In the cycle shown above, 1.2 moles of a diatomic ideal gas, g(gamma) = 7/5, goe
ID: 1487285 • Letter: I
Question
In the cycle shown above, 1.2 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 3.5
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 1450 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
Hi,
I will divide the problem in three parts, the first part will be the adiabatic expansion from the state 0 to the state 1, the second part will be the constant pressure compression from the state 1 to the state 2, and the third one will be the increase in pressure from the state 2 back to the state 0.
First we have to define what we know:
T0=1450K
V1=3.5V0
=7/5
n=1.2moles
Ok, let’s start
Part 1
An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. We have different relations for adiabatic processes. They are two:
1) P0*V0^=P1*V1^=K
2) T0*V0^( -1)=T1*V1^( -1)
We know that V1=3.5*V0 so we can obtain T1 from the second relation:
1450K*V0^(7/5-1)=T1*(3.5V0)^(7/5-1)
T1=1450K/(3.5^(7/5-1))
T1=878.5K
Part 2
In this part, we are looking for the final temperature at the end of the isobaric process. Since the pressure is a constant for this process, we can rearrange the ideal Gas Law putting the variables on one side of the equality and the constants on the other side, that is:
V1/T1=n*R/P1
V2/T2=n*R/P2
Then as we have and isobaric process, P1=P2, the relation turns:
V1/T1=V2/T2
Part 3
The final process in this problem is an isochoric one, where we increase the
pressure and in consequence, the temperature
The volume is constant, so we have that:
V2=V0
Using the relation of the part 2, we obtain:
V1/T1=V0/T2
And using the relation V1=3.5V0 we can calculate T2
T2= V0*T1/(3.5V0)
T2=T1/3.5
T2=878,5K/3.5
T2=251K The answer is logical because in the part 2 we are decreasing the volume without increasing the pressure therefore, the only way we can do that is decreasing the temperature of the system .
I hope this will help :)