One mole of a monatomic ideal gas is taken through the cycle shown in the figure: `A to B` adiabatic expansion

`B to C:` cooling at constant volume
`C to D` adiabatic compression `D to C: ` heating at constant volume.
The pressure and temperature at `P_A, P_B` etc. `T_A , T_B` etc. are denoted etc. respectively. Given that
`T_A = 1000 K, P_B =(2 //3)P_A` and `P_C = (t //3)P_A`
Calculate the following quantities.
The heat lost by the gas in process `B to C` (in J)

Answer

`-5297.63`

Explanation

Heat lost by the gas in process `B to C`
`Q_(BC )=C_v (T_B - T_C) = (3R)/2 (T_B - T_C)`
From gas laws `P_B/P_c = T_B/T_C ` (at constant volume)
or `T_C = (T_B T_C)/(P_B) = (T_B(P_A/3))/((2P_A/3))=T_B/2 = 850/2 = 425 K" ":." Heat lost during the process "BtoC`
`Q_(BC) = 3/2 xx8.31xx(425 - 850) = -5297 . 63 J`