# Calculating the mass of air in a pressurised air tank

• Engineering
• Gixer1127
In summary, the conversation discusses a problem involving determining the mass of air in a pressurized air tank using the ideal gas law equations. The main focus is on calculating the temperature of the air in the tank when the pressure drops from 320 bar to 280 bar. The conversation also raises questions about the accuracy of the calculated mass and what could cause the pressure to drop. After further checking, it is confirmed that the calculated mass is correct and the only variables that change in the problem are the temperature and mass.
Gixer1127
Homework Statement
Looking for some pointers in where I am going wrong with this if possible please
Relevant Equations
pV=nRT pV = (m/Mr)RT)
Hey folks, any help will be greatly appreciated. I have, what I thought, was a fairly simple equation to follow to determine the mass of air in a pressurised air tank. See below question and my attempt at solving using the pV = (m/Mr)RT) equation.
A pressurised air tank supplies compressed air to a new hybrid air engine. When the hybrid engine is not running the absolute pressure in the tank is 320 bar and the temperature of the air is 50°C.
Calculate 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 280 bar. Volume remains constant.

Using the General Gas Law: P1xV1/T1 = P2xV2/T2

Simplify the equation to P1/T1 = P2/T2 as there is no change to the volume.

Step 1) Convert the known temperature to Kelvin = T1 = 50⁰C + 273 = 323K

Insert values into the equation:
320/323 = 280/T2

T2 = (280/320) x 323 =282.62K

The temperature of the air in the high pressure air tank at 280 bar is 282.62K
Calculate the mass of air in the pressurised air tank using the temperature and pressure from part (a) and the information given here below.

Using the Characteristic Gas Law equation pV=nRT which can also be written as pV = (m/Mr)RT)

P is the pressure in Pa

V is the volume in m³

N is the amount of gas (in moles)

T is the temperature in K

R is the universal gas constant and is always 8.314J·K‾1

M is the mass of gas in g

Mr is the molecular mass of the gas (Air = 28.96g/mol)

Tank volume = 450 litres ( 0.45m³)

pV = (m/Mr)RT Change equation to find the mass m = (pV Mr)/RT

((280x10⁵)x0.45x28.96) / (8.314 x 282.62) = 15529g = 155.3kg

I'm pretty sure the mass of air in a 450litre air tank is not 155.3kg's but this is the only answer I keep getting so what have I missed?

Hi,

Pretty weird scenario. What causes the pressure to drop by 40 Bar ?

Gixer1127 said:
Simplify the equation to P1/T1 = P2/T2 as there is no change to the volume.
So what, exactly do you think DOES change ?

And: what mass do you get for the 320 Bar case ?

##\ ##

Gixer1127 said:
Homework Statement: Looking for some pointers in where I am going wrong with this if possible please
Relevant Equations: pV=nRT pV = (m/Mr)RT)

Hey folks, any help will be greatly appreciated. I have, what I thought, was a fairly simple equation to follow to determine the mass of air in a pressurised air tank. See below question and my attempt at solving using the pV = (m/Mr)RT) equation.
A pressurised air tank supplies compressed air to a new hybrid air engine. When the hybrid engine is not running the absolute pressure in the tank is 320 bar and the temperature of the air is 50°C.
Calculate 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 280 bar. Volume remains constant.

Using the General Gas Law: P1xV1/T1 = P2xV2/T2

Simplify the equation to P1/T1 = P2/T2 as there is no change to the volume.

Step 1) Convert the known temperature to Kelvin = T1 = 50⁰C + 273 = 323K

Insert values into the equation:
320/323 = 280/T2

T2 = (280/320) x 323 =282.62K

The temperature of the air in the high pressure air tank at 280 bar is 282.62K
Calculate the mass of air in the pressurised air tank using the temperature and pressure from part (a) and the information given here below.

Using the Characteristic Gas Law equation pV=nRT which can also be written as pV = (m/Mr)RT)

P is the pressure in Pa

V is the volume in m³

N is the amount of gas (in moles)

T is the temperature in K

R is the universal gas constant and is always 8.314J·K‾1

M is the mass of gas in g

Mr is the molecular mass of the gas (Air = 28.96g/mol)

Tank volume = 450 litres ( 0.45m³)

pV = (m/Mr)RT Change equation to find the mass m = (pV Mr)/RT

((280x10⁵)x0.45x28.96) / (8.314 x 282.62) = 15529g = 155.3kg

I'm pretty sure the mass of air in a 450litre air tank is not 155.3kg's but this is the only answer I keep getting so what have I missed?

BvU said:
Hi,

Pretty weird scenario. What causes the pressure to drop by 40 Bar ?So what, exactly do you think DOES change ?

