Thermodynamics

JEE Main level, single correct, medium

117 of 190

Question 117 of 190

Statement - I : When μ\mu amount of an ideal gas undergoes adiabatic change from state (P1,V1,T1)\left({P}_{1},{V}_{1},{T}_{1}\right) to state (P2,V2,T2)\left({P}_{2},{V}_{2},{T}_{2}\right), then work done is W=μR(T2−T1)1−γW=\dfrac{\mu R\left({T}_{2}-{T}_{1}\right)}{1-\gamma }, where γ=CPCV\gamma =\dfrac{{C}_{P}}{{C}_{V}} and R=R= universal gas constant.
Statement - II : In the above case, when work is done on the gas, the temperature of the gas would rise.

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Answer: (A) Both statement-I and statement-II are true.

The work done in an adiabatic process is given by,
Wadiabatic=nR(Tf−Ti)1−γ→{W}_{adiabatic}=\dfrac{nR\left({T}_{f}-{T}_{i}\right)}{1-\gamma }\rightarrow statement I is true.
For adiabatic process
∆Q=0∆Q=0. From the first law of thermodynamics,
∆Q=W+ΔU∆Q=W+\Delta U
0=W+ΔU0=W+\Delta U
ΔU=−W\Delta U=-W
If work is done on the gas, i.e. work is negative

∴ΔU∴\Delta U is positive.
∴∴ The temperature will increase. Hence, statement II is also true.

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