In the cycle shown above, 1.0 moles of a diatomic ideal gas, g(gamma) = 7/5, goe
ID: 1487600 • Letter: I
Question
In the cycle shown above, 1.0 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 1300 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.
Explanation / Answer
we know
PV^y = constant ( for adiabatic process )
TV^y-1 = constant
T1V1^y-1 = T2V2^y-1
T2 = T1 * (V1/V2)^y-1
now given
V1/V2 = 1/4
y = 7/5 = 1.4
T1 = 1300
T2 = 1300 * (1/4)^.4
T2 = 746.65
now from 2 to 3 at ( a constant pressure )
V2/T2 = V3/T3
T3 = V3 * T2 / V2
now V3/V2 = 4
T3 = 4 * T2
T3 = 2986.6