Question

Find the final temperature of one mole of an ideal gas which has an initial temperature t K. The gas does $9\,R\,joules$of work adiabatically. The ratio of specific heats of this gas at constant pressure and at constant volume is $\dfrac{4}{3}$.

Answer 1

Hint: The gaseous state is a less ordered condition in which molecules are separated by a considerable distance and intermolecular interactions are weak. The ideal gas notion is beneficial because it obeys the ideal gas law and has all three types of motion-translatory, rotatory, and vibratory. It also has high kinetic particles that are not prone to interparticle interactions.

Complete step-by-step solution:
One mole gas is stated to work adiabatically in the question. We'll determine the final temperature using the formula for work done in adiabatic operations.
For one mole of ideal gas, calculate the end temperature.
Initial temperature of the ideal gas $({T_1}) = t\,K$
When the temperature is $T\,K$ is does work $ = \,9R$(work done)
In adiabatic process
$\gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{4}{3}$
In adiabatic process work done is given by:
\[W = \dfrac{{R({T_2} - {T_1})}}{{1 - \gamma }}\]
As we know from the above discussion so we can write:
\[W = \dfrac{{R({T_2} - {T_1})}}{{1 - \dfrac{{{C_p}}}{{{C_v}}}}}\]
Now,
$9R = \dfrac{{R({T_2} - T)}}{{1 - \dfrac{4}{3}}}$
$R$ will be cancel on both sides
So, we can write:
$({T_2} - {T_{}}) = 9 \times \left( {\dfrac{{ - 1}}{3}} \right)$
${T_2} = (T - 1)K$
As a result, for one mole of an ideal gas, the final temperature is $(T - 1)K$

Note:If there is no heat exchange between the system and the environment at any point during the process, it is considered to be adiabatic. The system is insulated from the environment in an adiabatic process, and \[dq\]is equal to zero.