Thermodynamics (work, pressure, volume)

[Jgoshorn1]

Homework Statement

A heat engine does work by using a gas at an initial pressure of 1000 Pa and volume .1m 3. Step-by-step, it then increases the pressure to 10,000 Pa (at constant volume), increases the volume to .15m3 (at constant pressure), decreases the pressure back to 1,000 Pa (at constant volume) and returns the volume back to .1m3 (at constant pressure).

1)How much work is done by the gas during the first step?

2)How much work is done by the gas during the second step?

3)How much work is done by this heat engine in one complete cycle?

4) What is the change in internal energy during the first step?

5)What is the change in internal energy during the second step?

6)What is the change in internal energy over one complete cycle?

7)How much heat is added to the gas in the first step?

8)How much heat is added to the gas in the second step?

9) How much heat is added to the gas in one complete cycle?

Homework Equations

dW=P*dV
dU=Q-W

U=internal energy, Q= heat, w=work, p=pressure, v=volume

The Attempt at a Solution

1) W = 0 J because volume is constant)
2)dW=P(dV) = 10000(.15-.1) = 500
3)third step dW=P(dV) = 1000(.15-.1) = 50 so I just did work=500-50=450?

4)-5) I'm not sure how to find Q so I can use the dU=Q-W equation. Is there another equation that I don't know about? I don't have enough information for Q = ((kAdT)t)/L
Any hints here??
6) I think it would be 0J because the internal energy is a state function and it starts and ends the same.

7)-9) will be a piece of cake once I get 4)-5)

 

[ehild]

Nothing was given about the kind of the gas? Is it an ideal gas? If so, the internal energy is proportional to T: U=Cv* n* T where Cv is the specific heat capacity and n is the number of moles of the gas. For an ideal gas, Cv=f/2 R where f is the degrees of freedom of its molecules.

ehild

 

[Jgoshorn1]

I suppose I can assume it is an ideal gas because all of the previous problems we have done have been dealing with ideal gases only. But even if it were implied, how would I find the number of moles of gas or the degree of freedom??

 

[ehild]

You get n*R from the ideal gas law, and f=3 for mono-atomic gas molecules, 5 for two-atomic and 6 for three or more-atomic ones. Nothing was said about the kind of gas? Try f=5. The molecules of air, N2 and O2 are two-atomic. Or just give the result in terms of parameter f.

ehild

 

[rwisz]

Hello! I would like to provide some guidance on how to approach this problem.

Firstly, let's define some variables for easier understanding:
- P1 = initial pressure (1000 Pa)
- V1 = initial volume (0.1 m^3)
- P2 = final pressure (10,000 Pa)
- V2 = final volume (0.15 m^3)

1) The first step is an isochoric process (constant volume), therefore no work is done by the gas.
2) The second step is an isobaric process (constant pressure), therefore work done is dW = PdV = P(V2-V1) = 10,000(0.15-0.1) = 500 J.
3) In one complete cycle, the gas goes through 4 steps, with 2 isochoric processes and 2 isobaric processes. Therefore, the total work done by the gas in one cycle is: W = 0 + 500 + 0 + 500 = 1000 J.
4) To find the change in internal energy during the first step, we can use the equation dU = Q - W. As the volume is constant, there is no work done, therefore dU = Q. To find Q, we can use the first law of thermodynamics: Q = dU + W. As the initial and final volumes are the same, the change in internal energy is 0, therefore Q = 0 + W = 0 + 0 = 0 J.
5) To find the change in internal energy during the second step, we can use the same equation dU = Q - W. As the pressure is constant, there is no work done, therefore dU = Q. To find Q, we can use the first law of thermodynamics: Q = dU + W. As the initial and final pressures are the same, the change in internal energy is 0, therefore Q = 0 + W = 0 + 500 = 500 J.
6) As you correctly stated, the change in internal energy over one complete cycle is 0 J because it starts and ends at the same state.
7) To find the amount of heat added to the gas in the first step, we can again use the first law of thermodynamics: Q = dU + W. As we found in 4), the