Work done when 2 mol of an ideal gas is compressed from a volume of $5 \mathrm{~m}^{3}$ to $2 \cdot 5…
Work done when 2 mol of an ideal gas is compressed from a volume of $5 \mathrm{~m}^{3}$ to $2 \cdot 5 \mathrm{~m}^{3}$ at $300 \mathrm{~K}$, under a pressure of $100 \mathrm{k}$ pa is
$497.5 \mathrm{~kJ}$
$99.50 \mathrm{~kJ}$
$248.7 \mathrm{~kJ}$
$49 \cdot 75 \mathrm{~kJ}$
Solution
Given,
Number of moles (n) \(=2\)
Volume \(\left(\mathrm{V}_1\right)=5 \mathrm{~m}^3=5000 \mathrm{dm}^3\)
Volume \(\left(\mathrm{V}_2\right)=2.5 \mathrm{~m}^3=2500 \mathrm{dm}^3\)
Pressure \((\mathrm{p})=100 \mathrm{kPa}\)
We have, work done \((W)=-P_{\text {ext }} \cdot d V\)
\(\begin{aligned}
& =-100 \mathrm{kPa}(2500-5000) \\
& =-100 \times-2500 \mathrm{dm}^3 \\
& =250000 \mathrm{~J} \\
& =250 \mathrm{~kJ} \simeq 248.9 \mathrm{~kJ}
\end{aligned}\)