Question

16 g oxygen gas expands at STP to occupy double of its original volume. The work done during the process is

• A. -260.2 cal
• B. 260.2 kcal
• C. -272.8 cal
• D. 272.8 kcal

Answer 1

The correct option is C -272.8 cal

At STP, 16 g $O_2$ or $\frac{1}{2}$ mole $O_2$ will occupy $\left(V_1\right)$ 11.2 litre. When the volume is doubled $\left(V_2\right)=22.4L$
The change in volume = $(V_2-V_1)=22.4-11.2$ litre
Now, $W=-P_{ext}\times (V_2-V_1)$
$=-1\times 11.2$ L-atm
$=-\frac{1\times 11.2L.atm\times 2cal/K.mol}{0.0821L.atm/K.mol}$
$\Rightarrow W=-272.84$ cal