And: what mass do you get for the 320 Bar case ?

##\ ##
No idea what causes the pressure to drop, this is all the info I was given in the question. However, I have done what you have suggested with checking the 320bar and original temp and got a final mass of 155291g. In my original post I had missed a 4 (155294). So the only things that have changed are the temperature and the mass?
Chestermiller said:
Sorry Chester, I had missed a 4 from the final figure It should have been 155294. Is this anything like you got?

Gixer1127 said:
No idea what causes the pressure to drop, this is all the info I was given in the question. However, I have done what you have suggested with checking the 320bar and original temp and got a final mass of 155291g. In my original post I had missed a 4 (155294). So the only things that have changed are the temperature and the mass?
Yes. Dividing by 1000 gives 155

Gixer1127 said:
final mass of 155291g
No way you can quote six digit precision for mass when you are given two (or at best three) digit precision to start with.

So you get 155 kg mass at 320 Bar (absolute pressure).
When you apply
Gixer1127 said:
Simplify the equation to P1/T1 = P2/T2 as there is no change to the volume
you don't just assume the volume is the same, but you also assume the amount of air is the same. No wonder you get back the same 155 kg !

BvU said:
Pretty weird scenario. What causes the pressure to drop by 40 Bar ?
Well, it seems the scenario is that the cylinder is left to cool down from 50 ##^\circ##C to 10 degrees; perhaps overnight? What that has to do with
Gixer1127 said:
when the air engine starts and the absolute pressure in the tank drops to 280 bar
is a mystery to me.

BvU said:
you don't just assume the volume is the same, but you also assume the amount of air is the same. No wonder you get back the same 155 kg !

I think that's the issue. The problem states that the volume of the tank remains constant (the tank is rigid) but it doesn't say the mass of air inside such tank remains constant. In fact, that tank SUPPLIES air to the engine. That mass of air leaving the tank would be the cause behind the pressure drop.
Gixer1127 said:
A pressurised air tank supplies compressed air to a new hybrid air engine.

I will try to solve the problem without considering the amount of air in the tank to be constant. If that's right there should be enough information in the problem statement to solve it that way.

Last edited:
Juanda said:
If that's right there should be enough information in the problem statement to solve it that way
Seems to me there isn't.
What is the complete problem statement, verbatim ?

##\ ##

Juanda said:
I will try to solve the problem without considering the amount of air in the tank to be constant. If that's right there should be enough information in the problem statement to solve it that way.

Are we getting the original problem statement with all the information? I feel like there must be something we are missing. For example, I thought the problem statement to just be:

Gixer1127 said:
A pressurised air tank supplies compressed air to a new hybrid air engine. When the hybrid engine is not running the absolute pressure in the tank is 320 bar and the temperature of the air is 50°C.
Calculate 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 280 bar. Volume remains constant.

I was struggling to define State 1 and later a little bit of additional key information was dropped.
Gixer1127 said:
Tank volume = 450 litres ( 0.45m³)

Is there any additional information? For example, the amount of mass that left the pressurized tank when it started supplying air to the engine. Assuming air is not constant in the tank but the tank volume is, this is as far as I can get.

Known information is:
##P_1 = 320 bar = 32 \times 10^6 Pa ; \ T_1 = 50^{\circ}C=323K; \ V_1=V_2=0.45m^3 ; \ P_2 = 280bar=28 \times 10^6Pa##

State 1:
##P_1 V_1 = m_1 R_{air} T_1 \rightarrow m_1 = \frac{P_1T_1}{R_{air} T_1}=155.33kg##

State 2:
##P_2 V_2 = m_2 R_{air} T_2##

I got 1 equation but two unknowns (##m_2; \ T_2##). I feel like the problem statement is incomplete because the mass being conserved in the tank doesn't make much sense. Regarding the fact ##m_1=155kg## all I can say is that the math seems right. I don't know how realistic such an engine would be but that's not really the point of the exercise I believe.

BvU said:
Seems to me there isn't.
What is the complete problem statement, verbatim ?

##\ ##
There is enough information given if you assume that the tank is insulated and the gas expansion within the tank is adiabatic (even with gas leaving the tank). I'm just not sure that the OP is far enough advanced to address this.

BvU
Chestermiller said:
There is enough information given if you assume that the tank is insulated and the gas expansion within the tank is adiabatic (even with gas leaving the tank). I'm just not sure that the OP is far enough advanced to address this.

Anything else? Maybe adiabatic expansion up to atmospheric pressure? Amount of energy leaving the tank?

I would love to see the process to solve it. Problems with limited available information are not very common in books since they need to have close solutions but they certainly are common in the real world where sometimes it is necessary to complete the problem with "guesses" about the environment where the phenomenon is happening.

