internal energy

(viii) Reversible process

(h)P_{ext}=0

(ix) Free expansion

(i) At constant pressure

(x) \Delta H=q

(j) Infinitely slow process which
proceeds through a series of
equilibrium states.

(xi) Intensive property

(k) Entropy

(xii) Extensive property

(l) Pressure


(m) Specific heat

Answer:

(i) - (e); (ii)- (d); (iii)- (f); (iv)- (a); (v)- (g), (k), (l); (vi)- (b); (vii)- (c); (viii)- (j); (ix) (h); (x)- (i); (xi)- (a), (l), (m); (xii)- (g), (k)

Question:52

Match the following processes with entropy change:

Reaction

Entropy change

(i) A liquid vapourises

(a) \Delta S = 0

(ii)Reaction is non-spontaneous at all temperatures and \Delta H is positive

(b) \Delta S = Positive

(iii) Reversible expansion of an ideal gas

\Delta S = negative

Answer:

(i) - b
(ii) - c
(iii) -a
Explanation:

Reaction

Entropy change

(i) A liquid vaporises

As a liquid change to gaseous state, the movement of molecules will increase and hence the entropy will also increase.

ii) Reaction is non-spontaneous at all temperatures and AH is positive.

ΔG is positive as the process in non-spontaneous. Also, ΔH is positive. Hence, ΔS is negative.

(iii) Reversible expansion of an ideal
gas

This process at every stage is always in equilibrium

Question:53

Match the following :

\Delta(Parameters) \Delta _{r}H^{\ominus } \Delta _{r}S^{\ominus } \Delta _{r}G^{\ominus }

Description

(i) +\; \; \; \; \; \; \; -\; \; \; \; \; \; \; +

(a) Non-spontaneous at high temperature.

(ii) -\; \; \; \; \; \; \; -\; \; \; \; \; \; \; +at\; high\; T

(b) Spontaneous at all temperatures

(iii) -\; \; \; \; \; \; \; +\; \; \; \; \; \; \; -

(c) Non-spontaneous at all temperatures

Answer:

(i) - c
(ii) - a
(iii) -b
Explanation:

Parameters \Delta _{r}H^{\ominus } \Delta _{r}S^{\ominus } \Delta _{r}G^{\ominus }

Description

i) + - +

ΔG is positive, and ΔS is negative when we have a positive enthalpy change. So, at all temperatures, the process remains non-spontaneous

ii) - - + at a hight T

when we have a positive enthalpy change, ΔG is positive. The process will be non-spontaneous at high temperatures.

iii) - + -

When enthalpy change is negative, ΔG is negative, and ΔS is positive. Hence the process will be spontaneous at all temperatures.

Question:54

Match the following :

(i) Entropy of vapourisation

(a) decreases

(ii) K for spontaneous process

(b) is always positive

(iii) Crystalline solid state

(c) lowest entropy

(iv) \Delta U in adiabatic expansion of ideal gas

(d) \frac{\Delta H_{vap}}{T_{b}}

Answer:

(i) - b, d
(ii) - b
(iii) -c
(iv) -a
Explanation:

Col I

Col II

(i) Entropy of vaporisation

ΔH vap / T

(ii) K for spontaneous process

Since\; \; \Delta ,G^{\ominus}=\Delta , H^{\ominus}-T\Delta ,S^{\ominus}=-RT

(iii) Crystalline solid state

Molecules in the solid state of matter are the most stable and ordered and hence have the least entropy

(iv) ΔU in adiabatic expansion of ideal gas

ΔU decreases as in the situation of expansion, the system is the one that does work. so, q =0 and ΔU = -w

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Assertion and Reason Type

Question:55

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R) : The enthalpies of all elements in their standard state are zero.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is option (ii) Both A and R are true but R is not the correct explanation of A.
Explanation: the reactant enthalpies are always greater than the product enthalpies when it comes to the combustion reaction.

