MHT CET · Physics · Thermodynamics
In an adiabatic expansion of a gas initial and final temperatures are \(T_1\) and \(T_2\) respectively then the change in internal energy of the gas is \(
\text { [R = gas constant, } \mathrm{Y}=\text { adiabatic ratio }]
\)
- A zero
- B \(\frac{n R}{\gamma-1}\left(T_1-T_2\right)\)
- C \(\frac{\mathrm{nR}}{\mathrm{Y}-1}\left(\mathrm{~T}_2-\mathrm{T}_1\right)\)
- D \(\mathrm{nR}\left(\mathrm{T}_1-\mathrm{T}_2\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{nR}}{\mathrm{Y}-1}\left(\mathrm{~T}_2-\mathrm{T}_1\right)\)
Step-by-step Solution
Detailed explanation
The correct option is (C).
Concept: In adiabatic process work done is equal to change in internal energy as there is not enough time for heat transfer.
\(\Delta \mathrm{W}=\Delta \mathrm{U}\)
Work done can be calculated by relation: \(\Delta \mathrm{W}=\int \mathrm{p} \mathrm{dV}\).
Concept: In adiabatic process work done is equal to change in internal energy as there is not enough time for heat transfer.
\(\Delta \mathrm{W}=\Delta \mathrm{U}\)
Work done can be calculated by relation: \(\Delta \mathrm{W}=\int \mathrm{p} \mathrm{dV}\).
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