Question

Two vessels A and B of equal volume V₀ are connected by a narrow tube that can be closed by a valve. The vessels are fitted with pistons that can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (C_p/C_v = γ) at atmospheric pressure p₀ and atmospheric temperature T₀. The walls of vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and pressure.

Answer 1

Initial pressure of the gas in both the vessels = P₀
Initial temperature of the gas in both the vessels = T₀
Initial volume = V₀
$\frac{\mathrm{C}p}{\mathrm{C}v}=\gamma$

(a) The temperature inside the diathermic vessel remains constant. Thus,
P₁V₁ = P₂V₂
⇒ P₀V₀ = P₂ × 2V₀
$\Rightarrow P_2=\frac{P_0}{2}$
Temperature = T₀
The temperature inside the adiabatic vessel does not remain constant.
So, for an adiabatic process,
T₁V₁^γ−1 = T₂V₂^γ−1
⇒ T₀V₀^γ−1 = T × (2V₀)^γ−1
⇒ T₂ = T₀ × 2^1−γ
⇒ P₁V₁^γ = P₂V₂^γ
P₀V₀^γ = P₂ × (2V₀)^γ
$\Rightarrow P_2=\left(\frac{P_0}{2^{\gamma }}\right)$

(b) When the valves are open, the temperature remains T₀ throughout, i.e. T₁ = T₂ = T₀ .

Also, pressure will be same throughout. Thus, P₁ = P_2. As, temperature has not changed on side 1, so pressure on this side will also not change (volume is also fixed due to fixed piston) and will be equal to $P_0$ (pressure is an intrinsic variable).
On side 2, pressure will change to accommodate the changes in temperature on this side.
So, P₀ = P₁ + P₂
⇒ 2P₁ = 2P₂
So, P₁ = P₂ = $\frac{{\mathrm{P}}_0}{2}$