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One mole of a diatomic ideal gas (gamma=...

One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle.

Text Solution

Verified by Experts

The correct Answer is:
`791.4`
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`
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