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Old Oct 2nd 2014, 10:13 AM   #1
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Thermodynamics

Get could anyone help me with an engineering problem-

A quantity of oxygen is stored for the air tanks of the submarine and occupies a volume of 5meters cubed. The pressure of the gas is at atmospheric pressure, at an ambient temperature of 4degrees Celsius. What will be the pressure of the gas be if it is compressed to half it's volume and heated to a temperature of 42degrees Celsius. Calculate the mass of the oxygen using the density of oxygen and using the specific heat at constant volume ,calculate the characteristic gas constant and the specific heat capacity at constant pressure ,when Cv=660KJ/KgK.

I'd appreciate any help on this question, I have no idea where to start on this question.
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Old Oct 3rd 2014, 08:09 AM   #2
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Use the Ideal gas law: PV=nRT. Here n (number of moles of gas) is constant, as is R, so:

PV/T = nR = constant.

Thus:

P1V1/T1 = P2V2/T2

Where P1, V1, and T1 are the initial values of pressure, volume and temp (in kelvin) and P2, T2, and V2 are the final values. Solve for P2.

For the second part, use PV=nRT to solve for n. You can use either initial values of P, V and T or final values - it doesn't matter, just don't mix them up. Then from n calculate total mass of gas. Post back with your answers, and if still having issues with it or the third part we'll help you along.
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Old Oct 4th 2014, 06:11 AM   #3
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Part 1
PV=nRT
PV/T=nR
P1V1/T1=P2V2/T2

P1- 101325 Pascal
V1- 5m3
T1- 4oc -> 1092K

P2- 463.9
V2- 5m3/ 2= 2.5m3
T2- 42oc -> 11 466K
Using the equation i will solve the final pressure:
101325x 5/1092= 463.9 x 2.5/ 11 466
Final pressure= 463.9

Part 2
Use-PV=nRT
I will be using the final values.
Pressure=
463 Pascal
0.0045783370 Atmosphere
Volume=
2.5m3
2500 L
Temp=
11 466 K
n = 0.01216
PV=nRT
463 x 2.5 = 0.01216 x 0.08206 x 11 466
1125.5 = 11.44
1125.5/ 11.44 = 0.1048080 (Mass of gas)

Part 3
Specific gas constant-
S= R/MW
MW= R/S
S= 0.08206/ 0.01216 = 6.74835 (Specific gas constant)

Thanks for the reply, these are my workings out and i would appriciate any feedback and help you could offer.
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