JEE Advanced 2025 Paper 2, Physics Section 3 Q3: Thermodynamic Processes

Given data: $V_W=64$, $V_X=125$, $V_Y=250$ (all in cm${}^3$), and $nR{T}_W=P_W{V}_W=1$ J, so $P_W=1/(64\times 10^{-6})$ Pa $=10^6/64$ Pa.

$WX$ is adiabatic with $\gamma =5/3$ for a monatomic gas, so $P_X{V}_X^{\gamma }=P_W{V}_W^{\gamma }$, giving

$P_X=P_W{\left(\frac{V_W}{V_X}\right)}^{\gamma }$

$XY$ is isobaric: $P_Y=P_X$. Heat absorbed in an isobaric process for a monatomic gas:

$Q_{XY}=n{C}_P\Delta T=\frac{5}{2}(P_Y{V}_Y-P_X{V}_X)=\frac{5}{2}{P}_X(V_Y-V_X)$

Substitute $P_X=P_W(V_W/V_X{)}^{5/3}$:

$Q_{XY}=\frac{5}{2}{P}_W{V}_W\cdot {\left(\frac{V_W}{V_X}\right)}^{2/3}\cdot \frac{V_Y-V_X}{V_W}\cdot$
a cleaner form:

$Q_{XY}=\frac{5}{2}\cdot \frac{P_W{V}_W^{5/3}}{V_X^{5/3}}\cdot (V_Y-V_X)$

Plug in: $V_W^{5/3}=64^{5/3}=(2^6{)}^{5/3}=2^{10}=1024$; $V_X^{5/3}=125^{5/3}=(5^3{)}^{5/3}=5^5=3125$. With $P_W{V}_W=1$ J:

$Q_{XY}=\frac{5}{2}\cdot \frac{1024}{3125\cdot V_W}\cdot 1\cdot (250-125)\cdot 10^{-6}$
rearranging using $P_W{V}_W=1$:

$Q_{XY}=\frac{5}{2}\cdot (P_W{V}_W)\cdot {\left(\frac{V_W}{V_X}\right)}^{5/3}\cdot \frac{V_Y-V_X}{V_W}=\frac{5}{2}\cdot 1\cdot \frac{1024}{3125}\cdot \frac{125}{64}$

$=\frac{5}{2}\cdot \frac{1024\times 125}{3125\times 64}=\frac{5}{2}\cdot \frac{16}{25}=\frac{80}{50}=1.60$ J

$\therefore Q_{XY}=1.60$ J.