Question

Three samples A, B and C of the same gas (γ = 1.5) have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for A, adiabatic for B and isobaric for C. If the final pressures are equal for the three samples, find the ratio of the initial pressures.

Answer 1

There are three gases A, B and C.
It is given that initially,
V_A = V_B = V_C and T_A = T_B = T_C .

For A, the process is isothermal and for an isothermal process, PV = constant.
P_AV_A = P'_A2V_A
$\Rightarrow P{\mathit{\text{'}}}_{\mathrm{A}}\mathit{=}{P}_{\mathrm{A}}\mathit{\times }\frac{1}{2}$

For B, the process is adiabatic. So,
P_BV_B^γ = P_B'(2V_B)^γ
$\Rightarrow P_{\mathrm{B}}\text{'}\mathit{=}\frac{P_{\mathrm{B}}}{2^{1.5}}$

For C, the process is isobaric, which implies that the pressure will remain constant.
So, using the ideal gas equation
P$=\frac{\mathit{n}\mathit{R}\mathit{T}}{\mathit{V}}$, we get
$\frac{V_{\mathrm{C}}}{T_{\mathrm{C}}}\mathit{=}\frac{V{\mathit{\text{'}}}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathrm{C}}}\mathit{\Rightarrow }\frac{V_{\mathrm{C}}}{T_{\mathrm{C}}}\mathit{=}\frac{2V_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathit{C}}}$
⇒ T_C' = 2T_C

As the final pressures are equal,
$$\begin{gathered} \frac{P_{\mathit{A}}}{2}=\frac{P_{\mathit{B}}}{2^{1.5}}=P_{\mathit{C}} \\ \Rightarrow P_{\mathit{A}}\mathit{}\mathit{:}\mathit{}{P}_{\mathit{B}}\mathit{}\mathit{:}\mathit{}{P}_{\mathit{C}}=2:2^{1.5}:1 \end{gathered}$$