Question

Work done in reversible adiabatic process is numerically not equal to:

• A. $-P_{ext}\Delta V$
• B. $n{C}_v\Delta T$
• C. $\frac{nR}{\gamma -1}\Delta T$
• D. $\Delta U$

Answer 1

The correct option is A $-P_{ext}\Delta V$

In adiabatic process, $\Delta Q=0$
From first law of thermodynamics,
$\Delta U=Q+W$
$\Rightarrow \Delta U=W$
$\Rightarrow W=\Delta U=n{C}_v\Delta T=\frac{nR}{\gamma -1}\Delta T$
Also, since the process is reversible, $w=-\int PdV$ and not $-P\Delta V$