Question

$16\text{g}$ oxygen gas expands at $STP$ to occupy double of its original volume. Calculate the work done during the process is:

• A. $260\text{kcal}$
• B. $180\text{kcal}$
• C. $130\text{kcal}$
• D. $-272.8\text{kcal}$

Answer 1

The correct option is D $-272.8\text{kcal}$

At $STP,16\text{g}{O}_2$ or 1/2 mole $O_2$ will occupy 11.2 litre
Thus, if volume is doubled, it means
$(V_2-V_1)=22.4-11.2=11.2\text{litre}$
Now, $W=-P\times (V_2-V_1)=-1\times 11.2$ litre atm
$=-\frac{1\times 11.2\times 2}{0.0821}=-272.84\text{kcal}$