Fig. shows a double chambered vessel with thermally insulated walls and partitions. On each side there are n-moles of an ideal mono atomic gas. Initially the pressure, volume and temperature in each side is $P_{0}, V_{0}, T_{0}$ . The heater in first chamber supplies heat very slowly till the gas in the first chamber expands such that the pressure, volume and temperature of the gas on the left side is $P_{1}, V_{1}, T_{1}$ respectively and on right chamber is $P_{2} = \frac{27 P_{0}}{8}, V_{2} a n d T_{2}$ respectively

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Volume of first chamber is $[2 - {(\frac{8}{27})}^{3 / 5}] V_{0}$

Temperature in second chamber is ${(\frac{27}{8})}^{2 / 5} T_{0}$

Work done on the gas in second chamber in terms of molar heat capacity at constant volume and $T_{0} i s n C_{V} T_{0} [{(\frac{27}{8})}^{2 / 5} - 1]$

Work done on the gas in first chamber in terms of molar heat capacity at constant volume and $T_{0} i s n C_{V} T_{0} [{(\frac{27}{8})}^{2 / 5} - 1]$

answer is A, B, C.

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Detailed Solution

Since the process in chamber 2 is adiabatic
$∴ P_{0} V_{0}^{γ} = P_{2} V_{2}^{γ} ∴ P_{0} V_{0}^{5 / 3} = \frac{27}{8} P_{0} V_{2}^{5 / 3} ∴ V_{2} = {(\frac{8}{27})}^{3 / 5} V_{0}$
$∴$ Volume of chamber
$1 = 2 V_{0} - V_{2} = [2 - {(\frac{8}{27})}^{3 / 5}] V_{0} P_{0}^{1 - γ} T_{0}^{γ} = C ∴ T_{2} = {(\frac{27}{8})}^{2 / 5} T_{0}$

Work by the gas = $\frac{P_{0} V_{0} - P_{2} V_{2}}{γ - 1}$