Q59 / 180 · KCET 2022 · Chemistry
The work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of $1 \mathrm{~L}$ to $10 \mathrm{~L}$ at $300 \mathrm{~K}$ is $(R=0.0083 \mathrm{~kJ} \mathrm{~K} \mathrm{~mol}=1)$
- A$5.8 \mathrm{~kJ}$
- B$0.115 \mathrm{~kJ}$
- C$58.5 \mathrm{~kJ}$
- Correct answerD$11.5 \mathrm{~kJ}$
Solution
Given, initial volume $V_1=1 \mathrm{~L}$
Final volume, $V_2=2 \mathrm{~L}$
Temperature $=300 \mathrm{~K}$
$$
R=0.0083 \mathrm{~kJ} \mathrm{~K} \mathrm{~mol}^{-1}
$$
$\operatorname{Moles}(n)=2$ moles
We know that,
Work done, $W=2.303 n R T \log \frac{V_2}{V_1}$
$$
=2.303 \times 2 \times 0.0083 \times 300 \log 10=11.5 \mathrm{~kJ}
$$
Hence, the correct answer is option: 'D'.