Question

One mole of a gas occupying $3$ litres volume is expanded against constant external pressure of one atm to a volume of $15$ litre. The work done by the system is:

• A. $-1.215\times {10}^3$ J
• B. $+12.15\times {10}^3$ J
• C. $+121.5\times {10}^3$ J
• D. $+1.215\times {10}^3$ J

Answer 1

The correct option is A $-1.215\times {10}^3$ J

Solution:- (A) $-1.215\times {10}^{3}J$

As we know that, work done by a system against an external pressure is given as :

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

Given:-

$P_{ext.}=1atm$

$V_f=15L$

$V_i=3L$

$\therefore W=-1(15-3)$

$\Rightarrow W=-12L-atm=-1215.96J=1.215\times {10}^{3}J$

[1 L atm = 101.325 J]

Hence, the work done by the system is $-1.215\times {10}^{3}J$.