0.2 moles of an ideal gas is taken round the cycle ABC as shown in the figure. The path B rarr C is an adiabatic process A rarr B is an isochoric process and C rarr A is an isobaric process. The temperature at A and B are T_(A)300 K and T_(B) = 500 k and pressure at A is 1 atm and volume at A is 4.91. The volume at C is (given : gamma = (C_(P))/(C_(V)) = (5)/(3), R = 8.205 xx 10^(-2) L "atm mol"^(-1) K^(-1), ((3)/(2))^(2//5) = 0.81)

0.2 moles of an ideal gas is taken round the cycle ABC as shown in the

1.	0.2 moles of an ideal gas is taken round the cycle ABC as shown in the figure. The path B rarr C is an adiabatic process A rarr B is an isochoric process and C rarr A is an isobaric process. The temperature at A and B are T_(A)300 K and T_(B) = 500 k and pressure at A is 1 atm and volume at A is 4.91. The volume at C is (given : gamma = (C_(P))/(C_(V)) = (5)/(3), R = 8.205 xx 10^(-2) L "atm mol"^(-1) K^(-1), ((3)/(2))^(2//5) = 0.81)
Answer» 6.9 L 6.6 L 5.5 L 5.8 L Solution :For `A RARR B`, volume is CONSTANT `:. (P_(A))/(T_(A)) = (P_(B))/(T_(B))` or `P_(B) = (T_(B))/(T_(A)) xx P_(A) = (500)/(300) xx 1 = (5)/(3)` ATM For `B rarr C`, adiabatic process `:. (T_(C )^(gamma))/(P_(C )^(gamma - 1)) = (T_(B)^(gamma))/(P_(B)^(gamma - 1))` or `T_(C ) = ((P_(C ))/(P_(B)))^((gamma - 1)/(gamma)) xx T_(B) = [(1)/(5//3)]^(((5//3)-1)/((5//3))) xx 500` (Using (i)) `= ((3)/(5))^(2//5) xx 500` For `C rarr A`, pressure is constant `:. (V_(C ))/(T_(C )) = (V_(A))/(T_(A))` or `V_(C ) = V_(A) xx (T_(C ))/(T_(A)) = 4.9 xx ((3)/(5))^(2//5) xx 500 xx (1)/(300)` (Using (ii)) `= 4.9 xx 0.81 xx (5)/(3) = 6.6 L`