Search any question & find its solution
Question:
Answered & Verified by Expert
One mole of an ideal gas at initial temperature $\mathrm{T}$, undergoes a quasi-static process during which the volume $\mathrm{V}$ is doubled. During the process the internal energy $\mathrm{U}$ obeys the equation $\mathrm{U}=\mathrm{aV}^{3}$, where $a$ is a constant. The work done during this process is-
Options:
Solution:
1166 Upvotes
Verified Answer
The correct answer is:
$7 \mathrm{RT} / 3$
$U=a V^{3}$
$\frac{f n R T}{2}=a V^{3}$
$P V=R T$
$\frac{P V}{R}=a V^{3}$
$P=C V^{2}$
$W=\int_{V}^{2 V} P d V=\int_{V}^{2 V} C V^{2} d V=\frac{C}{3}\left(8 V^{3}-V^{3}\right)=\frac{7 V^{3} C}{3}$
$\frac{f R T}{2}=a V^{3}$
$P V=R T$
$\frac{f P V}{2}=a V^{3}$
$P=\frac{2 a}{f} V^{2}$
$C=\frac{2 a}{f}$
$W=\frac{7}{3} \times V^{3} \times \frac{2 a}{f}=\frac{7}{3 f}$ f $n R T=\frac{7 R T}{3}$
$\frac{f n R T}{2}=a V^{3}$
$P V=R T$
$\frac{P V}{R}=a V^{3}$
$P=C V^{2}$
$W=\int_{V}^{2 V} P d V=\int_{V}^{2 V} C V^{2} d V=\frac{C}{3}\left(8 V^{3}-V^{3}\right)=\frac{7 V^{3} C}{3}$
$\frac{f R T}{2}=a V^{3}$
$P V=R T$
$\frac{f P V}{2}=a V^{3}$
$P=\frac{2 a}{f} V^{2}$
$C=\frac{2 a}{f}$
$W=\frac{7}{3} \times V^{3} \times \frac{2 a}{f}=\frac{7}{3 f}$ f $n R T=\frac{7 R T}{3}$
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.