Search any question & find its solution
Question:
Answered & Verified by Expert
An ideal gas expands isothermally and reversibly from $10 \mathrm{~m}^{3}$ to $20 \mathrm{~m}^{3}$ at $300 \mathrm{~K}$ performing $5 \cdot 187 \mathrm{~kJ}$ of work on surrrounding. Calculate number of moles of gas undergoing expansion? $\left(\mathrm{R}=8 \cdot 314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.$ )
Options:
Solution:
2737 Upvotes
Verified Answer
The correct answer is:
3
$\mathrm{V}_{1}=10 \mathrm{~m}^{3}, \mathrm{~V}_{2}=20 \mathrm{~m}^{3}, \mathrm{~T}=300 \mathrm{~K}$
$\mathrm{W}_{\max }=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}, \mathrm{n}=?$
$\mathrm{W}_{\max }=-2.303 \mathrm{~nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
$\therefore-5187=-2.303 \times n \times 8.314 \times 300 \log _{10} \frac{20}{10}$
$\therefore \mathrm{n}=\frac{5187}{2.303 \times 8.314 \times 300 \times \log _{10} 2}=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}$
$\therefore \mathrm{n}=3 \mathrm{~mol}$
$\mathrm{W}_{\max }=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}, \mathrm{n}=?$
$\mathrm{W}_{\max }=-2.303 \mathrm{~nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
$\therefore-5187=-2.303 \times n \times 8.314 \times 300 \log _{10} \frac{20}{10}$
$\therefore \mathrm{n}=\frac{5187}{2.303 \times 8.314 \times 300 \times \log _{10} 2}=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}$
$\therefore \mathrm{n}=3 \mathrm{~mol}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.