Question

An ideal gas expands from $1\times {10}^{-3}{m}^3$ to $1\times {10}^{-2}{m}^3$ at $300K$ against a constant pressure of $1\times {10}^{5}N{m}^{-2}$. The work done is:

• A. $900kJ$
• B. $-900kJ$
• C. $270kJ$
• D. $-900J$

Answer 1

The correct option is D $-900J$

Solution:- (D) $-900J$

Given:-

$P=1\times {10}^{5}N/m^2$

$V_i=1\times {10}^{-3}{m}^3$

$V_f=1\times {10}^{-2}{m}^3$

$\Delta V=V_f-V_i={10}^{-2}-{10}^{-3}=9\times {10}^{-3}{m}^3$

As we know that,

$W=-P\Delta V$

$W=-1\times {10}^5({10}^{-2}-{10}^{-3})$

$\Rightarrow W=-{10}^5\times (9\times {10}^{-3})$

$\Rightarrow W=-9\times {10}^2=-900J$

Hence the work done is $-900J$.