Question

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

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

Answer 1

The correct option is D $272.8\text{cal}$

At STP, $16\text{g}$ of $O_2$ will occupy $11.2\text{L}$
Thus, if volume is doubled, it means $(V_2-V_1)=22.4-11.2=11.2\text{L}$
Now, $W=P\times (V_2-V_1)=1\times 11.2\text{L atm}=\frac{1\times 11.2\times 2}{0.0821}=272.8\text{cal}$