InterviewSolution
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InterviewSolution
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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 volmek `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 ? |
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Answer» Correct Answer - A::B In equililbrium position, , `P_(1)+F_("agent")=P_(2)A` `F_("agent")=(P_(2)-P_(1))A` Elementary work done by the agent `F_("agent")dx=(P_(2)-P_(1))A xx dx=(P_(2)-P_(1))dV`……(i) Applying `PV`=constant for tow parts, we have `P_(1)(V_(0)+Ax)=P_(0)V_(0)and P_(0)(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_0V_(0)(2Ax))/(V_(0)^(2)-A^(2)x^(2))=(2P_(0)V_(0)V)/(V_(0)^(2)-V^(2)) , (V=Ax)` When the volume of the left end is `eta` times the volume of right end, we have `(V_(0)+V)=mu(V_(0)-V)` `V=((eta-1)/(eta+1))V_(0)`......(ii) The work done by the agent is given by `W=int_(0)^(V)(P_(2)-P_(1))dT=int_(0)^(V)(2P_(0)V_(0))/((V_(0)^(2)-V^(2)))dV` `=-P_(0)V_(0)[ln(V_(0)^(2)-V^(2))]_(0)^(V)=-P_(0)V_(0)[ln(V_(0)^(2)-V^(2))-lnV_(0)^(2)]` `=-P_(0)V_(0)[ln{V_(0)^(2)-((eta-1)/(eta+1))^(2)V_(0)^(2)}-lnV_(0)^(2)]` |
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