Q140 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics

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

  • A
    2
  • Correct answer
    B
    4
  • C
    6
  • D
    8

Solution

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.

Hence, the correct answer is option: 'B'.

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