Question

Following figure shows an adiabatic cylindrical container of volume $V_o$ divided by an adiabatic smooth piston (area of cross section = $A$) in two equal parts. An ideal gas $(\frac{C_P}{C_V}=\gamma )$ is at pressure $P_1$ and temperature $T_1$ in left part and gas at pressure $P_2$ and temperature $T_2$ in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts can be (Suppose $x$ is displacement of the piston)

• A. $P_2$
• B. $P_1$
• C. $\frac{P_1{\left(\frac{V_o}{2}\right)}^{\gamma }}{{(\frac{V_o}{2}+Ax)}^{\gamma }}$
• D. $\frac{P_2{\left(\frac{V_o}{2}\right)}^{\gamma }}{{(\frac{V_o}{2}+Ax)}^{\gamma }}$

Answer 1

Answer: (C), (D)

$\frac{P_2{\left(\frac{V_o}{2}\right)}^{\gamma }}{{(\frac{V_o}{2}+Ax)}^{\gamma }}$

Let the initial volume of each container be $\frac{V_o}{2}$. Let $P_f$ be the final pressure on both sides(in equilibrium).
It is not specified which of the two pressures are bigger initially. So, we have to consider both the possibilities.

Case 1: $P_1>P_2$
If the displacement of the piston is $x$, then the final volumes will be different.
Final volume of left part, $V_{fl}=\frac{V_o}{2}+Ax$
Final volume of right part, $V_{fr}=\frac{V_o}{2}-Ax$

The given process is adiabatic as there is no exchange of heat.
$P_i{V}_i^{\gamma }=P_f{V}_f^{\gamma }$
Applying this to the left part we get,
$P_1{\left(\frac{V_o}{2}\right)}^{\gamma }=P_f{(\frac{V_o}{2}+Ax)}^{\gamma }$
$P_f=\frac{P_1{\left(\frac{V_o}{2}\right)}^{\gamma }}{{(\frac{V_o}{2}+Ax)}^{\gamma }}$
This is option (c)

Case 2:$P_1<P_2$
Final volume of left part, $V_{fl}=\frac{V_o}{2}-Ax$
Final volume of right part, $V_{fr}=\frac{V_o}{2}+Ax$

$P_i{V}_i^{\gamma }=P_f{V}_f^{\gamma }$
Applying this to the right part we get,
$P_2{\left(\frac{V_o}{2}\right)}^{\gamma }=P_f{(\frac{V_o}{2}+Ax)}^{\gamma }$
$P_f=\frac{P_2{\left(\frac{V_o}{2}\right)}^{\gamma }}{{(\frac{V_o}{2}+Ax)}^{\gamma }}$
This is option (d)