Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2024 — Paper 1 — Question 38

Consider the following volume - temperature (V - T) diagram for the expansion of 5 moles of an ideal monoatomic gas. Consider only P - V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence $\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z}$ is $\_\_\_\_\_$ -. [Use the given data: Molar heat capacity of the gas for the given temperature range, $\mathrm{C}_{\mathrm{v}, \mathrm{m}}=12 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and gas constant, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

Answer: 8120

Numerical answer — enter this value.

Step-by-step solution

$\quad$ Since $X \rightarrow Y$ is an isothermal process ( $d T=0$ ).

So, $\Delta H_{x-y}=\mathrm{nC}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}=0$

$\mathrm{Y} \rightarrow \mathrm{Z}$ is isochoric. $\Delta \mathrm{U}_{\mathrm{y} \rightarrow \mathrm{z}}=\mathrm{nC}_{\mathrm{v}, \mathrm{m}} \mathrm{dT}=5 \times 12 \times(415-335)$

$\Delta U_{y \rightarrow z}=4800 \mathrm{~J}$

$\Delta \mathrm{H}_{\mathrm{y} \rightarrow \mathrm{z}}=\Delta \mathrm{U}_{\mathrm{y} \rightarrow \mathrm{z}}+\Delta(\mathrm{PV})$

$=\Delta \mathrm{U}_{\mathrm{y} \rightarrow \mathrm{z}}+\mathrm{nR} \Delta \mathrm{T}=4800+5 \times 8.3 \times 80=8120 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Enthalpy and Calorimetry