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

In summary: For a compressor experiment, the mass transferred would be the difference between the initial and final mass.
  • #1
SherlockOhms
310
0

Homework Statement


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


Homework Equations


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


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?
 
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  • #2
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
As in, find V using PV = mRT and PV^n =PV^n?
 
  • #4
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
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
Simon Bridge said:
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?

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
I thought that air was an ideal gas?
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
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
I think I've been concentrating too much on a side issue.
Your observation about mass is probably on the right track.

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
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 sort of 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
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
Simon Bridge said:
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.

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

1. How is the volume of compressed air calculated?

The volume of compressed air can be calculated by using the Ideal Gas Law, which states that the volume is directly proportional to the number of moles of gas, the temperature, and the pressure. The formula is V = nRT/P, where V is the volume, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and P is the pressure.

2. What is the gas constant and how is it used in the calculation?

The gas constant, denoted by the symbol R, is a constant value that relates the properties of a gas to its physical state. It is used in the Ideal Gas Law to calculate the volume of compressed air, and its value depends on the units used for pressure and volume. In SI units, R has a value of 8.314 J/mol·K.

3. How does temperature affect the volume of compressed air?

According to the Ideal Gas Law, an increase in temperature will result in an increase in the volume of compressed air. This is because an increase in temperature causes the gas molecules to move faster and exert more pressure on the container, leading to an increase in volume.

4. Can the volume of compressed air change if the pressure remains constant?

No, according to Boyle's Law, the volume of a gas is inversely proportional to its pressure when the temperature is constant. This means that if the pressure remains constant, the volume of compressed air will also remain constant. However, if the temperature changes, the volume will also change.

5. Are there any other factors that can affect the volume of compressed air?

Yes, in addition to temperature and pressure, the composition of the gas can also affect its volume. Different types of gases have different molecular weights and properties, which can impact the volume of compressed air. Additionally, the size and shape of the container and any external forces can also affect the volume of compressed air.

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