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QJEEAdv24Ch232a: JEE Advanced 2024 Chemistry - Class 11 JEEAdv24Ch232a Solution: Consider the following volume-temperature diagram for the…

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

JEE Advanced 2024 Paper 1 Online Chemistry - Thermodynamics Question 2 English

Considering only P−V\mathrm{P}-\mathrm{V} work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z} is ____________.

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

Step-by-Step Verified Solution

For ideal gas

amp;ΔH=nCPΔTamp;∵CP=CV+R=12+8.3=20.3 J/K-mole amp;∴ΔH=5×20.3×(415−335)amp;ΔH=8120 Joule \begin{aligned} & \Delta H=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T} \\\\ & \because \mathrm{C}_{\mathrm{P}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}=12+8.3=20.3 \mathrm{~J} / \mathrm{K} \text {-mole } \\\\ & \therefore \Delta \mathrm{H}=5 \times 20.3 \times(415-335) \\\\ & \Delta \mathrm{H}=8120 \text { Joule } \end{aligned}

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