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posted by  chemlady on 7/17/2008 11:43:45 AM  |  status: Live  

thermo help.......will rate lifesaver

Course Textbook Chapter Problem
Thermodynamics Fundamentals of Engineering Thermodynamics (5th) by Moran, Shapiro 3 N/A
Question Details:
      
Problem 3.92
Determine the volume, in m3, occupied by 40 kg of nitrogen (N2) at 5 MPa, 180 K.

m3
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posted by rooya on 7/17/2008 12:18:40 PM  |  status: Live
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chemlady's comment:
"thanks...... we needed to find z using the tc and pc in order to calculate tr and pr to use a chart....."
Response Details:
we know that 
Pressure x Volume = Moles x Universal Gas Constant x Temperature
PV=nRT ------- (1)
n=mass/mol. mass =40/28    (for nitrogen)
P=5*10^6 N/m2
T=180K
M=40Kg
R =8.314 N.m/mol.K
substitute in EQ 1    to get Volume in m3
 
Plz feel free if u still have any doubt
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posted by zsm28 on 7/17/2008 3:07:38 PM  |  status: Live
Asker's Rating: Helpful   
chemlady's comment:
"thanks...... we needed to find z using the tc and pc in order to calculate tr and pr to use a chart....."
Response Details:
molar mass M = 28 g/mol = 28*10-3 kg/mol
mass m = 40 kg
pressure P = 5*106 Pa
temperature T = 180 K
PV = mRT/M
∴V = mRT/(PM) = 40*8.31*180/(5*106*28*10-3) = 0.43 m3
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