Search any question & find its solution
Question:
Answered & Verified by Expert
The mean kinetic energy of monoatomic gas molecules under standard conditions is $\left\langle E_1\right\rangle$. If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to $\left\langle E_2\right\rangle$. The ratio $\frac{\left\langle E_2\right\rangle}{\left\langle E_1\right\rangle}$ is
Options:
Solution:
2086 Upvotes
Verified Answer
The correct answer is:
4
In adiabatic process,
$$
T V^{\gamma-1}=\text { constant }
$$
In given conditions,
$$
V_1=V\left(\text { say), so } V_2=\left(\frac{V}{8}\right)\right.
$$
and $\gamma=\frac{5}{3}$ (gas is monoatomic)
So, $T_1 V^{\frac{5}{3}-1}=T_2\left(\frac{V}{8}\right)^{\frac{5}{3}-1} \Rightarrow \frac{T_2}{T_1}=\frac{V^{\frac{2}{3}}}{\left(\frac{V}{8}\right)^{\frac{2}{3}}}=8^{\frac{2}{3}}=4$
So, temperature of gas increases by 4 times.
As mean KE of gas molecule is proportional to its temperature, then $\mathrm{KE}$ also increases by 4 times.
$$
T V^{\gamma-1}=\text { constant }
$$
In given conditions,
$$
V_1=V\left(\text { say), so } V_2=\left(\frac{V}{8}\right)\right.
$$
and $\gamma=\frac{5}{3}$ (gas is monoatomic)
So, $T_1 V^{\frac{5}{3}-1}=T_2\left(\frac{V}{8}\right)^{\frac{5}{3}-1} \Rightarrow \frac{T_2}{T_1}=\frac{V^{\frac{2}{3}}}{\left(\frac{V}{8}\right)^{\frac{2}{3}}}=8^{\frac{2}{3}}=4$
So, temperature of gas increases by 4 times.
As mean KE of gas molecule is proportional to its temperature, then $\mathrm{KE}$ also increases by 4 times.
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.