List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process. .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top List-I List-II (I) 10^-3 ~kg of water at 100^\circ C is converted to steam at the same temperature, at a pressure of 10^5 ~Pa . The volume of the system changes from 10^-6 ~m^3 to 10^-3 ~m^3 in the process. Latent heat of water =2250\, ~kJ / kg . (P) 2 ~kJ (II) 0.2 moles of a rigid diatomic ideal gas with volume V at temperature 500 ~K undergoes an isobaric expansion to volume 3 ~V . Assume R=8.0 Jmol^-1 ~K^-1 . (Q) 7 k J (III) One mole of a monatomic ideal gas is compressed adiabatically from volume V=\frac13 ~m^3 and pressure 2 kPa to volume \fracV8 . (R) 4 ~kJ (IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 ~kJ of heat and undergoes isobaric expansion. (S) 5 ~kJ (T) 3 ~kJ Which one of the following options is correct?

I arrow P, II arrow R, III arrow T, IV arrow Q

I arrow T, II arrow R, III arrow S, IV arrow Q

I arrow S, II arrow P, III arrow T, IV arrow P

I arrow Q, II arrow R, III arrow S, IV arrow T

Heat and Thermodynamics PYQ — JEE Advanced 2022 · Shift 1 · Q53 with Solution

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I	List-II
(I) $10^{-3} \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature, at a pressure of $10^{5} \mathrm{~Pa}$ . The volume of the system changes from $10^{-6} \mathrm{~m}^{3}$ to $10^{-3} \mathrm{~m}^{3}$ in the process. Latent heat of water $=2250\, \mathrm{~kJ} / \mathrm{kg}$ .	(P) $2 \mathrm{~kJ}$
(II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 \mathrm{~K}$ undergoes an isobaric expansion to volume $3 \mathrm{~V}$ . Assume $R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ .	(Q) $7 k J$
(III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{3} \mathrm{~m}^{3}$ and pressure $2 \mathrm{kPa}$ to volume $\frac{V}{8}$ .	(R) $4 \mathrm{~kJ}$
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \mathrm{~kJ}$ of heat and undergoes isobaric expansion.	(S) $5 \mathrm{~kJ}$
	(T) $3 \mathrm{~kJ}$

Which one of the following options is correct?

A
I $\rightarrow$ T, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ Q
B
I $\rightarrow$ S, II $\rightarrow$ P, III $\rightarrow$ T, IV $\rightarrow$ P
C
I $\rightarrow$ P, II $\rightarrow$ R, III $\rightarrow$ T, IV $\rightarrow$ Q
D
I $\rightarrow$ Q, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ T

View written solutionFree

Correct answer: C

We compute the change in internal energy $\Delta U$ for each process.

1. Process I

Given:

Mass of water, $m=10^{-3}\,\text{kg}$
Latent heat, $L=2250\,\text{kJ/kg}=2.25\times 10^6\,\text{J/kg}$
Pressure, $P=10^5\,\text{Pa}$
Volume change: $\Delta V=10^{-3}-10^{-6}=0.999\times 10^{-3}\,\text{m}^3\approx 10^{-3}\,\text{m}^3$

Heat supplied: $Q=mL=10^{-3}\times 2.25\times 10^6=2250\,\text{J}$

Work done: $W=P\Delta V=10^5\times 0.999\times 10^{-3}\approx 100\,\text{J}$

By first law: $\Delta U=Q-W=2250-100=2150\,\text{J}\approx 2\,\text{kJ}$

So, $\boxed{\text{I} \rightarrow \text{P}}$

2. Process II

Given:

$n=0.2$ moles of rigid diatomic gas
Initial temperature $T_1=500\,\text{K}$
Isobaric expansion from $V$ to $3V$

At constant pressure, for an ideal gas: $T\propto V$ So, $T_2=3T_1=1500\,\text{K}$ Thus, $\Delta T=1000\,\text{K}$

