Thermodynamics question - Ideal gases

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  • #1
eurekameh
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A rigid tank with a volume of 0.75 m3 initially contains air at 70 kPa and 25 degrees C. A small hole develops in the tank. The surrounding air at 100 kPa and 25 degrees C slowly leaks into the tank due to the hole. Heat transfer between the surroundings and the tank maintains a constant air temperature of 25 degrees C inside the tank. Determine the heat transfer.

When I did:
m = (PV)/(RT) = (70 x 0.75)/(0.2870 x 298) = 0.614 kg and
m = V / specific volume = 0.75 / 631.9 = 0.00119 kg (where the specific volume was found on table -17 of the Thermodynamics Tables http://imageshack.us/photo/my-images/265/unledky.png/ ),
the results don't match each other. I assumed air was an ideal gas because pressure was low, temperature was high, compared to its critical point values. The error that was involved is big. Anyone know why?
 

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  • #2
dacruick
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I'm not sure what type of question this is but are you sure that your units are right?
 
  • #3
I like Serena
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Hi weetabixharry! :smile:

Where did you get the number 0.2870 for R?
R has a different value.
The division (PV)/(RT) would get you the number of moles.
You need to multiply by the atomic mass number of air, to find the mass.

The specific volume you got is for standard pressure 100kPa, so you can't compare it to the mass you got for 70kPa.
 
  • #4
ehild
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You used wrong value for the specific volume. Search for air density. It is about 1 kg/m3 at standard conditions.

ehild
 
  • #5
eurekameh
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Units are right.
I got the number 0.2870 from table A-1. http://imageshack.us/photo/my-images/825/unledxxa.png/
The division (PV)/(RT) = m, where R is the specific gas constant for a gas, not the universal gas constant. It's different for every gas.
Even if you use (PV)/(RT) = N, where N is the number of moles, and multiplied by the molar mass, you still get the same mass.
"The specific volume you got is for standard pressure 100kPa, so you can't compare it to the mass you got for 70kPa."
If I use 100 kPa, I still get a different mass.

Ehild, why is that the wrong value for the specific volume?
 
  • #6
I like Serena
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Ah, I see, first time that I see R used for something other than the universal gas constant.
So we have PV=nRT and also PV=mRT?
How neat (not!! :yuck:).

Looking at your table I see that the heading says vr.
There is no unit mentioned, but it is next to the internal energy, which is in kJ/kg.
Do you have a key that explains what vr is?
I do not think it is the specific volume.

Wiki gives density of air as 1.1839 kg/m3 at 25°C and 101 kPa, which corresponds to a specific volume of 0.8447 m3/kg.
 
  • #7
ehild
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Ehild, why is that the wrong value for the specific volume?

Because it is in m^3/kmol units, instead of m^3/kg.. So you got the mass in kmol-s.

Always check the units! .

ehild
 
  • #8
eurekameh
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"Wiki gives density of air as 1.1839 kg/m3 at 25°C and 101 kPa, which corresponds to a specific volume of 0.8447 m3/kg."
Is there a value for the specific volume at 70 kPa?

Because it is in m^3/kmol units, instead of m^3/kg.. So you got the mass in kmol-s.

Always check the units! .

ehild

I don't think that's the problem... I just figured vr is not the specific volume, but something else which I've recently learned, and not at all related to this problem.
 
  • #9
ehild
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You got the initial amount of gas, from the ideal gas law. The other calculation was wrong because of the improper units, so the results can not be the same.
So what is your problem really?

ehild
 
  • #10
I like Serena
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"Wiki gives density of air as 1.1839 kg/m3 at 25°C and 101 kPa, which corresponds to a specific volume of 0.8447 m3/kg."
Is there a value for the specific volume at 70 kPa?

Air is close enough to an ideal gas that the specific volume at 70 kPa is 101/70 x 0.8447 m3/kg.
 

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