One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$ at $300 \mathrm{~K}. \Delta \mathrm{U}$, q and work done in the process respectively are :
Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1}$ and $\mathrm{mol}^{-1}$
$\begin{aligned}
& \text { In } 10=2.3 \\
& \log 2=0.30 \\
& \log 3=0.48
\end{aligned}$

Question

Aicharya · Accepted Answer

$\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{V}_2}{\mathrm{~V}_1} \ & =-8.3 	imes 300 	imes \ln \left(\frac{20}{10}ight) \ & =-1.718 \mathrm{~kJ} \ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \ & \mathrm{w}=-1.718 \mathrm{~kJ} \ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0)\end{aligned}$

One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$…

Solution