Question

An adiabatic vertical cylinder fitted with and adiabatic frictionless piston contains $n$ moles of an ideal diatomic gas. The initial volume and temperature of the gas are $V_{0}$ and $T_{0}$ , respectively, with the piston fixed tightly. The atmospheric pressure is $P_{0}$ . If the piston be released, find the equilibrium temperature and volume of the gas, if the heat capacity of the cylinder and piston is negligibly small. Weight of the piston and cross-sectional area of the cylinder are $W$ and $A$ . respectively.

Accepted Answer

The system is isolated from the surrounding, the work done by the gas will be at the cost of its own internal energy. Let the final temperature and volume be $T$ and $V$ , respectively. Let us suppose that the final pressure $P$ of the gas will be just enough to hold the piston in equilibrium against its weight and the atmospheric pressure.

i.e., $P A = P_{0}^{A} + W$

$\Rightarrow P = P_{0} + \frac{W}{A}$

Change in internal energy of the gas,

$Δ U = n C_{V} Δ T = \frac{n R Δ T}{(γ - 1)} = \frac{n R [T - T_{0}]}{(\frac{7}{5} - 1)} = \frac{5 n R (T - T_{0})}{2}$

Work done by the gas against a constant pressure, $W = P Δ V$

$\Rightarrow W = (P_{0} + \frac{W}{A}) (V - V_{0})$

Using equation of state $P V = n R T$ , we have

$(P_{0} + \frac{W}{A}) V = n R T$ $. . . (ii)$

Since $W + Δ U = 0$ $[∵ Q = 0]$

$\Rightarrow (P_{0} + \frac{W}{A}) (V - V_{0}) + \frac{5 n R (T - T_{0})}{2} = 0$ $. . . (iii)$

Solving Eqs. $(ii)$ and $(iii)$ , for $V$ and $T$ ,we get $T = \frac{5 T_{0}}{7} + (P_{0} + \frac{W}{A}) \frac{2 V_{0}}{7 n R}$ and $V = \frac{2 V_{0}}{7} + \frac{5 n R T_{0}}{7 (P_{0} + \frac{W}{A})}$

Hence, the equilibrium Temperature and Volume of gas are $T = \frac{5 T_{0}}{7} + (P_{0} + \frac{W}{A}) \frac{2 V_{0}}{7 n R}$ and $V = \frac{2 V_{0}}{7} + \frac{5 n R T_{0}}{7 (P_{0} + \frac{W}{A})}$ respectively.

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