Work Done in reversible adiabatic process is given by:

2.303 R T log \frac{V_{2}}{V_{1}}

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\frac{n R}{(γ - 1)} (T_{2} - T_{1})

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2.303 R T log \frac{V_{1}}{V_{2}}

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None of these

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Solution

The correct option is B $\frac{n R}{(γ - 1)} (T_{2} - T_{1})$
We know, from first law, $Δ U = q + w$ .
For adibatic process, $q = 0$ .
Also $Δ U$ = $n C_{v} d T$ and w = $Δ U$ = $\frac{n R}{(γ - 1)} (T_{2} - T_{1})$ .
These are the derived formula for $W_{r e v}$ in adiabatic process.

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\frac{n R (T_{2} - T_{1})}{γ - 1}

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