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`)

Options

A. 6.9 L
B. 6.6 L
C. 5.5 L
D. 5.8 L

Answer

B

Explanation

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`