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JEE Advanced > Thermodynamics

Explore popular questions from Thermodynamics for JEE Advanced. This collection covers Thermodynamics previous year JEE Advanced questions hand picked by experienced teachers.

Q 1.

Correct4

Incorrect-1

The difference between heats of reaction at constant pressure and constant volume for the reaction:
{tex} 2 \mathrm { C } _ { 6 } \mathrm { H } _ { 6 } ( l ) + 15 \mathrm { O } _ { 2 ( \mathrm { g } ) } \rightarrow 12 \mathrm { CO } _ { 2 } ( \mathrm { g } ) + 6 \mathrm { H } _ { 2 } \mathrm { O } ( l ) {/tex} at {tex} 25 ^ { \circ } \mathrm { C } {/tex} in {tex} \mathrm { kJ } {/tex} is

{tex} - 7.43 {/tex}

B

{tex} + 3.72 {/tex}

C

{tex} - 3.72 {/tex}

D

{tex} + 7.43 {/tex}

Explanation



Q 2.

Correct4

Incorrect-1

For which change {tex} \Delta \mathrm { H } \neq \Delta \mathrm { E }: {/tex}

A

{tex} \mathrm { H } _ { 2 ( g ) } + \mathrm { I } _ { 2 ( g ) } \rightarrow 2 \mathrm { HI } ( g ) {/tex}

B

{tex} \mathrm { HCl } + \mathrm { NaOH } \rightarrow \mathrm { NaCl } {/tex}

C

{tex}\mathrm {C}_{(s)} +\mathrm {O}_{2(g)}\rightarrow\mathrm {CO}_{2(g)} {/tex}

{tex} \mathrm { N } _ { 2 } ( \mathrm { g } ) + 3 \mathrm { H } _ { 2 } ( \mathrm { g } ) \rightarrow 2 \mathrm { NH } _ { 3 } ( \mathrm { g } ) {/tex}

Explanation

Q 3.

Correct4

Incorrect-1

The {tex} \Delta \mathrm { H } _ { \mathrm { f } } ^ { 0 } {/tex} for {tex} \mathrm { CO } _ { 2 } ( g ) , \mathrm { CO } ( \mathrm { g } ) {/tex} and {tex} \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) {/tex} are {tex} - 393.5 , - 110.5 {/tex} and {tex} - 241.8 \mathrm { kJ } mol^{- 1}{/tex} respectively. The standard enthalpy change (in {tex} \mathrm { kJ } {/tex} for the reaction {tex} \mathrm { CO } _ { 2 } ( \mathrm { g } ) + \mathrm { H } _ { 2 } ( \mathrm { g } ) \rightarrow \mathrm { CO } ( \mathrm { g } ) + {/tex} {tex} \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) {/tex} is

A

{tex}524.1{/tex}

{tex}41.2{/tex}

C

{tex} - 262.5 {/tex}

D

{tex} - 41.2 {/tex}

Explanation

Q 4.

Correct4

Incorrect-1

In thermodynamics, a process is called reversible when

A

surroundings and system change into each other.

B

there is no boundary between system and surroundings.

the surroundings are always in equilibrium with the system.

D

the system changes into the surroundings spontaneously.

Explanation

Q 5.

Correct4

Incorrect-1

Which one of the following statements is false?

Work is a state function.

B

Temperature is a state function.

C

Change in the state is completely defined when the initial and final states are specified.

D

Work appears at the boundary of the system.

Explanation

Q 6.

Correct4

Incorrect-1

One mole of a non-ideal gas undergoes a change of state {tex} ( 2.0 \mathrm { atm } , 3.0 \mathrm { L } , 95 ( \mathrm { K } ) \rightarrow ( 4.0 \mathrm { atm } , 5.0 \mathrm { L } , 245 \mathrm { K } ) \mathrm { with } {/tex} a change in internal energy, {tex} \Delta U = 30.0 \mathrm { Latm } . {/tex} The change in enthalpy {tex} ( \Delta \mathrm { H } ) {/tex} of the process in {tex} \mathrm { L } {/tex} atm is

A

40.0

B

42.3

44.0

D

not defined, because pressure is not constant

Explanation

Q 7.

Correct4

Incorrect-1

Which of the reaction defines {tex} \Delta \mathrm { H } _ { \mathrm { f } } ^ { \circ } ? {/tex}

A

{tex} \mathrm { C } _ { ( \text {diamond) } } + \mathrm { O } _ { 2 ( g ) } \longrightarrow \mathrm { CO } _ { 2 ( g ) } {/tex}

{tex} \frac { 1 } { 2 } \mathrm { H } _ { 2 ( g ) } + \frac { 1 } { 2 } \mathrm { F } _ { 2 ( g ) } \longrightarrow \mathrm { HF } _ { ( g ) } {/tex}

C

{tex} \mathrm { N } _ { 2 ( g ) } + 3 \mathrm { H } _ { 2 ( g ) } \longrightarrow 2 \mathrm { NH } _ { 3 ( g ) } {/tex}

D

{tex} \mathrm { CO } _ { ( g ) } + \frac { 1 } { 2 } \mathrm { O } _ { 2 ( g ) } \longrightarrow \mathrm { CO } _ { 2 ( g ) } {/tex}

Explanation



Q 8.

Correct4

Incorrect-1

Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is

A

{tex} 11.4 \mathrm { kJ } {/tex}

B

{tex} - 11.4 \mathrm { kJ } {/tex}

{tex} 0 \mathrm { kJ } {/tex}

D

{tex} 4.8 \mathrm { kJ } {/tex}

Explanation

Q 9.

Correct4

Incorrect-1

The enthalpy of vapourization of liquid is {tex} 30 \mathrm { kJ } \mathrm { mol } ^ { - 1 } {/tex} and entropy of vapourization is {tex} 75 \mathrm { J } \mathrm { mol } ^ { - 1 } \mathrm { K } . {/tex} The boiling point of the liquid at {tex} 1 \mathrm { atm } {/tex} is

A

{tex} 250 \mathrm { K } {/tex}

{tex} 400 \mathrm { K } {/tex}

C

{tex} 450 \mathrm { K } {/tex}

D

{tex} 600 \mathrm { K } {/tex}

Explanation

Q 10.

Correct4

Incorrect-1

The direct conversion of A to B is difficult, hence it is carried out by the following shown path:

Given {tex} \Delta S _ { ( A \rightarrow C ) } = 50 {/tex} eu., {tex} \Delta S _ { ( C \rightarrow D ) } = 30 {/tex} eu., {tex} \Delta S _ { ( B \rightarrow D ) } = 20 {/tex} e.u. where e.u. is the entropy unit, then {tex} \Delta \mathrm { S } _ { ( \mathrm { A } \rightarrow \mathrm { B } ) } {/tex} is

{tex} + 60 {/tex} e.{tex} { u }{/tex}

B

{tex} + 100 \mathrm { e } . { u } {/tex}

C

{tex} - 60 {/tex} e. {tex}{ u } {/tex}

D

{tex} - 100 \mathrm { e } . { u } {/tex}

Explanation

Q 11.

Correct4

Incorrect-1

The value of {tex} \log _ { 10 } \mathrm { K } {/tex} for a reaction {tex} \mathrm { A } \rightleftharpoons \mathrm { B } {/tex} is

(Given {tex} : \Delta _ { \mathrm { r } } \mathrm { H } _ { 298 \mathrm { K } } ^ { \circ } = - 54.07 \mathrm { kJ } \mathrm { mol } ^ { - 1 } , \Delta _ { \mathrm { r } } \mathrm { S } _ { 298 \mathrm { K } } ^ { \circ } {/tex} {tex} = 10 \mathrm { JK } ^ { - 1 } \mathrm { mol } ^ { - 1 } {/tex} and {tex} \mathrm { R } = 8.314 \mathrm { JK } ^ { - 1 } \mathrm { mol } ^ { - 1 } ; {/tex} {tex} 2.303 \times 8.314 \times 298 = 5705 ) {/tex}

A

5

10

C

95

D

100

Explanation

Q 12.

Correct4

Incorrect-1

For the process {tex} \mathrm { H } _ { 2 } \mathrm { O } ( l ) ( 1 \text { bar, } 373 \mathrm { K } ) \rightarrow \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) ( 1 \text { bar, } 373 {/tex} {tex} \mathrm { K } ) , {/tex} the correct set of thermodynamic parameters is

{tex} \Delta G = 0 , \Delta S = + v e {/tex}

B

{tex} \Delta G = 0 , \Delta S = - \mathrm { ve } {/tex}

C

{tex} \Delta G = + \mathrm { ve } , \Delta S = 0 {/tex}

D

{tex} \Delta G = - \mathrm { ve } , \Delta S = + \mathrm { ve } {/tex}

Explanation

Q 13.

