QJEEM25Phf402: JEE Main 2025 Physics - Class 11 JEEM25Phf402 Solution: A poly-atomic molecule , where is gas constant) goes from phase …

As A to B is can adiabatic path,

So, $\Delta {Q_{A \to B}} = o$ .... (1)

So, now we need to find the heat absorbed for the isothermal path (B to C)

We know, change in internal energy

$\Delta u = m{c_v}\Delta T$

$\Rightarrow \Delta u = o$ (for isothermal as $\Delta T = o$ )

Now, using Ist law of thermodynamics

$\Delta Q = \Delta u + w$

$\Rightarrow \Delta Q = o + w \Rightarrow \Delta Q = w$

So, $\Delta {Q_{B \to C}} = {W_{B \to C}}$

$= nRT\ln \left( {{{{v_C}} \over {{v_B}}}} \right)$

$\Rightarrow \Delta {Q_{B \to C}} = nR(450)\ln \left( {{4 \over 3}} \right)$

$\Rightarrow \Delta {Q_{B \to C}} = 450nR\ln \left( {{4 \over 3}} \right)$ .... (2)

So net heat absorbed,

$Q = \Delta {Q_{A \to B}} + \Delta {Q_{B \to C}}$

$\Rightarrow Q = 450nR\ln \left( {{4 \over 3}} \right)$ ... (from 1 and 2)

Net heat absorbed per mole,

$\Rightarrow {Q \over n} = 450R(\ln 4 - \ln 3)$ .

Option 2 is correct.