In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T \Delta X where T is temperature of the system and \Delta X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas, X = 3 \over 2R\,\ln ( T \over T_A ) + R\,\ln ( V \over V_A ) Here, R is gas constant, V is volume of gas, TA and VA are constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities. If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P_0V_0 = 1 \over 3RT_0 , the correct match is,

I \to S, II \to T, III \to Q, IV \to U

I \to P, II \to R, III \to T, IV \to S

I \to P, II \to T, III \to Q, IV \to T

I \to P, II \to R, III \to T, IV \to P

Heat and Thermodynamics PYQ — JEE Advanced 2019 · Shift 2 · Q53 with Solution

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Correct answer: C

Identify the thermodynamic quantity $X$

For one mole of a monatomic ideal gas, $X=\frac{3}{2}R\ln\left(\frac{T}{T_A}\right)+R\ln\left(\frac{V}{V_A}\right)$

Hence, $dX=\frac{3}{2}R\frac{dT}{T}+R\frac{dV}{V}$

Given infinitesimal heat absorbed: $\delta Q=T\,dX$ So, $\delta Q=T\left(\frac{3}{2}R\frac{dT}{T}+R\frac{dV}{V}\right)=\frac{3}{2}R\,dT+RT\frac{dV}{V}$

Now for one mole of ideal gas, $PV=RT \Rightarrow RT=P V$ Thus, $RT\frac{dV}{V}=P\,dV$ So, $\delta Q=\frac{3}{2}R\,dT+P\,dV$

But for one mole monatomic gas, $dU=C_V dT=\frac{3}{2}R dT$ Hence, $\delta Q=dU+P\,dV$ which is exactly the first law. Therefore, $dX=dS$ So, $X$ is entropy (up to an additive constant).

Therefore, quantity I corresponds to entropy, i.e. I $\to$ S.

Use the given $T$ - $V$ diagram relation

From the options, only option C has $\text{I} \to \text{S}$ . So likely the correct option is C, but let us verify further.

The condition given is $P_0V_0=\frac{1}{3}RT_0$ For one mole gas, $PV=RT$ Hence at a state with $(T_0,V_0)$ , $P=\frac{RT_0}{V_0}=3P_0$ So the pressure corresponding to $(T_0,V_0)$ is $3P_0$ .

This kind of matching question typically uses the process line in the $TV$ diagram and asks about values of heat/work/internal energy/pressure etc. Since the exact figure and List-I/List-II entries are not visible in the text, the only unambiguous derivation possible from the given expression is that $X=S$ .

Eliminate options using this result

Options given:

A: I $\to$ P
B: I $\to$ P
C: I $\to$ S
D: I $\to$ P

Since only option C has $\text{I} \to \text{S}$ , the correct answer must be: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: D

My derived answer: C

They do not agree.

The reason is straightforward: from $\delta Q = T dX$ and the explicit form of $X$ , $dX=\frac{3}{2}R\frac{dT}{T}+R\frac{dV}{V}$ which is exactly the entropy differential for one mole of a monatomic ideal gas: $dS=C_V\frac{dT}{T}+R\frac{dV}{V}=\frac{3}{2}R\frac{dT}{T}+R\frac{dV}{V}$ Therefore, $X=S$ , not pressure. So any option with I $\to$ P cannot be correct.

Thus the stored answer D appears inconsistent with the thermodynamic identity given in the question.

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