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One mole of a monatomic ideal gas is tak...

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 A: ` 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 = (1 //3)P_A`
Calculate the following quantities.
The temperature `T_D` (in K) [Given : `(2/5)^(2//5) = 0.85` ]

Text Solution

Verified by Experts

The correct Answer is:
`500`
As A & D are one the same isochors
`P_D/P_A = T_D/T_A "i.e. "P_D = P_A T_D/T_A rArr" But are on the same adiabatic, "`
`(P_D/P_C)^(gamma-1) = (T_D/T_C)^(gamma) =((P_A T_D)/(P_C T_A))^(gamma-1)" or "T_D^(3//5) = T_C (P_A/(P_CT_A))^(1-1/gamma)`
i.e. `T_D^(3//5)=T_B/2 (P_A/((L//3)P_A xx1000))^(2/5) = 500 K`
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