One mole of a diatomic ideal gas `(gamma = 1.4)` is taken through a cyclic process starting from point A. The process `A to B` is an adiabatic compression. `B to C` is isobaric expansion, `D to A` an adiabatic expansion and is isochoric.
The volume ratio are `V_A //V_B - 16` and `V_C //V_B - 2` and the temperature at `T_A = 300k` is
Find the temperature of the gas at the point (in K).

Answer

`791.4`

Explanation

Now, the process `C to D` is adiabatic.
Therefore,
`T_D =T_C =(V_C/V_D)^(gamma -1) = (1818)(2/16)^(1.4 -1)`
`T_D = 791.A K`