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ChemistryInteger+4Question 42 of 51

JEE Advanced 2024 · Paper 1, Chemistry question 8

Consider the following volume−temperature (V−T) diagram for the expansion of 5 moles of an ideal monoatomic gas.

Considering only P-V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z is _____.

[Use the given data: Molar heat capacity of the gas for the given temperature range, CV,m=12 J K−1 mol−1C_{V,m} = 12\,\mathrm{J\,K^{-1}\,mol^{-1}} and gas constant, R=8.3 J K−1 mol−1R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}}]

V-T diagram: volume V (L) on the vertical axis, temperature T (K) on the horizontal axis. Point X is at (T = 335 K, V = 10 L), Y at (335 K, 20 L) and Z at (415 K, 20 L). The path goes X→Y vertically (isothermal expansion at 335 K, volume 10 L to 20 L, arrow upward) and then Y→Z horizontally (isochoric heating at 20 L, 335 K to 415 K, arrow to the right). Dashed guide lines mark V = 10 and 20 and T = 335 and 415.
Answer and worked solution

Official answer: 8000

Enthalpy is a state function, and for an ideal gas it depends only on TT, so only TX=335 KT_X = 335\,\mathrm{K} and TZ=415 KT_Z = 415\,\mathrm{K} matter.

Cp,m=CV,m+R=12+8.3=20.3 J K−1 mol−1C_{p,m} = C_{V,m} + R = 12 + 8.3 = 20.3\,\mathrm{J\,K^{-1}\,mol^{-1}}.

ΔH=nCp,mΔT=5×20.3×80=8120 J.\Delta H = nC_{p,m}\Delta T = 5\times20.3\times80 = 8120\ \mathrm{J}.

(The official key also accepts 8000 and 8300.)

Topics:First law and enthalpy
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