Search any question & find its solution
Question:
Answered & Verified by Expert
Work done during isothermal expansion of one mole of an ideal gas from 10 atm. to $1 \mathrm{~atm}$ at $300 \mathrm{~K}$ is
Options:
Solution:
2981 Upvotes
Verified Answer
The correct answer is:
$5744.1 \mathrm{~J}$
Given that
$P_1=10 \mathrm{~atm}, P_2=1 \mathrm{~atm}, T=300 \mathrm{~K}, n=1$
$R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$
Now, by using
$W=2.303 n R T \log _{10} \frac{P_2}{P_1}$
$=2.303 \times 1 \times 8.314 \times 300 \log _{10} \frac{1}{10}$
$W=5744.1 \mathrm{Joule}$
$P_1=10 \mathrm{~atm}, P_2=1 \mathrm{~atm}, T=300 \mathrm{~K}, n=1$
$R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$
Now, by using
$W=2.303 n R T \log _{10} \frac{P_2}{P_1}$
$=2.303 \times 1 \times 8.314 \times 300 \log _{10} \frac{1}{10}$
$W=5744.1 \mathrm{Joule}$
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.