Question

One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\dfrac{5}{3}$, the final temperature of the gas will be:-
A.$\left( {T - 2.4} \right)K$
B.$\left( {T + 4} \right)K$
C.$\left( {T - 4} \right)K$
D.$\left( {T + 2.4} \right)K$

Answer 1

Gaseous state is a less ordered state where the molecules are separated from each other by large distances and have weak intermolecular forces. The gases have all the three types of motion –translatory, rotatory and vibratory and possess high kinetic energy. An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law.

Complete step by step answer:
In the question it is said that one mole gas works adiabatically (the system where heat remains constant). We will apply the formula for work done in adiabatic processes and calculate the final temperature.
Calculating the final temperature for one mole of ideal gas.
Here the initial temperature of the ideal gas $\left( {{T_1}} \right)$ $=$ $T\,K$
At initial temperature ($T\,K$) it does work $=$ $6R$ (work done)
Since the process is adiabatic
$\gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{5}{3}$ (Given)
For an adiabatic process work done is
$W = \dfrac{{R\left( {{T_2} - {T_1}} \right)}}{{1 - \gamma }}$
Since we know that $\gamma = \dfrac{{{C_p}}}{{{C_v}}}$ we will put $\dfrac{{{C_p}}}{{{C_v}}}$ instead
$\Rightarrow W = \dfrac{{R\left( {{T_2} - {T_1}} \right)}}{{1 - \dfrac{{{C_p}}}{{{C_v}}}}}$
We have attained the above mentioned equation for W.
Now we will put the value of $\dfrac{{{C_p}}}{{{C_v}}}$ , initial temperature $\left( {{T_1}} \right)$ and work done by the ideal gas in the above attained equation. So, it can be written as follows:
$\Rightarrow 6R = \dfrac{{R\left( {{T_2} - T} \right)}}{{1 - \dfrac{5}{3}}}$
(R will be cancelled on both sides)
So, the equation becomes as follows:
$\Rightarrow 6\, = \dfrac{{\left( {{T_2} - T} \right)}}{{\dfrac{{ - 2}}{{\,\,3}}}}$
$\Rightarrow \left( {{T_2} - T} \right) = 6 \times (\dfrac{{ - 2}}{{\,\,3}})$
And hence on doing the multiplication,we have
$\Rightarrow \left( {{T_2} - T} \right) = - 4$
$\Rightarrow {T_2} = \left( {T - 4} \right)K$
(After calculation we will get the value of final temperature of the one of ideal gas)
Therefore, the final temperature for the one mole of an ideal gas is $\left( {T - 4} \right)K$.
Hence the correct option is (C).

Note:
A process is said to be adiabatic if no heat exchange takes place between the system and the surroundings during any step of the process.
In the case of adiabatic process, the system is insulated from the surroundings and $dq$ is equal to zero.