Question

A quantity of air $ \left( {\gamma = 1.4} \right) $ at $ 27^\circ {\text{C}} $ is compressed suddenly, the temperature of the air system will
(A) Fall
(B) Rise
(C) Remain unchanged
(D) First rise and then fall

Answer 1

Hint
In this question the concept of the ideal gas is used as we know that at a certain temperature and pressure limit air acts as an ideal gas. As the air compressed suddenly, the amount of heat released from the system will be negligible.

Complete step by step answer
In the question, the specific heat ratio of the air is $ 1.4 $ and the initial temperature of the air is $ 27^\circ {\text{C}} $ .
As we know that the first law of thermodynamics is given as,
 $ {Q_{1 - 2}} = \Delta U + {W_{1 - 2}} $
Here, the amount of heat transfer from state $ 1 - 2 $ is $ {Q_{1 - 2}} $ , the change in the internal energy is $ \Delta U $ , and the amount of work done from state $ 1 - 2 $ is $ {W_{1 - 2}} $ .
As we know that if the air is compressed suddenly, the amount of heat release from the air will be negligible due to very short interval of time, so $ {Q_{1 - 2}} \approx 0 $ and hence from the equation of the first law of thermodynamics,
 $ {Q_{1 - 2}} = \Delta U + {W_{1 - 2}} $
Now, we substitute the value of the heat transfer in the above equation as,
 $ 0 = \Delta U + {W_{1 - 2}} $
By simplifying we get,
 $ \Rightarrow - {W_{1 - 2}} = \Delta U $
From the above expression we conclude that the amount of work done will be totally converted into the change in the internal energy of the air.
We know that the formula for the change in internal energy is given by,
 $ \Delta U = m{c_v}\left( {{T_2} - {T_1}} \right) $
Here, the mass of the air is $ m $ , the specific heat at constant volume is $ {c_v} $ , the initial temperature of the air is $ {T_1} $ , and the final temperature of the air is $ {T_2} $ .
As we know that the work is done on the system to compress the air, so the magnitude of the work done should be negative that is $ - {W_{1 - 2}} $ . Now, substitute the expression of the change in the internal energy in the equation of the first law of thermodynamics to obtain,
 $ - \left( { - {W_{1 - 2}}} \right) = m{c_v}\left( {{T_2} - {T_1}} \right) $
By simplification we get,
 $ \Rightarrow {W_{1 - 2}} = m{c_v}\left( {{T_2} - {T_1}} \right) $
From the above result, the value of change in internal energy is positive that means,
 $ \Rightarrow {T_2} - {T_1} > 0 $
 $ \therefore {T_2} > {T_1} $
The final temperature of the air is greater than the initial temperature. Hence, if a quantity of air is compressed suddenly, the temperature of the air system will rise.
Therefore, the correct option is B.

Note
As per the sign convention for the work done, if the work is done on the system, the magnitude of the work done should be negative and if the work is done by the system then the magnitude will be positive.