Juanda said:
Anything else? Maybe adiabatic expansion up to atmospheric pressure? Amount of energy leaving the tank?

I would love to see the process to solve it. Problems with limited available information are not very common in books since they need to have close solutions but they certainly are common in the real world where sometimes it is necessary to complete the problem with "guesses" about the environment where the phenomenon is happening.
Are you familiar with the equation ##Pv^{\gamma}=Constant## for adiabatic expansion of an ideal gas, where v is the molar volume and ##\gamma=C_p/C_v##?

Chestermiller said:
Are you familiar with the equation ##Pv^{\gamma}=Constant## for adiabatic expansion of an ideal gas, where v is the molar volume and ##\gamma=C_p/C_v##?
I didn't consider that could be used in open systems. I always saw it used in piston-like systems. When moving to open systems enthalpy is usually king. I need to recheck my books about this.

If it can indeed be used then I can establish a relation between the initial mass in the tank and the final mass of air in the tank after some of it goes into the engine.

Using the information from #9
##v_1 = \frac{V_1}{m_1}=0,002896906 \ m^3/kg##

Using the adiabatic expansion as suggested by @Chestermiller in #12. (##\gamma = 1.4##)
##P_1v_1^\gamma =P_2v_2^\gamma \rightarrow v_2=e^\frac{\ln\left ( \frac{P_1v_1^\gamma}{P_2} \right ) }{\gamma} =0,003186818 \ m^3/kg##

The mass that remains in the tank will then be
##m_2 = \frac{V_2}{v_2}=141,20 \ kg##

It is now possible to find the temperature of the air remaining in the tank with
##P_2 V_2 = m_2 R_{air} T_2 \rightarrow T_2=\frac{P_2 V_2}{R_{air} m_2 }=310,9 \ K##

So the math at least seems consistent. We assumed air leaves the tank which causes a drop in pressure. The results agree with this. The mass decreases and temperature decreases as well.

I assume it must also be possible to know the energy that left the system to go into the engine with
##\Delta U_{tank} = c_{v_{air}}\left ( m_2T_2-m_1T_1 \right )##
(It'll be negative because the tank decreases its potential energy to transfer work to the engine)

Is that correct?

Last edited:
Juanda said:
I didn't consider that could be used in open systems. I always saw it used in piston-like systems. When moving to open systems enthalpy is usually king. I need to recheck my books about this.

If it can indeed be used then I can establish a relation between the initial mass in the tank and the final mass of air in the tank after some of it goes into the engine.

Using the information from #9
##v_1 = \frac{V_1}{m_1}=0,002896906 \ m^3/kg##

Using the adiabatic expansion as suggested by @Chestermiller in #12. (##\gamma = 1.4##)
##P_1v_1^\gamma =P_2v_2^\gamma \rightarrow v_2=e^\frac{\ln\left ( \frac{P_1v_1^\gamma}{P_2} \right ) }{\gamma} =0,003186818 \ m^3/kg##

The mass that remains in the tank will then be
##m_2 = \frac{V_2}{v_2}=141,20 \ kg##

It is now possible to find the temperature of the air remaining in the tank with
##P_2 V_2 = m_2 R_{air} T_2 \rightarrow T_2=\frac{P_2 V_2}{R_{air} m_2 }=310,9 \ K##

So the math at least seems consistent. We assumed air leaves the tank which causes a drop in pressure. The results agree with this. The mass decreases and temperature decreases as well.

I assume it must also be possible to know the energy that left the system to go into the engine with
##\Delta U = c_{v_{air}}\left ( m_1T_1-m_2T_2 \right )##

Is that last statement correct?
You can do this analysis using either the open system or the closed system version of the 1st law of thermodyamics. The open system version would read $$dU=d(un)=hdn$$which leads to $$nC_vdT=RTdn$$.

Using the closed system version of the 1st law of thermodynamics involves recognizing that the final mass of air remaining in the tank has undergone an adiabatic reversible expansion, pushing the air ahead of it (i.e., doing work on it) into the engine, and also recognizing that, during the process, it exchanges no heat with the air ahead of it (since the temperature is uniform in the tank).

Both these methods give exactly the same answer(s).

Juanda
Chestermiller said:
Using the closed system version of the 1st law of thermodynamics involves recognizing that the final mass of air remaining in the tank has undergone an adiabatic reversible expansion, pushing the air ahead of it (i.e., doing work on it) into the engine, and also recognizing that, during the process, it exchanges no heat with the air ahead of it (since the temperature is uniform in the tank).
Yep, that's how I was visualizing it. Having your confirmation makes me sure of my understanding.
Thanks!

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