The mass of air in the pressurised air tank

You are still making mistakes. You forgot to divide the absolute pressure (270) by the atmospheric pressure (101,325). You also forgot to divide the volume (450) by the pressure (270). 437,955,831 / 270 = 1,429,312,000You are still making mistakes. You forgot to divide the absolute pressure (270) by the atmospheric pressure (101,325). You also forgot to divide the volume (450) by the pressure (270). 437,955,831 / 270 = 1,429,312,000f
  • #1
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Homework Statement


A pressurised air tank supplies compressed air to an air engine. When the air engine is not running the absolute pressure in the tank is 300 bar and the temperature of the air is 60°C.

Calculate:
(a) The temperature of the air in the high pressure air tank when the air engine starts and the absolute pressure in the tank drops to 270 bar. Volume remains constant.

(b) The mass of air in the pressurised air tank using the temperature and pressure from part 1 (a).

Homework Equations


M = pV / RT [/B]
p = 270
V = 450
R = 8.314
T = 299.7

The Attempt at a Solution


(a) is fine just need help with (b)
I found the temperature to be 26.7 for the answer to (a) and added 273 to covert into kelvins (299.7)
270 x 450 / 8.314 x 299.7 = 4,379,787 which seems way too high.

How do I convert this into a 3 figure answer from mol mass to j/kgk or can the mass be that high?
[/B]
 
  • #3
I've tried 4,379 / 273 = 16.04, would that be any closer?
Or am I confusing temperatures with weight units?
 
  • #4
270 x 450 / 8.314 x 299.7 = 4,379,787

As Chestermiller was trying to get you to notice, you calculated incorrectly. The right side does not equal the left side. I believe you forgot that the 299.7 is in the denominator.
 
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  • #5
I didn't realize that both sides had to equal each other.
Could you point me in the right direction with the formula for the mass, I believe it to be pV / RT?
 
  • #6
I didn't realize that both sides had to equal each other.
Could you point me in the right direction with the formula for the mass, I believe it to be pV / RT?
You are misunderstanding. You made a calculation error. You said 2+2 = 17. If you type what you wrote into a calculator exactly the way you wrote it you get your 4379787. That’s because the two numbers in the denominator should be in parentheses. Your calculator doesn’t know what you mean and it multiplies by 299.7 instead of dividing by 299.7 making your answer wrong by 299.7^2
 
  • #7
Even when you get the arithmetic right, it's still wrong. What are the units of P, V, R and T? Are they consistent? What are the units of your answer?
This is an excellent example of the rule that you should NEVER simply plug numbers into an equation without thinking about units. Get into the habit of always including the units of every quantity; it will save you a lot of mistakes.
E.g. P atm * V L/(R L-atm/mol/K * T K) = (PV/RT) mol
The answer to a question that asks for mass can never be a bare number. It must have units - is it kg, g, lb, amu etc?
And no, mass is not pV/RT. What is the basic ideal gas law?
 
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  • #8
Cutter Ketch & mjc 123 thanks for your replies.
The Ideal Gas Law. The volume (V) occupied by n moles of any gas has a pressure (P) at temperature (T) in Kelvin. The relationship for these variables, P V = n R T, where R is known as the gas constant, is called the ideal gas law.
I'm very new to physics, but I've found the mols to be 4,379787 from 270bar x 450l / 8.314J·K-1·mol- x 299.7K
which can't be right.
But I also tried 270bar x 450l / 8.314J·K-1·mol- / 299.7K = 48.76kg which sounds like the better answer but I know I've left out the air value 28.96Mr (g/mol) from the equation.
So I tried 28.96Mr x 8.314J·K-1·mol- x 299.7K / 270bar x 450l = 120,266
My tutor isn't much help and my notes don't seem cover everything asked in the assessment, but I'm probably going to struggle to understand a lot of what your explaining but do appreciate your help.
 
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  • #9
The units of R have to be consistent with the units of your other parameters. Are they?
 
  • #10
You're just guessing. Don't guess. Think.
You know that PV/RT gives you the number of moles (though you got both the calculation and the units wrong). How do you get from the number of moles to the mass?
And you really MUST get the units right. bar*L/(J/mol/K*K) doesn't give you moles. It gives you mol*bar*L/J. You must use consistent units. My preference is to convert everything into SI units before applying the equation, but you can use other units if you are consistent. There is a value of R in L*atm/mol/K - google "gas constant" and see if you can find it.
 
