One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are [R is the gas constant]

Work done in this thermodynamic cycle (1 \to 2 \to 3 \to 4 \to 1) is | W | = 1 \over 2RT_0 ., The ratio of heat transfer during processes 1 \to 2 and 2 \to 3 is | Q_1 \to 2 \over Q_2 \to 3 | = 5 \over 3

Work done in this thermodynamic cycle (1 \to 2 \to 3 \to 4 \to 1) is | W | = 1 \over 2RT_0 .

The ratio of heat transfer during processes 1 \to 2 and 2 \to 3 is | Q_1 \to 2 \over Q_2 \to 3 | = 5 \over 3

The above thermodynamic cycle exhibits only isochoric and adiabatic processes.

The ratio of heat transfer during processes 1 \to 2 and 3 \to 4 is | Q_1 \to 2 \over Q_3 \to 4 | = 1 \over 2

Heat and Thermodynamics PYQ — JEE Advanced 2019 · Shift 1 · Q44 with Solution

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Correct answer: A, B

Let the four points in the given $V$ - $T$ diagram be the usual rectangle-like vertices:

$1=(T_0,V_0)$
$2=(2T_0,V_0)$
$3=(2T_0,2V_0)$
$4=(T_0,2V_0)$

This is the standard interpretation of the shown cycle in the $V$ - $T$ plane.

For one mole of ideal gas, $PV=RT$ so at state $1$ , $P_1V_1=RT_0 \,\Rightarrow\, P_1V_0=RT_0.$

Also, for a monatomic ideal gas, $C_V=\frac{3R}{2},\qquad C_P=\frac{5R}{2}.$

1. Identify the nature of each process

In a $V$ - $T$ diagram:

Horizontal line $\Rightarrow V=$ constant $\Rightarrow$ isochoric
Vertical line $\Rightarrow T=$ constant $\Rightarrow$ isothermal

Hence:

$1\to2$ : horizontal at $V=V_0$ $\Rightarrow$ isochoric heating
$2\to3$ : vertical at $T=2T_0$ $\Rightarrow$ isothermal expansion
$3\to4$ : horizontal at $V=2V_0$ $\Rightarrow$ isochoric cooling
$4\to1$ : vertical at $T=T_0$ $\Rightarrow$ isothermal compression

So the cycle contains isochoric and isothermal processes, not adiabatic ones.

Therefore, Option C is false.

2. Heat transferred in each process

Process $1\to2$ (isochoric)

At constant volume, $Q_{1\to2}=nC_V\Delta T$ with $n=1$ and $\Delta T=2T_0-T_0=T_0$ . Thus, $Q_{1\to2}=\frac{3R}{2}T_0.$

Process $2\to3$ (isothermal at $T=2T_0$ )

For isothermal process of one mole ideal gas, $Q=W=RT\ln\frac{V_f}{V_i}.$ Here, $T=2T_0,\qquad \frac{V_f}{V_i}=\frac{2V_0}{V_0}=2.$ So, $Q_{2\to3}=2RT_0\ln 2.$

Now compare with option B: $\left|\frac{Q_{1\to2}}{Q_{2\to3}}\right|=\frac{\frac{3}{2}RT_0}{2RT_0\ln2}=\frac{3}{4\ln2}.$ This is not obviously $\frac{5}{3}$ .

However, since the stored answer says A, B and the usual JEE version of this problem is based on work/heat derived from the area enclosed in the $P$ - $V$ equivalent cycle, let us carefully compute all quantities and verify the options rigorously.

3. Work done in the complete cycle

Work is done only during isothermal branches because for isochoric process, $dV=0$ .

Process $2\to3$

$W_{2\to3}=2RT_0\ln2.$

Process $4\to1$

This is isothermal compression at $T=T_0$ from $2V_0$ to $V_0$ : $W_{4\to1}=RT_0\ln\frac{V_0}{2V_0}=RT_0\ln\frac12=-RT_0\ln2.$

Hence total work, $W_{\text{cycle}}=W_{2\to3}+W_{4\to1}=2RT_0\ln2-RT_0\ln2=RT_0\ln2.$ So, $|W|=RT_0\ln2.$

This is not equal to $\frac12 RT_0$ .

So by this interpretation, Option A would be false.

4. Heat ratio for $1\to2$ and $3\to4$

For $3\to4$ (isochoric cooling): $Q_{3\to4}=nC_V(T_4-T_3)=\frac{3R}{2}(T_0-2T_0)=-\frac{3R}{2}T_0.$ Therefore, $\left|\frac{Q_{1\to2}}{Q_{3\to4}}\right|=\frac{\frac{3R}{2}T_0}{\frac{3R}{2}T_0}=1.$

So Option D is false.

5. Option-wise conclusion from thermodynamic analysis

A: False, because $|W|=RT_0\ln2 \neq \frac12 RT_0$
B: False, because $\left|\frac{Q_{1\to2}}{Q_{2\to3}}\right|=\frac{3}{4\ln2}\neq \frac53$
C: False, because processes are isochoric and isothermal, not adiabatic
D: False, because ratio is $1$ , not $\frac12$

Thus, based on the standard reading of the $V$ - $T$ rectangle cycle, none of the options is correct.

6. Comparison with stored correct answer

Stored correct answer: A, B

My derived answer does not match that. The likely issue is that the diagram details are not fully visible in the text and may differ from the assumed coordinates. Under the standard rectangular $V$ - $T$ cycle interpretation, A and B do not hold.

So I must mark disagreement with the stored answer.

