3 mole of an ideal gas is taken through the process shown. BC is adiabatic and the total work done in the cycle is 560 J. (`gamma = 1.4, R = 8.31"Jm o1"^(-1) K^(-1)` , take `(4//3)^(0.4)= 1.125`) Choose correct Options.

Options

A. The work done in BC is 3255 J
B. The work done in BC is 3120 J
C. `T_B` is 450 K
D. `T_B` is 500 K

Answer

B, C

Explanation

For the process `AC, T prop V rArr " "` if `V_A = V_0 (=V_B)` then ` V_C =(4V_0)/3`
BC is adiabtic `rArr = T_B V_0^(gamma -1) = T_C V_C^(-1) " "rArr" "T_B V_0^(0.4) = 400 ((4V_0)/3)^(0.4) " "rArr" "T_B = 400 (4//3)^(0.4) = 450 K`
`rArr " "W_(BC) = - Delta U_(BC) = -(n R Delta T_(BC))/(gamma -1) = + ( 3xx 8.31 xx(450 - 400))/(0.4) = 3120 J`