For a rigid diatomic gas: $C_V=\frac{5}{2}R$

Hence, $\Delta U=nC_V\Delta T$ $\Delta U=0.2\times \frac{5}{2}\times 8 \times 1000$ $\Delta U=0.2\times 20\times 1000=4000\,\text{J}=4\,\text{kJ}$

So, $\boxed{\text{II} \rightarrow \text{R}}$

3. Process III

Given:

One mole monatomic ideal gas
Initial volume $V_1=\frac13\,\text{m}^3$
Initial pressure $P_1=2\,\text{kPa}=2000\,\text{Pa}$
Final volume $V_2=\frac{V_1}{8}$
Adiabatic compression

For monatomic gas: $\gamma=\frac53$

Using adiabatic relation: $P_1V_1^\gamma=P_2V_2^\gamma$ $P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=2000\times 8^{5/3}$ Now, $8^{5/3}=(2^3)^{5/3}=2^5=32$ So, $P_2=2000\times 32=64000\,\text{Pa}$

Initial internal energy: $U_1=\frac32 P_1V_1=\frac32\times 2000\times \frac13=1000\,\text{J}$

Final internal energy: $U_2=\frac32 P_2V_2=\frac32\times 64000\times \frac{1}{24}=4000\,\text{J}$ (since $V_2=\frac{1}{3}\times\frac18=\frac{1}{24}$ )

Thus, $\Delta U=U_2-U_1=4000-1000=3000\,\text{J}=3\,\text{kJ}$

So, $\boxed{\text{III} \rightarrow \text{T}}$

4. Process IV

Given:

$n=3$ moles diatomic gas with vibration active
Heat supplied $Q=9\,\text{kJ}$
Isobaric expansion

For diatomic gas with vibration active:

Degrees of freedom $f=7$

$C_V=\frac{7}{2}R, \qquad C_P=C_V+R=\frac{9}{2}R$

In isobaric process: $Q=nC_P\Delta T$ $\Delta U=nC_V\Delta T$ So, $\Delta U=Q\frac{C_V}{C_P}=9\times \frac{\frac72}{\frac92}=9\times \frac79=7\,\text{kJ}$

Thus, $\boxed{\text{IV} \rightarrow \text{Q}}$

5. Final matching

We obtained:

$\text{I} \rightarrow \text{P}$
$\text{II} \rightarrow \text{R}$
$\text{III} \rightarrow \text{T}$
$\text{IV} \rightarrow \text{Q}$

This matches: $\boxed{\text{Option C}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored answer.

More from Heat and Thermodynamics

List-I	List-II
(I) $10^{-3} \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature, at a pressure of $10^{5} \mathrm{~Pa}$ . The volume of the system changes from $10^{-6} \mathrm{~m}^{3}$ to $10^{-3} \mathrm{~m}^{3}$ in the process. Latent heat of water $=2250\, \mathrm{~kJ} / \mathrm{kg}$ .	(P) $2 \mathrm{~kJ}$
(II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 \mathrm{~K}$ undergoes an isobaric expansion to volume $3 \mathrm{~V}$ . Assume $R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ .	(Q) $7 k J$
(III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{3} \mathrm{~m}^{3}$ and pressure $2 \mathrm{kPa}$ to volume $\frac{V}{8}$ .	(R) $4 \mathrm{~kJ}$
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \mathrm{~kJ}$ of heat and undergoes isobaric expansion.	(S) $5 \mathrm{~kJ}$
	(T) $3 \mathrm{~kJ}$

Which one of the following options is correct?

A
I $\rightarrow$ T, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ Q
B
I $\rightarrow$ S, II $\rightarrow$ P, III $\rightarrow$ T, IV $\rightarrow$ P
C
I $\rightarrow$ P, II $\rightarrow$ R, III $\rightarrow$ T, IV $\rightarrow$ Q
D
I $\rightarrow$ Q, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ T

View written solutionFree

Correct answer: C

We compute the change in internal energy $\Delta U$ for each process.

