A massless piston divides a closed, thermally insulated cylinder into two equal parts. One part contains m=28 gram of nitrogen at temperature $T_{0}$ . At this temperature one- third of molecules are dissociated into atoms and the other part is evacuated. The piston is released and the gas fills the whole volume of the cylinder. Then the piston is slowly displaced back to its initial position. The increase in internal energy of the gas is: (Neglect further dissociation of molecules during motion of the piston)

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Solution

$Molar mass of N_{2} = M = 28 g {mol}^{- 1} Number of moles of N_{2}, n_{0} = \frac{m}{M} = \frac{28}{28} = 1 mol Since, One-third molecules are dissociated into atoms. N_{2} \to 2 N ∴ No. of moles of diatomic N_{2} = \frac{2}{3} mol and No. of moles of monatomic nitrogen =2× (\frac{1}{3}) = \frac{2}{3} mol C_{V mix} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}} = \frac{(\frac{2}{3}) (\frac{3}{2}) R + (\frac{2}{3}) (\frac{5}{2}) R}{\frac{2}{3} + \frac{2}{3}} = 2 R ∴ C_{P mix} = C_{V mix} + R = 2 R + R = 3 R ∴ γ_{mix} = \frac{C_{P mix}}{C_{V mix}} = 1.5 Now, Piston is diaplaced back to its initial position, and during the adiabatic compression, volume of the gas mixture is halved. ∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow T_{0} V_{0}^{γ - 1} = T_{2} {(\frac{V_{0}}{2})}^{γ - 1} \Rightarrow T_{2} = T_{0} \sqrt{2} Change in Internal energy = ΔU = (n_{1} + n_{2}) C_{V} (T_{2} - T_{0}) = (\frac{2}{3} + \frac{2}{3}) \times 2 R (T_{0} \sqrt{2} - T_{0}) = \frac{8}{3} {RT}_{0} (\sqrt{2} - 1)$

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