JEE Mains · Physics · STD 11 - 11. thermodynamics
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)
- A \(\frac{1}{2}\,R\Delta T\)
- B \(\frac{1}{2}\,KR\Delta T\)
- C \(\frac{3}{2}\,R\Delta T\)
- D \(\frac{{2K}}{3}\,\Delta T\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\,R\Delta T\)
Step-by-step Solution
Detailed explanation
\({\mathrm{VT}=\mathrm{K}} \) \({\Rightarrow \quad \mathrm{V}\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)=\mathrm{k}} \) \({\Rightarrow \quad \mathrm{PV}^{2}=\mathrm{K}} \)…
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