Question

The work done during the expansion of a gas from a volume of $4{\text{dm}}^3$ to $6{\text{dm}}^3$ against a constant external pressure of $3\text{atm}$ is $\left(1\text{L atm = 101.32 J}\right)$

• A. $-6\text{J}$
• B. $-608\text{J}$
• C. $+304\text{J}$
• D. $-304\text{J}$

Answer 1

The correct option is B $-608\text{J}$

$\text{Work}=-P_{ext}\times \text{Volume change}$
$=-3\times (6-4)\times 101.32=6\times 101.32$
$=-607.92\text{J}=-608\text{J}$