Question

1 mole of an ideal gas at initial temperature T was cooled is ochorically till the gas pressure decreased n times. Then by an isobaric process, the gas was restored to the initial temperature T. Find the net heat absorbed by the gas in the whole process.

Answer 1

Let the pressure and volume of the gas at $A$ be $P$ and $V$ respectively.

Using Ideal gas equation at A : $PV=nRT$ where $n=1$

$⟹$ $PV=RT$ .............(1)

Process A to B is isochoric i.e Volume is constant :

Temperature at state B $T_2=\frac{T}{n}$ (Given)

Using $\frac{P_A}{T_A}=\frac{P_2}{T_2}$

$\therefore$ $\frac{P}{T}=\frac{P_2}{\frac{T}{n}}$ $⟹P_2=\frac{P}{n}$

Process B to C is isobaric i.e pressure is constant : Thus $P_3=P_2=\frac{P}{n}$

Temperature at state C $T_3=T$ (Given)

Using $\frac{V_B}{T_B}=\frac{V_3}{T_3}$

$\therefore$ $\frac{V}{\frac{T}{n}}=\frac{V_3}{T}$ $⟹V_3=nV$

As the initial and final temperature of the gas is same, thus the change in internal energy after both the processes is zero i.e $\Delta U_{AC}=0$

Total work done in going from A to C, $W_{AC}=W_{isochoric}+W_{isobaric}$

$\therefore$ $W_{AC}=0+\frac{P}{n}\times (V_3-V_2)=\frac{P}{n}\times (nV-V)=PV(1-\frac{1}{n})$

Using 1st law of thermodynamics : $Q_{AC}=\Delta U_{AC}+W_{AC}$

$\therefore$ $Q_{AC}=W_{AC}$ $(\because \Delta U_{AC}=0)$

$⟹$ Net heat absorbed $Q_{AC}=PV(1-\frac{1}{n})=RT(1-\frac{1}{n})$ (from (1))