Question

The work done during the expansion of a gas from a volume of $4d{m}^3$ to $6d{m}^3$ against a constant external pressure of $3atm$ is:

• A. $-6J$
• B. $-608J$
• C. $+304J$
• D. $-304J$

Answer 1

The correct option is B $-608J$

Solution:- (B) $-608J$

As we know that, work done is given as-

$W=-P_{ext.}\Delta V$

Given:-

$P_{ext.}=3atm$

$V_i=4{dm}^3=4L$

$V_f=6{dm}^3=6L$

$\Delta V=V_f-V_i=6-4=2L$

$\therefore W=-3\times 2=-6L-atm=-608J$

Hence the work done during the expansion of a gas is $-608J$.