Question

One mole of a diatomic gas undergoes a process $TV$ = constant . If the temperature of gas increases by $\Delta t$, work done for this process is ?
A. $ - R\Delta t$
B. $ - 3R\Delta t$
C. $ - 5R\Delta t$
D. $ - 7R\Delta t$

Answer 1

Hint: Remember all the types of thermodynamic processes and try to link the given above equation with the process. Now we see that the given information of TV = constant is matching with the adiabatic process, which has the equation of $P{V^\gamma } = $ constant.

Complete step by step answer:
Adiabatic process: In thermodynamics, the adiabatic process is a type of thermodynamic process which occurs without any transfer of heat and mass between the system and the surroundings. The curve of the adiabatic process is similar to that of an isothermal process, but it has a slope higher than that of the adiabatic process. Remember that the adiabatic process transfers energy to its surroundings only as work.

The basic formula of an adiabatic process is
$P{V^\gamma } = $ constant
where, $P$ = Pressure and $V$ = volume .
The other formula of adiabatic process is
$T{V^{\gamma - 1}} = $ constant
Where $T$ = temperature.
Now if we compare the given equation in the question $TV = $ constant and $T{V^{\gamma - 1}} = $ constant,
We get $\gamma = 2$ . Thus we can now apply the value of $\gamma = 2$ to the formula of work in an adiabatic process and get the value needed.
Thus $W = - \dfrac{{nR\Delta t}}{{\gamma - 1}}$
Now placing the values , we get
$\therefore W = - R\Delta t$

Hence the correct answer is option A.

Note: The conditions of an adiabatic process are to be remembered :
-The system should be perfectly insulated from its surroundings.
-The process must be carried out very quickly, this is to ensure that there is not enough time for the heat transfer to have taken place.
A suitable example for an adiabatic process is a pendulum oscillating in a vertical plane, a quantum harmonic oscillator ,etc .