A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume `V_0` in which an ideal gas is contained under the same pressure `p_0` and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas `eta` times compared to that of the other by slowly moving the piston ?

Explanation

Let the agent move the piston as shown.
In equilibrium position
`P_(1)A+F_("agent")=P_(2)A`
`F_("agent")=(P_(2)-P_(1))A`
Work done by ht agent
`F_("agent")dx=(P_(2)-P_(1))Axxdx=(P_(2)-P_(1))dV`
Applying PV=constant for two parts, we have
`P_(1)(V_(0)+ax)=P_(0)V_(0)` and `P_(2)(V_(0)-Ax)=P_(0)V_(0)`
`P_(1)=(P_(0)V_(0))/((V_(0)+Ax))` and `P_(2)=(P_(0)V_(0))/((V_(0)-Ax))`
`P_(2)-P_(1)=(P_(0)V_(0)(2Ax))/(V_(0)^(2)-A^(2)X^(2))=(2P_(0)V_(0)V)/(V_(0)^(2)-V^(2))`
when the volume of the left end is `eta` times the volume of right end, We have
`(V_(0)+V)=eta(V_(0)+V)`
`V=((eta-1)/(eta+1))Vk_(0)`.......2
The work done is given by
`W=int_(0)^(v)(P_(2)-P_(1))dV=int_(0)^(v)(2P_(0)V_(0)V)/((V_(0)^(2)-V^(2)))dV=-P_(0)V_(0)["In"(V_(0)^(2)-V^(2))]_(0)^(v)`
`=-P_(0)V_(0)["In" (V_(0)^(2)-V^(2))-"In" V_(0)^(2)]=-P_(0)V_(0)["In"{V_(0)^(2)-((eta-1)/(eta+1))^(2)V_(0)^(2)}-"In" V_(0)^(2)]`
`=-P_(0)V_(0)["In" (V_(0)^(2)-V^(2)))-"In" V_(0)^(2)]=-P_(0)V_(0)["In"{{V_(0)^(2)-((eta-1)/(eta+1))^(2)V_(0)^(2)}-"In"V_(0)^(2)]`
`=-P_(0)V_(0)["In"{4 eta //(eta+1)^(2)}]=P_(0)V_(0)"In" [((eta+1)^(2))/(4n)]`