Question

$16g$ of oxygen gas expand at STP to occupy double of its original volume. The work done during the process is:
(1 L.atm = 101.3 J)

• A. 260 kcal
• B. 180 kcal
• C. $-130$ kcal
• D. $-271.17$ kcal

Answer 1

The correct option is D $-271.17$ kcal

At STP, $16gof{O}_2$ or $\frac{1}{2}$ mole of $O_2$ will occupy 11.2 L volume.
Now, if the volume is doubled, it means $V_2$ = 22.4 L
So, ($V_2-V_1$) = 22.4 - 11.2 = 11.2 L
At STP = P = 1 atm
Now, $w=-P\times (V_2-V_1)=-1\times 11.2L.atm$ = -11.2 L.atm = $-11.2\times 101.3$ J =
$w=-1134.56$ J
Since, 1 cal = 4.184 J
w = $\frac{-1134.56}{4.184}$ = $-271.17$ cal