Question

A diatomic gas initially at $18^\circ C$ is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be
(1) $10^\circ C$
(2) $887^\circ C$
(3) $668K$
(4) $144^\circ C$

Answer 1

Hint : Diatomic molecules are molecules made of only two atoms, the two atoms may be of the same or different elements. An adiabatic process is a process that occurs without exchanging heat or mass between the system and its surroundings.

Formula used:
$x^\circ C = (x + 273)K$
For the adiabatic process $T{V^{\gamma - 1}} = constant$
Where $T$ is the temperature, $V$ is the volume, $\gamma$ is the specific heat at constant pressure divided by specific heat at constant volume.

Complete step by step answer
Let, The initial volume of the gas is ${V_1}$ .
The initial temperature of the gas is ${T_1}$ .
The final volume of the gas is ${V_2}$ .
The final temperature of the gas is ${T_2}$ .
It is given in the question that,
${T_1} = 18^\circ C$
As $x^\circ C = (x + 273)K$
Therefore,
$\Rightarrow {T_1} = 18 + 273K$
$\Rightarrow {T_1} = 291K$
It is given ${V_2} = \dfrac{{{V_1}}}{8}$ .
We know that for the adiabatic process $T{V^{\gamma - 1}} = constant$
Where $T$ is the temperature, $V$ is the volume, $\gamma$ is the specific heat at constant pressure divided by specific heat at constant volume.
Hence,
${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)^{\gamma - 1}}$
For diatomic gas $\gamma = 1.4$
$\Rightarrow \dfrac{{291}}{{{T_2}}} = {\left( {\dfrac{{\dfrac{{{V_1}}}{8}}}{{{V_1}}}} \right)^{1.4 - 1}}$
$\Rightarrow \dfrac{{291}}{{{T_2}}} = {\left( {\dfrac{1}{8}} \right)^{0.4}}$
$\Rightarrow {T_2} = 291 \times {(8)^{0.4}}$
On solving further we get
$\Rightarrow {T_2} = 668.5K$
Hence the correct answer to our question is (C); $668K$ .

Additional Information
$\gamma$ is the specific heat at constant pressure divided by specific heat at constant volume.
Let the specific heat at constant pressure be ${C_P}$ and the specific heat at constant volume be ${C_V}$ .
${C_V} = \dfrac{{fR}}{2}$ where $f$ is the degree of freedom of the gas molecule which is $5$ for diatomic gas molecule and $R$ is the universal gas constant.
${C_P} - {C_V} = R$
Hence,
${C_P} = \dfrac{{(f + 2)R}}{2}$
Therefore,
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$
On substituting values we get,
$\Rightarrow \gamma = \dfrac{{f + 2}}{f}$
$\Rightarrow \gamma = \dfrac{7}{5}$
Hence,
$\Rightarrow \gamma = 1.4$ .

Note
The formula $T{V^{\gamma - 1}} = constant$ is only valid for a reversible adiabatic process as in the question it was not mentioned the process is reversible or irreversible. Unless and until the question says the process is irreversible consider it as a reversible process.