Question

One mole of $O_2$ gas having a volume equal to $22.4$ Litres at $0^{\circ }C$ and $1$ atmospheric pressure in compressed isothermally so that its volume reduces to $11.2$ litres. The work done in this process is -

• A. 1672.5 J
• B. 1728 J
• C. - 1728 J
• D. - 1572.5 J

Answer 1

Answer: (D)

Work done in adiabatic process,
$W=\mu RT{\log }_e\frac{V_e}{V_r}$
$W=1\times 8.314\times 273\times {\log }_e\left(\frac{11.2}{22.4}\right)$
$W=8.314\times 273\times {\log }_e\left(\frac{1}{2}\right)$
$W=8.314\times 273\times \left[{\log }_e(1)-{\log }_e(2)\right]$
$W=8.314\times 273\times (0-0.69)$
$W=-1566.1J$