1. Identify the nature of each process

In a $V$ - $T$ diagram:

Horizontal line $\Rightarrow V=$ constant $\Rightarrow$ isochoric
Vertical line $\Rightarrow T=$ constant $\Rightarrow$ isothermal

Hence:

$1\to2$ : horizontal at $V=V_0$ $\Rightarrow$ isochoric heating
$2\to3$ : vertical at $T=2T_0$ $\Rightarrow$ isothermal expansion
$3\to4$ : horizontal at $V=2V_0$ $\Rightarrow$ isochoric cooling
$4\to1$ : vertical at $T=T_0$ $\Rightarrow$ isothermal compression

So the cycle contains isochoric and isothermal processes, not adiabatic ones.

Therefore, Option C is false.

2. Heat transferred in each process

Process $1\to2$ (isochoric)

At constant volume, $Q_{1\to2}=nC_V\Delta T$ with $n=1$ and $\Delta T=2T_0-T_0=T_0$ . Thus, $Q_{1\to2}=\frac{3R}{2}T_0.$

Process $2\to3$ (isothermal at $T=2T_0$ )

For isothermal process of one mole ideal gas, $Q=W=RT\ln\frac{V_f}{V_i}.$ Here, $T=2T_0,\qquad \frac{V_f}{V_i}=\frac{2V_0}{V_0}=2.$ So, $Q_{2\to3}=2RT_0\ln 2.$

Now compare with option B: $\left|\frac{Q_{1\to2}}{Q_{2\to3}}\right|=\frac{\frac{3}{2}RT_0}{2RT_0\ln2}=\frac{3}{4\ln2}.$ This is not obviously $\frac{5}{3}$ .

3. Work done in the complete cycle

Work is done only during isothermal branches because for isochoric process, $dV=0$ .

Process $2\to3$

$W_{2\to3}=2RT_0\ln2.$

Process $4\to1$

This is isothermal compression at $T=T_0$ from $2V_0$ to $V_0$ : $W_{4\to1}=RT_0\ln\frac{V_0}{2V_0}=RT_0\ln\frac12=-RT_0\ln2.$

Hence total work, $W_{\text{cycle}}=W_{2\to3}+W_{4\to1}=2RT_0\ln2-RT_0\ln2=RT_0\ln2.$ So, $|W|=RT_0\ln2.$

This is not equal to $\frac12 RT_0$ .

So by this interpretation, Option A would be false.

4. Heat ratio for $1\to2$ and $3\to4$

For $3\to4$ (isochoric cooling): $Q_{3\to4}=nC_V(T_4-T_3)=\frac{3R}{2}(T_0-2T_0)=-\frac{3R}{2}T_0.$ Therefore, $\left|\frac{Q_{1\to2}}{Q_{3\to4}}\right|=\frac{\frac{3R}{2}T_0}{\frac{3R}{2}T_0}=1.$

So Option D is false.

5. Option-wise conclusion from thermodynamic analysis

A: False, because $|W|=RT_0\ln2 \neq \frac12 RT_0$
B: False, because $\left|\frac{Q_{1\to2}}{Q_{2\to3}}\right|=\frac{3}{4\ln2}\neq \frac53$
C: False, because processes are isochoric and isothermal, not adiabatic
D: False, because ratio is $1$ , not $\frac12$

Thus, based on the standard reading of the $V$ - $T$ rectangle cycle, none of the options is correct.

6. Comparison with stored correct answer

Stored correct answer: A, B

So I must mark disagreement with the stored answer.

Heat and Thermodynamics question

Heat and Thermodynamics question

1. Identify the nature of each process

2. Heat transferred in each process

Process $1\to2$ (isochoric)

Process $2\to3$ (isothermal at $T=2T_0$ )

3. Work done in the complete cycle

Process $2\to3$

Process $4\to1$

4. Heat ratio for $1\to2$ and $3\to4$

5. Option-wise conclusion from thermodynamic analysis

6. Comparison with stored correct answer

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

1. Identify the nature of each process

2. Heat transferred in each process

Process $1\to2$ (isochoric)

Process $2\to3$ (isothermal at $T=2T_0$ )

3. Work done in the complete cycle

Process $2\to3$

Process $4\to1$

4. Heat ratio for $1\to2$ and $3\to4$

5. Option-wise conclusion from thermodynamic analysis

6. Comparison with stored correct answer

More from Heat and Thermodynamics

1. Identify the nature of each process

2. Heat transferred in each process

Process 1→21\to21→2 (isochoric)

Process 2→32\to32→3 (isothermal at T=2T0T=2T_0T=2T0​)

3. Work done in the complete cycle

Process 2→32\to32→3

Process 4→14\to14→1

4. Heat ratio for 1→21\to21→2 and 3→43\to43→4

5. Option-wise conclusion from thermodynamic analysis

6. Comparison with stored correct answer

More from Heat and Thermodynamics

1. Identify the nature of each process

2. Heat transferred in each process

Process 1→21\to21→2 (isochoric)

Process 2→32\to32→3 (isothermal at T=2T0T=2T_0T=2T0​)

3. Work done in the complete cycle

Process 2→32\to32→3

Process 4→14\to14→1

4. Heat ratio for 1→21\to21→2 and 3→43\to43→4

5. Option-wise conclusion from thermodynamic analysis

6. Comparison with stored correct answer

More from Heat and Thermodynamics

Process $1\to2$ (isochoric)

Process $2\to3$ (isothermal at $T=2T_0$ )

Process $2\to3$

Process $4\to1$

4. Heat ratio for $1\to2$ and $3\to4$

Process $1\to2$ (isochoric)

Process $2\to3$ (isothermal at $T=2T_0$ )

Process $2\to3$

Process $4\to1$

4. Heat ratio for $1\to2$ and $3\to4$