In the cycle shown above, 0.8 moles of a diatomic ideal gas, g(gamma) = 7/5, goe
ID: 1488917 • Letter: I
Question
In the cycle shown above, 0.8 moles of a diatomic ideal gas, g(gamma) = 7/5, goes through the following processes:
1 -> 2: Adiabatic expansion of the volume by a factor of 4.0
2 -> 3: Constant pressure compression back to the initial volume
3 -> 1: Increase in pressure at constant volume back to the initial state.
The initial temperature of the gas is 1650 K.
What is the temperature of the gas at the end of the isobaric (constant pressure) process? Give your answer in Kelvin, but don't include units in your answer.
Q=0Explanation / Answer
1-->2
T1*V1^(gamma-1) = T2*v2^(gamma-1)
T1 = 1650 K
V1 = 4 v2
1650*4^(7/5 -1 ) = T2*1^(7/5 -1 )
T2 = 2872.8 K
for
2-->3
T2 = 2872.8 K
T2*v2^(gamma-1) = T3*v1^(gamma-1)
2872*4^(7/5 -1 )= T3
T3 = 5000 K <<-------answer