ThermodynamicsHard

Question

A portion of helium gas in a vertical cylindrical container is in thermodynamic equilibrium with the surroundings. The gas is confined by a movable heavy piston. The piston is slowly elevated by a distance H from its equilibrium position and then kept in the elevated position long enough for the thermal equilibrium to be re-established. After that, the container is insulated and then the piston is released. After the piston comes to rest, what is the new equilibrium position of the piston with respect to initial position?

Options

A.The piston ends up 0.4H above its initial position.
B.The piston ends up 0.6H above its initial position.
C.The piston ends at its initial position.
D.The piston ends up 0.4H below its initial position.

Solution

After 1st step, the process is irreversible adiabatic. Hence, ΔU = w

$n.C_{V,m}.\left( T_{2} - T_{1} \right) = - P_{ext}\left( V_{2} - V_{1} \right) $$${\text{Or, }n.\frac{3}{2}R\left( \frac{P_{1}V_{2}}{nR} - \frac{P_{1}V_{1}}{nR} \right) = - P_{1}\left\lbrack V_{2} - \left( V_{1} + A.H \right) \right\rbrack }{\therefore V_{2} = V_{1} + 0.4\text{ H.A} \Rightarrow x = 0.4H}$$

(The final pressure of gas after 2nd step will remain same as initial, beginning of processes.)

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