Question:56

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R) : Decrease in enthalpy is a contributory factor for spontaneity.

(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:

The answer is option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: In the case of a spontaneous process, the randomness should be positive, and ΔU should be negative.

Question:57

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R) : In crystals, molecules organise in an ordered manner.

(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:

The answer is option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: As the liquid starts to crystallize the randomness in the molecules decreases as they get arranged in an orderly fashion. Hence, the entropy decreases.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Long Answer Type

Question:58

Derive the relationship between \Delta H and \Delta U for an ideal gas. Explain each term involved in the equation.
Answer:

Now, the heat absorbed at constant volume is equal to change in the internal Energy.
\Delta U = q_{p} - p\; \Delta V
qp is the heat absorbed.
p \Delta V is the expansion work done by the system
the equation can be represented as U_2 - U_1 = p (V_2 - V_1)
so, we get,
q_{p} = (U_2 + Pv_2) - (U_1 + Pv_1) ----------------- (1)
now we take into consideration, the enthalpy
H = U + pV -------------------------- (2)
Hence, we can rewrite the first equation as,
q_{p} = H_2 - H_1 = \Delta H
now, in case of the state functions.
H is a state function because it depends on U, p and V, all of which are state functions.
So, \Delta H, as well as qp, is path independent.
We can rewrite equation 2 in case of finite changes as,
\Delta H = \Delta U + p\Delta V --------------------- (3)
We measure changes in enthalpy, when heat is being absorbed at a constant pressure.
ΔV = 0, for constant volume. So, equation 3 becomes ΔH = ΔU = qv
Let Va be the total volume of gaseous reactants and Vb be the total volume of gaseous products. Let nA be the number of moles of gas reactants and nB be the number of moles for the gaseous products. The ideal gas law is:
pVa= nA R T
pVb= nB R T
now,
pVb - pVa = (nB - nA) RT
hence,
p\Delta V = \Delta n_g RT ------------------- (4)
now, by using equation 3 and equation 4,
we can write
ΔH = ΔU + Δng RT

Question:59

Extensive properties depend on the quantity of matter, but intensive properties do not. Explain whether the following properties are extensive or intensive.
Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.

Answer:

Extensive properties Mass, internal energy, heat capacity.
Intensive properties Pressure, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.
Ratio of two extensive properties is always intensive.
Extensive/Extensive = Intensive
So, mole fraction and molarity are intensive properties.

Question:60

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s).
Answer:

When one mole of an ionic compound dissociates into its ions, the enthalpy change related to this process is called lattice enthalpy.
Na^{+} Cl^{-}\rightarrow Na^{+} + Cl^{-}
\Delta H_{lattice}= 788 kJ/mol
Lattice enthalpy for NaCl is calculated as follows:

1) Sublimation of sodium metal
Na(s)\rightarrow Na(g)
2) Ionization enthalpy: ionization of sodium atoms
Na(g)\rightarrow Na^{+}(g)+e^{-1}(g)
3) Bond dissociation enthalpy: dissociation of chlorine
\frac{1}{2}Cl_{2}(g)\rightarrow Cl^{-}(g)
4) Electron gain enthalpy: electron gain by chlorine
Cl^{-}(g)+e^{-1}(g)\rightarrow Cl(g)
5)Na^{+}(g)+Cl^{-}(g)\rightarrow Na^{+}Cl^{-}(s)
These steps collectively form the BORN-HABER Cycle, wherein the sum total of enthalpy changes round a cycle is equal to 0.

