# Ideal Gas Expansion - Finding final pressure and work done by gas

1. Oct 27, 2014

1. The problem statement, all variables and given/known data
a ideal monoatomic gas initally has a temperature of 315K and a pressure of 6.87atm . It then expands from a volume of 440cm^3 to volume 1550cm^3 .

If the expansion is isothermal, what is (A) the final pressure (in atm's) and (B) the work done by he gas.

2. Relevant equations
$PV^{\gamma}=$ constant

3. The attempt at a solution
For the final pressure I just did the following, no idea if this is correct mind you.

$P_iV_i^{\gamma}=P_fV_f^{\gamma} \\ P_f=\frac{P_iV_i^{ \gamma } } { V_f^{ \gamma } }=\frac{ (6.87 \times 1.01 \times 10^5)(0.44)^{ \frac{7}{5} } }{1.55^{\frac{7}{5}}} = 119027.5 Pa \\ \frac{119027.5}{1.01 \times 10^5} = 1.178 atm \\$

Then for the work done i did the below,

I first found the constant, c, the value of PV^7/5, which was 219842.6

then did
$W=\int_{V_i}^{V_f} P \,\, dV =\int_{V_i}^{V_f} \frac{c}{V^{\gamma}} = c \int_{V_i}^{V_f} V^{-\gamma}$

I forgot to mention, our lecturer told us to assume gamma is 7/5 for this case.
$W =c \int_{V_i}^{V_f} V^{-\gamma} = [c(\frac{5V^\frac{12}{5}}{12} ) ]_{V_i}^{V_f} \\ W = [(219842.6)\frac{5(1.178 \times 1.01 \times 10^5)^{\frac{12}{5}}}{12} ] - [(219842.6)\frac{5(6.87 \times 1.01 \times 10^5)^{\frac{12}{5}}}{12} ] \\ W= (1.39 \times 10^{17}) - (9.57 \times 10^{18} ) = -9.43 \times 10^{18} J$

Extremely new to this material and have not got my head around the material yet, so I would be suprised if this is correct. The value does seem rather low. What I am a little concerned with is the fact they explicitly state the temperature of the gas, but I have not used temperature in any of the equations/formulas used. The question does go on further to ask for the final volume and work if the process was adiabatic rather than isothermal, so it may be used there?

EDIT: Just realised a school boy error with the integration. Have gone back and already amended it as of now.

Last edited: Oct 27, 2014
2. Oct 27, 2014

### rude man

Bad start. That's the p-V relationship for a reversible adiabatic process, not an isothermal one.

3. Oct 27, 2014

Ah right ok, thanks. The question I posted up is Parts A and B. However, there is also a part C and D where it asks what the final pressure and work would be if it were an Adiabatic process instead. So it would be good to know if I have done that bit correct?

For isothermal expansion, is it simply $PV=nRT$ that would/could be used?

Thanks.

4. Oct 28, 2014

### rude man

Then you omitted a bunch of dV's in your integrals and you did not evaluate ∫VdV correctly. I haven't seen your emendation.

Also, be aware that in addition to being adiabatic the process also has to be quasi-static. (It is sufficient but not necessary for it to be reversible).
Yes.

5. Oct 28, 2014

Ok thanks. For parts D and C it just says if it were an Adiabatic process instead, it doesnt state anything else about it being reversible etc, will that make a difference? If I could change the OP and amend it to say Adiabatic rather than Isothermal I would, as I think that is the one I need more help with, so lets carry on with that one. I will try and amed the integration now below (as I can no longer edit the OP). Also apologies for forgetting some of the dV's, as I am (relatively) new to integration it is something I sometimes forget, in my OP where I stated I went back to amend an error, I did, but obviously not every error, just the one I could see; I just double checked something on my calculator and do not know what happened last night but I can see where I went wrong, I added 1 to 7/5 , not -7/5.

For the pressure, I hope what I did was correct.

$P_iV_i^{\gamma}=P_fV_f^{\gamma} \\ P_f=\frac{P_iV_i^{ \gamma } } { V_f^{ \gamma } }=\frac{ (6.87 \times 1.01 \times 10^5)(0.44)^{ \frac{7}{5} } }{1.55^{\frac{7}{5}}} = 119027.5 Pa \\ \frac{119027.5}{1.01 \times 10^5} = 1.178 atm \\$

Then for the work done i have amended the integration, without putting values in. Just to see if the integration itself is correct.

('c' is the constant that $P_iV_i^{\gamma}$ and $P_fV_f^{\gamma}$ are equal to.

$W=\int_{V_i}^{V_f} P \,\, dV =\int_{V_i}^{V_f} \frac{c}{V^{\gamma}} dV = c \int_{V_i}^{V_f} V^{-\gamma} dV = [c(\frac{V^{-\gamma +1}}{-\gamma +1}) ]_{V_i}^{V_f}$

Last edited: Oct 28, 2014
6. Oct 28, 2014

### rude man

How many cc's in 1 cubic meter? Not 1000. Otherwise looks OK. (Except why your prof wants you to use an incorrect value for monatomic gas gamma is beyond me ... )
This integration is still incorrect. What is ∫xadx?

7. Oct 28, 2014

Oh yes, silly me, for some reason was think cubic mm, so the pressure calculation is just a factor of ten out, I will change that later when I can as I am off in a min.

Isn't it $\frac{x^{a+1}}{a+1}$ ? Ignoring the constant of integration .

EDIT: oh I see you quoted before I amended something in my post, I forgot the minus signs but put them in now.

8. Oct 28, 2014

### rude man

I think you've got it now.

You should discuss the requirement of quasi-staticity in adiabatic processes, without which pv to the gamma = constant is invalid.

For example, a free expansion of gas in an insulated box (gas is initially on one side, then is allowed to expand to full extent of box) is adiabatic (no heat in or out) but is not quasi-static and therefore you can't use pv to the gamma = constant.

9. Oct 28, 2014

Ok, many thanks for your help. My lecturer did say to use 7/5 for gamma, is that a wrong value to use in this case?

10. Oct 28, 2014

### rude man

Yes, for a monatomic ideal gas it should be 5/3.

11. Oct 28, 2014

Ok thanks, I will double check with him about that.

12. Oct 28, 2014

Can I just double check, in the adiabatic case, regardless of the value for gamma, should the work done be a negative number?

Thanks

13. Oct 28, 2014

### rude man

If the piston is expanding what will be the sign of pdV? What's the sign of p? of dV?

14. Oct 28, 2014

The question doesnt state anything about a piston though, or is it the same for all cases? The pressure would decrease, if like in this case the volume is increased, so would P be negative? dV would be positive, so therefore PdV would be negative? If its that easy, I did not know you could treat the differentials like normal.

I think I have done it correct then. But the reason I asked was because I have had a go at the first two parts, regarding the final pressure and work done in the isothermal case, and using $PV=nRT$, I got the final pressure fine but then my final answer for the work done is a positive value, so does that mean it is definitely wrong?

15. Oct 28, 2014

### rude man

p may be decreasing but it's still positive. No such thing as negative pressure in a gas.
work = ∑ pdV = ∫pdV. p is +, dV is +, so W is +.

I think I have done it correct then. But the reason I asked was because I have had a go at the first two parts, regarding the final pressure and work done in the isothermal case, and using $PV=nRT$, I got the final pressure fine but then my final answer for the work done is a positive value, so does that mean it is definitely wrong?[/QUOTE]
as I said , W is +.

16. Oct 28, 2014