Correct4

Incorrect-1

The species which by definition has ZERO standard molar enthalpy of formation at {tex} 298 \mathrm { K } {/tex} is

A

{tex} \mathrm { Br } _ { 2 } ( \mathrm { g } ) {/tex}

{tex} \mathrm { Cl } _ { 2 } ( \mathrm { g } ) {/tex}

C

{tex} \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) {/tex}

D

{tex} \mathrm { CH } _ { 4 } ( \mathrm { g } ) {/tex}

Explanation

Q 14.

Correct4

Incorrect-1

The standard enthalpies of formation of {tex} \mathrm { CO } _ { 2 } ( \mathrm { g } ) , \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { l } ) {/tex} and glucose(s) at {tex} 25 ^ { \circ } \mathrm { C } {/tex} are {tex} - 400 \mathrm { kJ } / \mathrm { mol } , - 300 \mathrm { kJ } / \mathrm { mol } {/tex} and {tex} - 1300 {/tex} {tex} \mathrm { k } \mathrm { J } / \mathrm { mol } , {/tex} respectively. The standard enthalpy of combustion per gram of glucose at {tex} 25 ^ { \circ } \mathrm { C } {/tex} is

A

{tex} + 2900 \mathrm { kJ } {/tex}

B

{tex} - 2900 \mathrm { kJ } {/tex}

{tex} - 16.11 \mathrm { kJ } {/tex}

D

{tex} + 16.11 \mathrm { kJ } {/tex}

Explanation

Q 15.

Correct4

Incorrect-1

For the process {tex} \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightarrow \mathrm { H } _ { 2 } \mathrm { O } ( g ) {/tex} at {tex} T = 100 ^ { \circ } \mathrm { C } {/tex} and {tex}1{/tex} atmosphere pressure, the correct choice is

A

{tex} \Delta S _ { \text {system } } > 0 {/tex} and {tex} \Delta S _ { \text {surroundings } } > 0 {/tex}

{tex} \Delta S _ { \text {system } } > 0 {/tex} and {tex} \Delta S _ { \text {surroundings } } < 0 {/tex}

C

{tex} \Delta S _ { \text {system } } < 0 {/tex} and {tex} \Delta S _ { \text {surroundings } } > 0 {/tex}

D

{tex} \Delta S _ { \text {system } } < 0 {/tex} and {tex} \Delta S _ { \text {surroundings } } < 0 {/tex}

Explanation

Q 16.

Correct4

Incorrect-1

One mole of an ideal gas at {tex} 300 \mathrm { K } {/tex} in thermal contact with surroundings expands isothermally from {tex} 1.0 \mathrm { L } {/tex} to {tex} 2.0 \mathrm { L } {/tex} against a constant pressure of {tex} 3.0 \mathrm { atm } {/tex}. In this process, the change in entropy of surroundings {tex} \left( \Delta \mathrm { S } _ { \text {surr } } \right) {/tex} in {tex} \mathrm { JK } ^ { - 1 } {/tex} is {tex} ( 1 \mathrm { Latm } = 101.3 \mathrm { J } ) {/tex}

A

{tex}5.763{/tex}

B

{tex}1.013{/tex}

{tex} - 1.013 {/tex}

D

{tex} - 5.763 {/tex}

Explanation

Q 17.

Correct4

Incorrect-1

The standard state Gibbs free energies of formation of {tex}C(graphite){/tex} and {tex}C(diamond){/tex} at T = 298 K are
{tex}\triangle_{f}G^\circ[C(graphite)]=0 kJmol^{-1}{/tex}
{tex}\triangle_{f}G^\circ[C(diamond)]=2.9 kJmol^{-1}{/tex}
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite {tex}[C(graphite)]{/tex} to diamond {tex}[C(diamond)]{/tex} reduces its volume by {tex} 2 \times 10^{-6} m^3 mol^{-1}{/tex}. If {tex}C(graphite){/tex} is converted to {tex}C(diamond){/tex}isothermally at T = 298 K, thepressure at which {tex}C(graphite){/tex} is in equilibrium with {tex}C(diamond){/tex}, is
[Useful information :{tex} 1 J = 1 kgm^{-2}s^{-2}; 1 Pa= 1 kgm^{-1}s^{-2}, 1 bar = 10^5 Pa{/tex}]

14501 bar

B

58001 bar

C

1450 bar

D

29001 bar

Explanation