The work done in adiabatic process is given by:

n R \frac{{(T}_{1} - T_{2})}{γ}

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n R \frac{{(T}_{1} - T_{2})}{γ - 1}

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n γ {(T}_{1} - T_{2})

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γ \frac{{(T}_{1} - T_{2})}{n}

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Solution

The correct option is B $n R \frac{{(T}_{1} - T_{2})}{γ - 1}$
Work done in adiabatic process:
$Δ W = - Δ U = n C_{v} (T_{1} - T_{2}) = \frac{P_{1} V_{1} - P_{2} V_{2}}{(γ - 1)} = \frac{n R (T_{1} - T_{2})}{γ - 1}$

work done by system is $(+ v e)$ , if $T_{1} > T_{2}$ (hence expansion)
work done on the system is $(- v e)$ if $T_{1} < T_{2}$ (hence compression)

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The work done in adiabatic process is given by:

The correct option is B nR(T1−T2)γ−1Work done in adiabatic process:ΔW=−ΔU=nCv(T1−T2)=P1V1−P2V2(γ−1)=nR(T1−T2)γ−1work done by system is (+ve), if T1>T2 (hence expansion)work done on the system is (−ve) if T1<T2 (hence compression)