Question:61

\Delta G is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that \Delta G is a measure of free energy. Find the unit of \Delta G. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?
Answer:

We are aware of the equation that,
\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}
The temperature of the system and surroundings will be equal if both are in equilibrium. Also, the enthalpy changes are inversely proportional in both of them.
We can represent the entropy changes in the surroundings as:
\Delta S_{Surr}=\Delta H_{surr}/T=-(\Delta H_{sys}/T)
\Delta S_{tot}=\Delta S_{sys}-(\Delta H_{sys}/T)
Hence, we can rewrite this as:
T\Delta S_{tot}=T\Delta S_{sys}- \Delta H_{sys}
In case of a spontaneous process:
-(\Delta H_{sys}-T\Delta S_{sys})>0,since \Delta S _{tot}>0
We can write the above equation as:
-\Delta G > 0
\Delta G = \Delta H - T \Delta S > 0
ΔG is the measure of free energy as it is the net available energy for performing some useful work.
Criteria for spontaneity:
\Delta G < 0: spontaneous process
\Delta G > 0: non-spontaneous process

Question:62

Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (p_{i}, V_{i}) to (p_{f} ,V_{f}). With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure P_{f}.
Answer:


Let us assume that a cylinder has 1-mole gas without any weight or friction having area of cross-section A. Hence, the total volume is depicted by Vi and the initial pressure is assumed to be p.
Let us assume that pext is the external pressure.
Now, if pext> p, the piston moves down until pext = p,
Hence, the final volume is Vf and the distance moved by piston be ?l
Thus, ?V = ?l × A (eq-1)
?V = Vf - Vi
We know that, force = pressure ×area
Therefore, F = pext× A (eq-2)
If w is work done on the system
W = force × displacement
= pext × A ×?l
From eq-1
W = pext×(-?V)
W = -pext?V
W = -pext( vf – vi)
If vf> vi work is done by the system and w is negative. If vf< vi work is on the system and w is positive

The students preparing for Class XI examination can refer to the chapters and the topics that are covered in the chapter. The students can use the NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 download pdf feature to save solutions in their device. The solutions provided in the chapter will help the students in understanding the different topics to solve the questions.

What will Students Learn in NCERT Exemplar Class 11 Chemistry Solutions Chapter 6?

  • The chapter will help the students in understanding the different concepts related to the relationship between heat and the other forms of energy.

  • NCERT exemplar Class 11 Chemistry chapter 6 solutions will let students know the important aspects of Thermodynamics in a better way.

  • They will also learn about the relationship between heat and other forms of energy and their importance that is required in the field.

  • In NCERT exemplar Class 11 Chemistry solutions chapter 6 Thermodynamics, the students will learn about a closed system, that only energy can be exchanged with the surrounding.

  • The topics will help the students in learning about the surroundings of a system that is the part of the universe that does not contain the system.

  • Based on the exchange of energy and matter, thermodynamic systems can be classified into 3 types: closed system, open system, and an isolated system.

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Important Topics To Cover For Exams From NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 Thermodynamics

Mentioned below are the important topics that will help the students in studying for the examination:

  • It will help the students in learning about the relationship between heat and other forms of energy and their importance that is required in the field.

  • Class 11 Chemistry NCERT Exemplar solutions chapter 6 deals with the different forms of energy and how they are used in our day-to-day life. Learning from the solutions will also help the students in preparing for the competitive exam.

  • NCERT Exemplar solutions for Class 11 Chemistry chapter 6 explains about a closed system, only energy can be exchanged with the surrounding.

  • In open system energy as well as matter can be exchanged with the surrounding. In an isolated system, both energy and matter cannot be exchanged with the surrounding.

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Check Solutions of Textbook Chapters

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

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Frequently Asked Question (FAQs)

1. What are the topics that are covered in this chapter?

The topics that are covered in NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 are Thermodynamic Terms, State of the System, Internal Energy as a State Function, Measurement Of ΔU and ΔH: Calorimetry, etc. 

2. How many questions are covered in this chapter?

There are a total of 22 questions that are covered in this chapter.

3. Will the solutions from this help in qualifying the competitive examination?

Yes, the NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 helps build the foundation based on which you will be ready to solve any questions in the best possible manner.

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Yes, these solutions will help in board examinations.

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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