Q11 / 160 · AP EAMCET 2023 (16 May Shift 2) · Chemistry

If $100 \mathrm{~L}$ of an ideal gas at a pressure of $2 \mathrm{~atm}$ compressed isothermally and reversibly to a final volume of ' $\mathrm{X}$ ' $\mathrm{L}$ releases $-460.6 \mathrm{~L} \mathrm{~atm}$ heat, the final volume ' $\mathrm{X}$ ' (in $\mathrm{L}$ ) is

  • A
    $1$
  • B
    $20$
  • Correct answer
    C
    $10$
  • D
    $2$

Solution

$\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=0 \Rightarrow \mathrm{w}=-\mathrm{q}$ We have, $\mathrm{q}=-460.6 \mathrm{~L} \mathrm{~atm}$ $\Rightarrow \mathrm{w}=+460.6 \mathrm{~L} \mathrm{~atm}$ Now, $\begin{aligned} & \mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & =-2.303\left(\mathrm{P}_1 \mathrm{~V}_1\right) \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & =-2.303(2 \times 100) \log \left(\frac{\mathrm{V}_2}{100}\right) \\ & =-460.6\left[\log \left(\mathrm{V}_2\right)-\log (100)\right] \\ & -460.6\left[\log \left(\mathrm{V}_2\right)-2\right]=921.2-460.6 \log \left(\mathrm{V}_2\right)=460.6 \\ & \Rightarrow \log \left(\mathrm{V}_2\right)=\frac{921.2-460.6}{460.6}=1 \\ & \Rightarrow \mathrm{V}_2=10 \mathrm{~L}(\mathrm{X}) \end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions