# Calculating the volume that a compressed mass of air will occupy.

1. Oct 23, 2013

### SherlockOhms

1. The problem statement, all variables and given/known data
This question is to due with a compressor experiment. I've actually posted 2/3 questions on this write up in the past 2 days. Anyway, a compressor takes air from the atmosphere and compresses it before forcing it into a receiver volume. You are asked to calculate the mass of air that is initially in the receiver, the final mass in the receiver and subsequently the mass that has been transferred in. You are given the atmospheric conditions P0 and T0 as well as the final conditions PF and TF. The volume of the receiver is also given. Also n, the polytropic exponent was found to be ≈ 1 (1.014 to be exact). Finally, you are asked to find the volume that the transferred mass of gas occupies at atmospheric conditions using P1V1n = P2V2n = PVn

2. Relevant equations
Given equations:
P1V1n = P2V2n = PVn
PV = mRT.

3. The attempt at a solution
PV = mRT.
m = PV/RT.
minitial = P0V/RT0.
mfinal = PFV/RTF.
mtransferred = mfinal - minitial.
Correct so far?
This is where I get a little stuck. You are told to use P1V1n = P2V2n = PVn to calculate the volume that mtransferred will occupy at P0 and T0. Why is it not ok to use PV = mRT for this question? Subbing mtransferred, P0 and T0 into the ideal gas law and solving for V. I've actually asked the lecturer why this is and he told me that the ideal gas law could be used but it's use will be a little more involved. Anybody have any ideas where I may be going wrong? And how I could actually use P1V1n = P2V2n = PVn to solve for the volume occupied at atmospheric temp and pressure?

2. Oct 23, 2013

### Simon Bridge

Mass transferred equation is good.
The process of compression is important to the final state - in your case, there is an assumption that there is no energy transfer so the compression is adiabatic.

Try doing it both ways and see what the difference is.

3. Oct 24, 2013

### SherlockOhms

As in, find V using PV = mRT and PV^n =PV^n?

4. Oct 24, 2013

### SherlockOhms

What confuses me a that if you find V using PV = mRT and then sub it into the relation PV^n =PV^n using atmospheric pressure and your found V on the left as well as final pressure and the receiver volume on the left, it doesn't balance, i.e. The left doesn't equal the right. Would this just be because the values used will have some experimental error associated with them or have I made an incorrect assumption?

5. Oct 24, 2013

### Simon Bridge

It's an incorrect assumption :)

How does PV=mRT take account of the fact air is not an ideal gas?
How about the PV^n = constant?

6. Oct 24, 2013

### SherlockOhms

I thought that air was an ideal gas?
Using PV^n = PV^n doesn't take into account the actual mass of air in question though. Is that irrelevant? I mean, I used PV = mRT to find m(initial) and m(final), remember? Why is it valid in that case?

7. Oct 24, 2013

### Simon Bridge

You thought something about real life was "ideal"?!
Go look up the definition of an ideal gas. Compare with the properties of air that you know about.

But - you are basically right in that the air is being approximated by an ideal gas - since PV^n is for a polytropic process.

I want you to start thinking about the assumptions that go into the ideal gas equation vs what goes into the polytropic process equation - and where the constants come from.

The difference lies in the difference between the two equations.
Your process is actually quite close to an isothermal process - which is kinda interesting.

8. Oct 24, 2013

### SherlockOhms

I see. I'll have a look at that now. Again though, why was the ideal gas formula relevant to the mass balance? Air was being considered ideal in that instance but not here. I'll go have a look at the exact theory behind the two equations, but could you first explain the discrepancy? Thanks for the help, by the way.

9. Oct 24, 2013

### Simon Bridge

I think I've been concentrating too much on a side issue.

note: in the relation PV^n=constant - the mass is taken into account in the "constant" part.
If n is close to 1, then the process is close to isothermal.

The main trouble you have is that you are asked for the uncompressed volume of the transferred gas.

If the bottle starts with $m_0$ gas, and ends up with $m_1$ then the mass transferred is $m_t=m_1-m_0$ like you found (I'm just defining notation.)

Pretending it's an ideal gas...
The state equation reads: $P_1V_1=(m_0+m_t)RT_1$ starting from $P_0V_0=m_0T_0$
You know the process conforms to $PV^n=\text{const.}$

$P_1V_1=P_0V_0\frac{T_1}{T_0} +m_tRT_1$
... and solve for $m_t$: is probably what I'd have done.
... do you know the final temperature $T_1$?

Can you say that for the transferred $m_t$, that $P_1V_1^n=P_0V_0^n$?

10. Oct 25, 2013

### SherlockOhms

Thanks for the reply. Yeah, you're given the final temperature, do finding the transferred mass is very straightforward. Yeah, so using the relationship P1V1^n = P2V2^n, I can solve for the initial Volume, yes? That's also straightforward. You're given 3 variables and the exponent, so after some algebra using a few natural logs and exponential functions, you can solve for the V at atmospheric conditions. I mean, I understand that it's solvable. The only thing that's confusing me is the fact that V has to be solved this way and not using the Ideal gas formula using P atmospheric, T atmospheric and the transferred mass as your input. It just seems sorta inconsistent that you can use the ideal gas law for finding the transferred mass but not in this instance. Why is it that air can be considered an ideal gas when solving for m but then must be treated as real for solving for V?

Are we going in circles here. Maybe I need to read over your previous posts in more detail. Also, apologies for any typos. Typing on the move here.

11. Oct 25, 2013

### Simon Bridge

Investigate - work out what the initial volume has to be the "proper" way - then, using that value and the ideal gas equation, find m (treat it as unknown). Compare with the different values you have already.

12. Oct 27, 2013

### SherlockOhms

Will do. I'll see what happens. Thanks for that.