Question

Following figure shows on 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 is at pressure P₁ and temperature T₁ in left part and gas at pressure P₂ and temperature T₂ 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 will be (Suppose x = displacement of the piston)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

As finally the piston is in equalbrium, bith the gases must be at same presure $P_f$. It is given
that displacement of piston be final state X and if A is the area of cross-section of the piston.
Hence the final volumes of the left and right part finally
can be given by the figure as
$V_1=\frac{V_0}{2}+Ax$ and $V_R=\frac{V_0}{2}-Ax$

As it is given that the container walls and the piston are
adiabatic in left side and the gas undergoes adiabatic
expansion and on final state of gas on left side the gas undergoes adiabatic compressive Thus we have for
initial and final stste of gas on left side

$P_1{\left(\frac{V_0}{2}\right)}^{\gamma }=P_f{(\frac{V_0}{2}+Ax)}^{\gamma }$ ...(i)

Similarly for gas in right side, we have

$P_2{\left(\frac{V_0}{2}\right)}^{\gamma }=P_f{(\frac{V_0}{2}-Ax)}^{\gamma }$ ...(ii)
From eq. (i) and (ii)

$\frac{P_1}{P_2}=\frac{{(\frac{V_0}{2}+A_X)}^{\gamma }}{{(\frac{V_0}{2}-A_X)}^{\gamma }}\Rightarrow A_x=\frac{V_0}{2}\frac{[P_1^{1/\gamma }-P_2^{1/\gamma }]}{[P_1^{1/\gamma }+P_2^{1/\gamma }]}$