In the cycle shown above, 0.4 moles of a diatomic ideal ideal gas, g(gamma)= 7/5
ID: 1487520 • Letter: I
Question
In the cycle shown above, 0.4 moles of a diatomic ideal ideal gas, g(gamma)= 7/5, goes through the following processes: 1-> 2: Adiabatic expansion of the volume by a factor of 2.5 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 1850 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: Your Answer:Explanation / Answer
Given that T1 = 1850 K
T2 = ?
V2 = 2.5*V1
but in adiabatic expansion
gamma = 7/5 = 1.4
T1*V1^(gamma-1) = T2*V2^(gamma-1)
1850*V1^(1.4-1) = T2*2.5^(1.4-1)*V1^(1.4-1)
T2 = 1850/2.5^(0.4) = 1282.3 K
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2->3
V2/T2 = V3/T3
but V2 = 2.5*V3 = 2.5*V1
2.5*V3/1282.3 = 2.5*V3/T3
T3 = 1282.3*2.5 = 3205.75 K
So the temparature of the gas at the end of the isobaric process is 3205.75 K