1. Process I

Given:

Mass of water, $m=10^{-3}\,\text{kg}$
Latent heat, $L=2250\,\text{kJ/kg}=2.25\times 10^6\,\text{J/kg}$
Pressure, $P=10^5\,\text{Pa}$
Volume change: $\Delta V=10^{-3}-10^{-6}=0.999\times 10^{-3}\,\text{m}^3\approx 10^{-3}\,\text{m}^3$

Heat supplied: $Q=mL=10^{-3}\times 2.25\times 10^6=2250\,\text{J}$

Work done: $W=P\Delta V=10^5\times 0.999\times 10^{-3}\approx 100\,\text{J}$

By first law: $\Delta U=Q-W=2250-100=2150\,\text{J}\approx 2\,\text{kJ}$

So, $\boxed{\text{I} \rightarrow \text{P}}$

2. Process II

Given:

$n=0.2$ moles of rigid diatomic gas
Initial temperature $T_1=500\,\text{K}$
Isobaric expansion from $V$ to $3V$

At constant pressure, for an ideal gas: $T\propto V$ So, $T_2=3T_1=1500\,\text{K}$ Thus, $\Delta T=1000\,\text{K}$

For a rigid diatomic gas: $C_V=\frac{5}{2}R$

Hence, $\Delta U=nC_V\Delta T$ $\Delta U=0.2\times \frac{5}{2}\times 8 \times 1000$ $\Delta U=0.2\times 20\times 1000=4000\,\text{J}=4\,\text{kJ}$

So, $\boxed{\text{II} \rightarrow \text{R}}$

3. Process III

Given:

One mole monatomic ideal gas
Initial volume $V_1=\frac13\,\text{m}^3$
Initial pressure $P_1=2\,\text{kPa}=2000\,\text{Pa}$
Final volume $V_2=\frac{V_1}{8}$
Adiabatic compression

For monatomic gas: $\gamma=\frac53$

Using adiabatic relation: $P_1V_1^\gamma=P_2V_2^\gamma$ $P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=2000\times 8^{5/3}$ Now, $8^{5/3}=(2^3)^{5/3}=2^5=32$ So, $P_2=2000\times 32=64000\,\text{Pa}$

Initial internal energy: $U_1=\frac32 P_1V_1=\frac32\times 2000\times \frac13=1000\,\text{J}$

Final internal energy: $U_2=\frac32 P_2V_2=\frac32\times 64000\times \frac{1}{24}=4000\,\text{J}$ (since $V_2=\frac{1}{3}\times\frac18=\frac{1}{24}$ )

Thus, $\Delta U=U_2-U_1=4000-1000=3000\,\text{J}=3\,\text{kJ}$

So, $\boxed{\text{III} \rightarrow \text{T}}$

4. Process IV

Given:

$n=3$ moles diatomic gas with vibration active
Heat supplied $Q=9\,\text{kJ}$
Isobaric expansion

For diatomic gas with vibration active:

Degrees of freedom $f=7$

$C_V=\frac{7}{2}R, \qquad C_P=C_V+R=\frac{9}{2}R$

In isobaric process: $Q=nC_P\Delta T$ $\Delta U=nC_V\Delta T$ So, $\Delta U=Q\frac{C_V}{C_P}=9\times \frac{\frac72}{\frac92}=9\times \frac79=7\,\text{kJ}$

Thus, $\boxed{\text{IV} \rightarrow \text{Q}}$

5. Final matching

We obtained:

$\text{I} \rightarrow \text{P}$
$\text{II} \rightarrow \text{R}$
$\text{III} \rightarrow \text{T}$
$\text{IV} \rightarrow \text{Q}$

This matches: $\boxed{\text{Option C}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored answer.

Heat and Thermodynamics question

Heat and Thermodynamics question

1. Process I

2. Process II

3. Process III

4. Process IV

5. Final matching

6. Comparison with stored answer

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

1. Process I

2. Process II

3. Process III

4. Process IV

5. Final matching

6. Comparison with stored answer

More from Heat and Thermodynamics