Help with Thermodynamics process equations and units

[Brinkley23]

Homework Statement

A pressurised air tank supplies compressed air into a hybrid engine. When the engine is not running the absolute pressure in the tank is 320bar and the air temp is 50 deg celsius.

1.Calculate the temperature of the air in the high pressure air tank when the engine starts and the absolute pressure in the tank drops to 280bar. Volume remains constant.

2.Calculate the mass of air in the pressurised air tank using the temp and pressure from part 1

Relevant Equations

pV=nRT
T=pV/nR
m=pVMr/RT

V=0.45m3
R=8.314
P=280bar or 28000000pa (or 28000kpa?)
n=28.96 (air mol)

For part 1. I need to find the temperature, I rearranged the equation so that:

T = pV/nR

(28000000)(0.45) / (28.96)(8.314)

This gives me a value of 52331.4

I am unsure where the units I have used in the equation are correct.

I have used pa for pressure, or should I be using kpa? this would give me a value of 52.3, which being a temperature looks a lot more likely?

assuming that 52.3 is deg celsius and adding 273 to convert to kelvins this would give me 325.3k (which is needed for part 2)

For part 2. I need to find the mass, I rearranged the equation so that:

m = pVMr / RT

(28000)(0.45)(28.96) / (8.314)(325.3)

=134.92g = 0.13kg

this time is used 28000kpa in the equation instead of 28000000pa.

Hopefully I'm not too far off and someone can help me with using the correct units?

Thanks!

 

[Chestermiller]

Why are you not working out the units yourself? Where did the 0.45 m^3 come from?

 

[Brinkley23]

Why are you not working out the units yourself? Where did the 0.45 m^3 come from?

I believe I have used the correct units, my notes say that:
p = pa
v = m^3
n= moles
T= K
R=universal gas constant 8.314

original given pressure value given was 280bar, converted to 28000000pa
original volume was 450L, converted to 0.45m^3

using these figures to work out temperature the value, just doesn't seem right, but if I were to use 28000kpa, it gives me a much more realistic answer, which needs to be in K

 

[Chestermiller]

I believe I have used the correct units, my notes say that:
p = pa
v = m^3
n= moles
T= K
R=universal gas constant 8.314

original given pressure value given was 280bar, converted to 28000000pa
original volume was 450L, converted to 0.45m^3

using these figures to work out temperature the value, just doesn't seem right, but if I were to use 28000kpa, it gives me a much more realistic answer, which needs to be in K

The air that remains in the tank after some of the moles have escaped and the pressure has dropped has experienced an adiabatic reversible expansion.

With regard to the issue of units, you should be able to work this out yourself. What are the units of R?

 

[Brinkley23]

The air that remains in the tank after some of the moles have escaped and the pressure has dropped has experienced an adiabatic reversible expansion.

With regard to the issue of units, you should be able to work this out yourself. What are the units of R?

units for R=8.314 JK^-1mol^-1

 

[Chestermiller]

units for R=8.314 JK^-1mol^-1

In terms of N and m, what are the units of J? In terms of N and m, what are the units of Pa? In terms of Pa and m, what are the units of J?

What is the equation for the final temperature of the adiabatic reversible expansion in terms of the initial and final pressure and temperature, the final pressure, and the ratio gamma of the specific heats?

 

[Brinkley23]

In terms of N and m, what are the units of J? In terms of N and m, what are the units of Pa? In terms of Pa and m, what are the units of J?

What is the equation for the final temperature of the adiabatic reversible expansion in terms of the initial and final pressure and temperature, the final pressure, and the ratio gamma of the specific heats?

ok, so I've have completely started again,

for part 1 I have now used Gay-Lussac's law. p1/t1 = p2/t2

Therefore t2 = t1p2/p1
t2= (323)(28000000) / 32000000 = 282.6k

for part 2 I have rearranged the ideal gas law to solve for n.

n=pV / RT

P=28000000pa
V=0.45m^3
R=8.314
T=282.6K
dry air = 28.96g/mol

n = (28000000)(0.45) / (8.314)(282.6) = 5362.8 mole

5362.8*28.96 = 155305.5g = 155.3kg

Please say I am now on the right path?

 

[Chestermiller]

Guy-Lusssc’s law applies only of the number of moles is constant (which is not the case here).

 

[Brinkley23]

Guy-Lusssc’s law applies only of the number of moles is constant (which is not the case here).

Ok, so I am back to square 1. Is the Ideal Gas Law what I should be using then still?

 

[Chestermiller]

See what I said in post #4 about an adiabatic reversible expansion. Do you know what equations apply to that?

 

[Brinkley23]

See what I said in post #4 about an adiabatic reversible expansion. Do you know what equations apply to that?

No, what worries me with that is, my tutors have not provided any material for me to learn adiabatic reversible expansion, so I can't see them wanting me to use it in my answer (I have emailed them).

Im sure you are correct, but when I have an issue, they normally just reply "refer to your notes, everything you need is there for you", seeing as they have not provided any material for that Id be very surprised if they expect me to use it.

 

[Chestermiller]

No, what worries me with that is, my tutors have not provided any material for me to learn adiabatic reversible expansion, so I can't see them wanting me to use it in my answer (I have emailed them).

Im sure you are correct, but when I have an issue, they normally just reply "refer to your notes, everything you need is there for you", seeing as they have not provided any material for that Id be very surprised if they expect me to use it.

I’m anxious to hear their response.