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Thermo molar volume question

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Given T=572 ºF, pressure= 145 psi (g) using the steam tables and ideal gas law
    Molar Mass= 18 g/mol for water
    Find: Is the water vapour acting ideally?
    2. Relevant equations
    Ideal Gas Law: PV=nRT

    P is the pressure =145 psi (g)
    V is the molar volume
    n= mol
    R= Ideal gas constant 8.314 J(K^-1)(mol^-1)
    T is temperature in kelvin, so temperature in farenheit= 572 ºF =573.15 ºK

    How can I calculate the molar volume using the steam tables? and how can I know if the water vapour is acting ideally?

    3. The attempt at a solution

    Ideal gas law: PV= nRT
    so: V= nRT/P= (18 mol)(8.314 JK^-1mol^-1)(573.15 ºK)= 85773.04=8.5x10(^4) metres cube
    where: P= 145 psi (g)= 999739.80705 Pascals
    1 Psi = 6 894.75729 Pascals
  2. jcsd
  3. Nov 12, 2014 #2


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    Green is good, and red is not so good; the steam tables give you measured values of properties. Pay attention to your calculation of ideal properties, and you can compare those to table props at the same conditions.
  4. Nov 12, 2014 #3
    That (g) next to the pressure: is that gage?

    What units are used in your steam tables?

  5. Nov 13, 2014 #4
    It doesn´t say. I think it will be gauge pressure

    units of steam tables? vg= metres cube/kg

    If u want to have a look, Steam tables attached

    By the way, I always do everything different from everybody else, that why I´m different

    Attached Files:

  6. Nov 13, 2014 #5
    so how can I calculate the molar volume using the steam tables? and how can I know if the water vapour is acting ideally?
  7. Nov 13, 2014 #6
    You said that the units of specific volume in the steam table are m3/kg. You convert that to m^3/gram. How many grams are there in a g-mole of water?
    You compare the molar volume you calculate from the ideal gas law with the molar volume you determine from the steam tables. Do you know how to use the ideal gas law to calculate the molar volume?

  8. Nov 13, 2014 #7
    yes, of course. Ideal gas law equation: PV=nRT

    deal gas law: PV= nRT
    so: V= nRT/P= (18 mol)(8.314 JK^-1mol^-1)(573.15 ºK) ( ) = 85773.04=8.5x10(^4) metres cube
  9. Nov 13, 2014 #8
    This is an incorrect application of the ideal gas law. What happened to the pressure in your equation? The molar volume is defined as the volume occupied by 1 mole of the gas. If you have 1 mole (not 18), what is the volume of this mole of gas at the temperature and pressure of the system?

    Did it really make sense to you that the volume of one g-mole of water vapor could be 85000 cubic meters, given that, under standard conditions, the molar volume is only 22.4 liters = 0.0224 cubic meters? After you do a calculation, it helps to do a reality check.

  10. Nov 13, 2014 #9
    PV= nRT,

    V=nRT/P= (1 mol)(8.314 JK^-1mol^-1)(573.15 ºK)/(1000)= 4.76 metres cube

    145 psi= 10^5 Pa x 10 = 10^6Pa= (10^3 KPa)=1000 KPa= pressure

    To calculate the molar volume of steam:

    For steam: R= R/M= (8.31434)/18=0.4619 KJ/KG

    hence: PV=nRT
    So V=RT/P=(0.4619)(300+273.15)/1000 KPa=0.264 metres cube/kg
  11. Nov 13, 2014 #10
    Why are you dividing by pressure in Kpa? The gas constant is in Joules. You found the volume of 1 kg-mole, not the volume of 1 gm-mole. Your final answer for the specific volume is correct, however. Please pay more attention to units.
  12. Nov 13, 2014 #11

    the pressure is always in PA or Kpa. I diveded by pressure to calculate the Volume
    Yes, the gas constant is in KJ/kg like I wroted it.
  13. Nov 13, 2014 #12
    Well, it looks like you got the right answer in the end. But, again, please pay lots of attention to units when you do a problem.

  14. Nov 13, 2014 #13

    P= 1 MPA, V= 0.264 Vactual=0.25790 Z=PVactual/RT=0.9744

    SO: CONCLUSION: As seen the Ideal Gas Assumption is very good for this case. The water vapour is acting ideally
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