If the work done by $2 \mathrm{~mol}$ of an ideal gas during isothermal reversible expansion from $5…

If the work done by $2 \mathrm{~mol}$ of an ideal gas during isothermal reversible expansion from $5 \mathrm{~L}$ to $50 \mathrm{~L}$ is $-189.1 \mathrm{~L}$ atm at constant pressure, the temperature of the gas (in ${ }^{\circ} \mathrm{C}$ ) is
  1. $500$
  2. $227$
  3. $327$
  4. $127$

Solution

$\begin{aligned} & \mathrm{W}_{\text {rev }}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right) \\ & =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \end{aligned}$ We have, $\mathrm{n}=2 \mathrm{~mol}, \mathrm{~V}_1=5 \mathrm{~L}, \mathrm{~V}_2=50 \mathrm{~L}$, $\mathrm{W}=-189.1 \mathrm{~L} \mathrm{~atm}$ $\begin{aligned} \Rightarrow \mathrm{T} & =\frac{\mathrm{W}_{\mathrm{rev}}}{-2.303 \mathrm{nR} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)} \\ & =\frac{-189.1}{-2.303(2)(0.082) \log \left(\frac{50}{5}\right)} \\ & \cong 500 \mathrm{~K}=227^{\circ} \mathrm{C} \end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 2)

Practice more Chemical Thermodynamics questions on Aicharya