# Calculating work done in a piston

1. Sep 19, 2012

### Nuclear_eng

I come from a civil engineering background and I'm about to start a masters in nuclear energy. I have no experience in thermodynamics and I've been doing a bit of revision before the course commences to get me up to speed but I'm really stuck on the following problem although it should be pretty simple.

1. The problem statement, all variables and given/known data

Calculate the work done when 0.1kg of helium, confined in a cylinder by a piston, expands adiabatically from a volume of 0.1m3 and a pressure of 8 bar to a volume of 0.8m3 and a pressure of 0.4 bar.

2. Relevant equations

W=∫V2V1P(V)dV

3. The attempt at a solution

In order to do the integration I need to know how pressure varies with volume so I tried using the following ideal gas equation

P=nRT/V

I get to this stage and I worked out that n=24.98 mol but then I become stuck. R is the ideal gas constant but how do I deal with the temperature?

Thanks

Last edited: Sep 19, 2012
2. Sep 19, 2012

### voko

This is an adiabatic reversible process, also known as isentropic, for which there is a bunch of additional properties ( = equations). If you are not familiar with the process, you need to study it, you won't be able to solve this otherwise.

3. Sep 19, 2012

### Nuclear_eng

I thought in an isentropic process the entropy doesn't change? The second part of this question asks for the change in entropy? I'm confused.

Can you not provide me with some more useful information. I've been studying thermodynamics pretty solidly for the past two weeks but I'm really struggling with this question. It's not a homework question so help is allowed according to the forum rules.

4. Sep 19, 2012

### voko

The second law of thermodynamics states that $\Delta Q \le T \Delta S$. The process you have is reversible (there is no heat exchange and no energy loss except in mechanical work), so the stricter law holds: $\Delta Q = T \Delta S$. However, $\Delta Q = 0$ (no heat exchange), $T \Delta S = 0$. Temperature is not zero, so the change in entropy must be zero. Thus the process is isentropic.

5. Sep 19, 2012

### Nuclear_eng

Do you think something is missing from this question? I've really been struggling with it and according to the text book the change in entropy is 146.5 J/K, not zero.

Also have you actually been able to get the correct answer of 72 kJ for the work done?

Thanks for the help

6. Sep 19, 2012

### voko

Hmm. The adiabatic gas law is $pV^{\gamma} = const$. This means, in particular, $p_1V_1^{\gamma} = p_2V_2^{\gamma}$, so $\frac {p_1} {p_2} = (\frac {V_2} {V_1})^{\gamma}$. For helium, a monoatomic gas, $\gamma = \frac 5 3$. $\frac {p_1} {p_2} = 8/0.4 = 20$, $(\frac {V_2} {V_1})^{\gamma} = (0.8/0.1)^{(5/3)} = 32$. These are not equal, so I fail to see how this expansion can be adiabatic.

7. Sep 19, 2012

### Nuclear_eng

When I first attempted the question I came to the same conclusion as you have above but I didn't have enough confidence in my own understanding of the subject to be sure I was right. I think I'll leave this question now and move on to some other topics.

Thank you very much for the help, much appreciated.

8. Sep 19, 2012

### voko

What this could be, however, is a quasi-adiabatic process, also known as polytropic, i.e., $pV^n = k = const$, where $1 < n < \gamma$. Indeed, in this case, as seen in my previous message, $8^n = 20$, or $n = \ln 20 / \ln 8 = 1.44 < \frac 5 3$.

9. Sep 19, 2012

### Nuclear_eng

hmmm I tried using n=1.44 and I got a value of 109kJ. I used the equations on the following link

http://www.roymech.co.uk/Related/Thermos/Thermos_Thermodynamics.html

Thanks for the help though. I think you've helped me gain more of an understanding although I still haven't quite got the correct answer. I'll keep working on it tonight

10. Sep 19, 2012

### voko

Somehow the page does not load for me. Anyway, the relevant equations are fairly standard. I think there is some mistake in the way the problem was formulated. Does it come from a textbook or was it just given to you by your instructor?

11. Sep 19, 2012

### Staff: Mentor

You can do this problem by starting from scratch, using the first law. For an adiabatic reversible process, dU = -pdV. For an ideal gas, the internal energy is a function of temperature only, such that dU = CvdT, where Cv is the heat capacity at constant volume. For an monatomic gas, the heat capacity is temperature independent. So, for an ideal monatomic gas,

CvdT = -RTdV/V (this equation uses the molar heat capacity).

Divide both sides by T to get exact differentials on both sides. Then integrate to get ratio of final- to initial temperature. Substitute back into ideal gas law to get ratio of final- to initial pressures. Use this to solve for Cv/R from the data. Cv/R should come out close to 3/2.

Chet

12. Sep 19, 2012

### voko

If it did, then, by Mayer's relation, $c_p = c_v + R \approx \frac 5 2 R$, and we would have $\gamma = \frac {c_p} {c_v} \approx \frac 5 3$. However, it is clearly not in this case. See message #6 in the thread.

Am I missing something?

13. Sep 20, 2012

### Nuclear_eng

The problem was given to me by a lecturer to do over summer before I start the course. I'm still having no luck with this question. Has anybody actually managed to get 72kJ as their answer?

14. Sep 20, 2012

### Staff: Mentor

I get 7.2 kJ if I use γ = 1.44. Of course, as voko noted above, 1.44 is not the correct value for Helium. The stated pressures and volumes are simply not consistent with adiabatic reversible expansion of Helium.

Also, it looks like there is a factor of 10 missing somewhere. The conversion factors I used were:

1 bar = 1/1.01325 atm
1 m3= 1000 liters
1 liter-atm = 101.325 joules

Chet

15. Sep 20, 2012

### Nuclear_eng

Thanks chester. That sounds much more promising but I wonder why it's a factor of 10 out. Could I ask how you got 7.2kJ? What were the equations you used?

16. Sep 20, 2012

### Staff: Mentor

Uh oh. I made a mistake in my previous post calculation. I used γ = 5/3 in the (γ - 1) of the denominator, but used γ = 1.44 in the volume ratio term. With using γ = 1.44 throughout, I get 10.9 kJ for the work, and with 5/3, I get 9.0 kJ.

17. Sep 20, 2012

### Nuclear_eng

I used the following equation to calculate work done and when I used n=5/3 I got 72kJ.

Work done = (P1V1-P2V2)/(n-1)

Work done = ((0.1 x 8e5)-(0.8 x 0.4e5))/((5/3)-1) = 72kJ

My understanding really isn't great on this. I think I might leave it till I start the course in October and ask a lecturer to explain it to me.

Thanks for all the help!

18. Sep 20, 2012

### Staff: Mentor

I found the missing factor of 10. My mistake was that I used 0.8 bars for the initial pressure, rather than 8 bars.

Chet