Question

Two moles of an ideal gas are allowed to expand from a volume of $10d{m}^3$ 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 $\left(V_1\right)=10d{m}^3=0.01m^3$

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

Final volume $\left(V_2\right)=2m^3$

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

Temperature $(T)=300K$

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

$W=-p_{\text{ext }}\left(V_2-V_1\right)$

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

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

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