Question

Two moles of an ideal gas are allowed to expand from a volume of $10 \, dm^3$ to $ 2m^3$ at 300 K against a pressure of 101.325 KPa. Calculate the work done.

• A. - 20.16 kJ
• B. 13.22 kJ
• C. - 810.6 J
• D. -18.96 kJ

Answer 1

Answer: (A)

Given,

Number of moles $(n)=2$

Initial volume $\left(V_{1}\right)=10\, dm ^{3}=0.01 \,m ^{3}$

$\left(\because 1 \,dm ^{3}=\frac{1}{1000} m ^{3}\right)$

Final volume $\left(V_{2}\right)=2 \,m ^{3}$

$p_{\text {ext }}=101.325 \,kPa$

Temperature $(T)=300 \,K$

$\because$ Work done due to change in volume against a constant pressure is given by

$W=-p_{\text {ext }}\left(V_{2}-V_{1}\right)$

$\therefore W=-101.325 kPa \times(2-0.01) m ^{3}$

$=-201.636 kPa m ^{3}$ or $-201.6 kJ$

$\left(\because 1 kJ =1 kPa m ^{3}\right)$