Question

Work Done in reversible adiabatic process is given by:

• A. $2.303RT\log \frac{V_2}{V_1}$
• B. $\frac{nR}{(\gamma -1)}(T_2-T_1)$
• C. $2.303RT\log \frac{V_1}{V_2}$
• D. None of these

Answer 1

The correct option is B $\frac{nR}{(\gamma -1)}(T_2-T_1)$

We know, from first law, $\Delta U=q+w$.
For adibatic process, $q=0$.
Also $\Delta U$ = $n{C}_{v}dT$ and w = $\Delta U$ = $\frac{nR}{(\gamma -1)}(T_2-T_1)$.
These are the derived formula for $W_{rev}$ in adiabatic process.