AP EAMCET · PHYSICS · Thermodynamics
One mole of a gas expands such that its volume ' \(V\) ' changes with absolute temperature ' \(T\) ' in accordance with the relation \(V=K T^2\) where ' \(K\) ' is a constant. If the temperature of the gas changes by \(60^{\circ} \mathrm{C}\), then work done by the gas is ( \(R\) is universal gas constant).
- A \(K R \ln 60\)
- B \(R \ln 60\)
- C \(40 K R\)
- D \(120 R\)
Answer & Solution
Correct Answer
(D) \(120 R\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=K T^2 \Rightarrow d V=2 K T d T\) So, \(\quad W=\int P d V=R \int \frac{T}{v} d V\) \[ =R \int \frac{T}{K T^2} 2 K T=2 R \int_0^{60} d T=120 R \]
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