JEE Advanced 2024 Paper 1 · Q08 · First Law / Enthalpy in V-T Process

Question

Consider the volume–temperature ($V$–$T$) diagram for the expansion of $5$ moles of an ideal monoatomic gas, where the $V$-axis (in litres) is plotted against the $T$-axis (in kelvin). State $X$ is at $T = 335\,\text{K}$, $V = 10\,\text{L}$. State $Y$ is at $T = 335\,\text{K}$, $V = 20\,\text{L}$ — the process $X \to Y$ is isothermal at $335\,\text{K}$. State $Z$ is at $T = 415\,\text{K}$, $V = 20\,\text{L}$ — the process $Y \to Z$ is isochoric at $V = 20\,\text{L}$. Considering only $P$-$V$ work, the total change in enthalpy (in Joule) for the transformation $X \to Y \to Z$ is ______.

[Given: molar heat capacity $C_{V,m} = 12\,\text{J\,K}^{-1}\text{mol}^{-1}$; gas constant $R = 8.3\,\text{J\,K}^{-1}\text{mol}^{-1}$]

[Given: molar heat capacity $C_{V,m} = 12\,	ext{J\,K}^{-1}	ext{mol}^{-1}$; gas constant $R = 8.3\,	ext{J\,K}^{-1}	ext{mol}^{-1}$]

SolveKar AI · Accepted Answer

Step $X 	o Y$ is isothermal (T = 335 K constant) for an ideal gas: $\Delta U_{XY} = 0$ and $\Delta H_{XY} = nC_{p,m}\Delta T = 0$ since $\Delta T = 0$.
Step $Y 	o Z$ is isochoric (V = 20 L constant): $\Delta T = 415 - 335 = 80$ K.
$C_{p,m} = C_{V,m} + R = 12 + 8.3 = 20.3 \, J\,K^{-1}\,mol^{-1}$.
$\Delta H_{YZ} = n C_{p,m} \Delta T = 5 	imes 20.3 	imes 80 = 8120 \, J$.
Total $\Delta H_{X 	o Y 	o Z} = \Delta H_{XY} + \Delta H_{YZ} = 0 + 8120 = \mathbf{8120 \, J}$.

JEE Advanced 2024 Paper 1 · Q08 · First Law / Enthalpy in V-T Process

Solution