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.