JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2019
- Hard
- MCQ
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation \(VT = K\), where \(I\) is a constant. In this process the temperature of the gas is increased by \(\Delta T\). The amount of heat absorbed by gas is (\(R\) is gas constant)
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Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\,R\Delta T\)
Step-by-step Solution
Detailed explanation
\({\mathrm{VT}=\mathrm{K}} \)
\({\Rightarrow \quad \mathrm{V}\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)=\mathrm{k}} \)
\({\Rightarrow \quad \mathrm{PV}^{2}=\mathrm{K}} \)
\({\because \quad \mathrm{C}=\frac{\mathrm{R}}{1-\mathrm{x}}+\mathrm{C}_{\mathrm{v}} \text { (For polytropic process) }} \)
\({\mathrm{C}=\frac{\mathrm{R}}{1-2}+\frac{3 \mathrm{R}}{2}=\frac{\mathrm{R}}{2}} \)
\({\therefore \quad \Delta \mathrm{Q}=\mathrm{nC} \Delta \mathrm{T}}\)
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