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  • #11
I found the value 0.082057 L atm mol-1 K-1 instead of 8.314
I've converted 270 bar to 266.469 atm
266.469 x 450 / 0.082057 x 299.7 = 437,955,831
Number of moles = 437,955,831
 
  • #12
I found the value 0.082057 L atm mol-1 K-1 instead of 8.314
I've converted 270 bar to 266.469 atm
266.469 x 450 / 0.082057 x 299.7 = 437,955,831
Number of moles = 437,955,831
You still messed up the arithmetic (making the same mistake as before). Also, I assume the volume is 450 liters and not 450 m^3?
 
  • #13
It's good also to get into the habit of writing parentheses when appropriate, e.g.
266.469 x 450 / (0.082057 x 299.7)
Otherwise you will often make mistakes. Make sure, too, that you know how to enter an expression like that into your calculator or spreadsheet so that it does the right calculation.
 
  • #14
The volume is 450 litres (not m^3) and I don't know where I'm going wrong with the arithmetic, I've tried the different values for R as mjc123 asked me too, and converted 270 bar into atm.
I hope your not sick of me getting it wrong but its all new to me.
 
  • #15
Okay, do it in two steps. Calculate X = P x V, and Y = R x T. Then divide X by Y. Can you see how you weren't doing that?
 
  • #16
(0.082057 x 299.7) / 266.469 x 450 = 41.530
I understand to do the bracket sum first now.
 
  • #17
You are still making exactly the same error, but this time with the expression the wrong way up! I said divide X by Y; that means X/Y, not Y/X. (Perhaps I should have put brackets round the numerator as well, but I didn't think you were going to put it on the bottom!) You need to calculate
(266.469 x 450) / (0.082057 x 299.7)
 
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  • #18
I did X / Y and I'm getting 121,500 / 2,491.7058 = 48.761kg which sounds right
266.469 x 450 / 0.082057 x 299.7 = 437,955,831
 
  • #19
You have used the value 8.314 for R instead of 0.082!
And for the umpteenth time, PV/RT does NOT give you kg! What does it give you?
 
  • #21
No it does not! If pV = nRT, what is PV/RT?
 
  • #22
$$n=\frac{PV}{RT}=\frac{(266.49)(450)}{(0.082157)(299.7)}=4876\ gm-moles$$
What is the molar mass of air?
How many grams of air is 4876 gm-moles?
How many kg is that?
 
  • #23
Molar mass of air = 28.96
4876 = 87841.1g or 87.84kg
But I get 437,457,232 from your equation.
 
  • #24
Molar mass of air = 28.96
4876 = 87841.1g or 87.84kg
But I get 437,457,232 from your equation.
You need to learn how to do arithmetic. My equation does not give that. You multiplied by 299.7 rather than dividing by it. And your multiplication of 28.96 times 4876 should not give 87841.1.
 
  • #25
Yeah I multiplied the 299.7 and dividing it got me to your answer.
I didn't multiply 28.96 by 4876, I had to google the grams of 4876 moles and I found 87841.1g and take 87.84kg to be my answer.
 
  • #26
Yeah I multiplied the 299.7 and dividing it got me to your answer.
I didn't multiply 28.96 by 4876, I had to google the grams of 4876 moles and I found 87841.1g and take 87.84kg to be my answer.
That would be the result if you had 4876 moles of water (of molar mass 18). But you have 4876 moles of air (of molar mass 28.96).
 
  • #27
4879moles x 28.96 = 141,295.85

141,295.85 / 4879 = 28.96

Molecular mass = 28.96kg
 
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  • #29
141,295.85g so 141kg which again seems so high
 
  • #30
141,295.85g so 141kg which again seems so high
Really? In a tank with a (large) volume of 0.45 m^3 and a (high) pressure of nearly 300 atm, you feel that that seems high?
 
  • #31
It's just the picture I have makes it look fairly small, but with the equations I've done it has to be the correct answer.
Sorry it's taken so long but in my notes there's no examples with different values to compare and my tutor isn't easy to get hold of.
Thanks again.
 
  • #32
It's just the picture I have makes it look fairly small, but with the equations I've done it has to be the correct answer.
Sorry it's taken so long but in my notes there's no examples with different values to compare and my tutor isn't easy to get hold of.
Thanks again.
Most of your difficulty has not been with the physics of the problem. It has been with your difficulty with mathematics.
 

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