Question

One mole of an ideal gas at an initial temperature of $T$ $K$ does $6\;R$ joule of work adiabatically. The ratio of specific heats ${C_P}$ and ${C_V}$ is $\dfrac{5}{3}$ . The final temperature of the gas is
(A) $\left( {T + 2.4} \right)K$
(B) $\left( {T - 2.4} \right)K$
(C) $\left( {T + 4} \right)K$
(D) $\left( {T - 4} \right)K$

Answer 1

Hint: Use the equation of state for an adiabatic process made on gas. You will find that the relation between the temperature and volume of the gas is constant. Then substitute the given values and find the final temperature.
Formula used:
Change in the internal energy
$\Delta U = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
According to the first law of thermodynamics:
$\Delta U = W + \Delta Q$

Complete Step-by-step solution:
We have considered that in adiabatic processes there is no transmission of heat and mass between the surrounding and system.
The first law of thermodynamics gives relation in the change of internal energy $\Delta U$ to the work done $W$ by the system and the added heat $\Delta Q$ into it.
$\Delta U = W + \Delta Q$
Now we know that in the adiabatic process the net heat used to be zero so the process may be adiabatic.
$ \Rightarrow \Delta Q = 0$
Put it in thermodynamics equation we get,
$ \Rightarrow 0 = W + \Delta U$
$ \Rightarrow \Delta U = - W$ .............. $\left( 1 \right)$
Now, write down the expression for the change in internal energy $\Delta U$ for adiabatic processes.
$\Delta U = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Here, $n$ is the number of moles and $R$ is the gas constant.
Put the value of $\Delta U$ from the equation $\left( 1 \right)$ we get,
$ - W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$ ............... $\left( 2 \right)$
Given:
number of moles $n = 1$ ,
Initial temperature ${T_1} = T$ ,
Work done by an ideal gas $W = 6R$ ,
The ratio of specific heats $\gamma = \dfrac{5}{3}$
Substitute all the known values in the equation $\left( 2 \right)$ we get,
$ \Rightarrow - 6R = \dfrac{{1 \times R \times \Delta T}}{{\dfrac{5}{3} - 1}}$
$ \Rightarrow - 6 = \dfrac{{\Delta T}}{{\dfrac{2}{3}}}$
On further solving the above equation we get,
$ \Rightarrow \Delta T = \dfrac{2}{3} \times - 6$
$ \Rightarrow \Delta T = - 4$ ............ $\left( 3 \right)$
As we know,
$\Delta T = {T_2} - {T_1}$
where ${T_2}$ is the final temperature and ${T_1}$ is the initial temperature
Substitute all the given values in the above equation,
$ \Rightarrow - 4 = {T_2} - T$
Solve the equation to find the value of ${T_2}$ ,
$ \Rightarrow {T_2} = \left( {T - 4} \right)K$
As a result, the final temperature of the gas is $\left( {T - 4} \right)K$ .

Hence, the correct answer is option (C).

Additional information:
$\gamma $ is the fraction of specific heat at constant pressure to specific heat at constant volume. It is also known by a relation called Mayer’s formula.

Note: We have used the ideal gas law here because it was mentioned in the question that the gas is ideal. Otherwise, we would have to use real gas laws. If in the question it was mentioned that the process is sudden but not mentioned adiabatic, then also the process would be adiabatic. The adiabatic process does not involve any transfer of heat outside or inside the system.