Question

Following figure shows on adiabatic cylindrical container of volume $V_0$ 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 equlibrium. The final common pressure of the two parts will be
[Suppose $x$ = displacement of the piston]

• A. $P_2$
• B. $P_1$
• C. $\frac{P_1\left(\frac{V_0}{2}\right)}{{(\frac{V_0}{2}+Ax)}^{\gamma }}$
• D. $\frac{P_2{\left(\frac{V_0}{2}\right)}^{\gamma }}{{(\frac{V_0}{2}+Ax)}^{\gamma }}$

Answer 1

The correct option is C $\frac{P_1\left(\frac{V_0}{2}\right)}{{(\frac{V_0}{2}+Ax)}^{\gamma }}$

As finally the piston is in equilibrium, both the gases must be at same pressure $P_f$. It is given that displacement of piston be in final state $x$ and $A$ is the area of cross-section of the piston.
The final volumes of the left and right parts are as shown in the figure below
Volume of the gas in the left part$V_L=\frac{V_0}{2}+Ax$
and
Volume of the gas in the right part $V_R=\frac{V_0}{2}-Ax$
As it is given that the container walls and the piston are adiabatic, in left side the gas undergoes adiabatic expansion and on the right side the gas undergoes adiabatic compression.
Thus, from the equation of state of an ideal gas undegoing an adibatic process interms of $P$ and $V$ we can write that,
For gas on left side,
$P_1{\left(\frac{V_0}{2}\right)}^{\gamma }=P_f{(\frac{V_0}{2}+Ax)}^{\gamma }..............\left(1\right)$
Similarly for gas on right side, we have
$P_2{\left(\frac{V_0}{2}\right)}^{\gamma }=P_f{(\frac{V_0}{2}-Ax)}^{\gamma }.................\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{P_1}{P_2}=\frac{{(\frac{V_0}{2}+Ax)}^{\gamma }}{{(\frac{V_0}{2}-Ax)}^{\gamma }}$
${\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma }}=\frac{(\frac{V_0}{2}+Ax)}{(\frac{V_0}{2}-Ax)}$
Using componendo - dividendo rule we get,
$\frac{P_1^{\frac{1}{\gamma }}+p_2^{\frac{1}{\gamma }}}{P_1^{\frac{1}{\gamma }}-P_2^{\frac{1}{\gamma }}}=\frac{V_0}{2Ax}$
\(\Rightarrow Ax = \dfrac{V_0}{2}\left[\dfrac{[P_1\dfrac{ 1^{1/\gamma-P_2^{1/\gamma}}]}{[P_1^{}1/\gamma+P_2^{1/\gamma}]}\right]\)