ideal gas law?

We consider an ideal gas of molar mass M = 16 g/mol at +10oC. What is the pressure of the gas if its density is 2.4 kg/m3 ?

Express the result in the unit [Pa] and to three significant figures. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4.

Please show all the calculation steps so that I understand how the answer was determined

1 Answer

  • 1 decade ago
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    I assume that it is 10 degrees C and that you meant that the molar mass is 16 and that the gas is not oxygen.


    Since the density is 2.4 kg/m^3 and you know that the molar mass is 16 g/mol, which is a lot lighter than air whose molar mass is 0.79*(28) +0.21*(32)= 29 g/mol. Then:

    (2.4 X 10^3 g/m^3)/(16 g/mol) =24 X 10^2/16(mol/m^3)=

    1.5 X 10^2 mol/m^3

    (1500 mol/m^3)X(1m^3/1000 L) = 1500 mol/1000 L = 1.5 mol/L

    Now according to the Idea gas law, what is the volume of one mole of ideal gas at Standard Temperature and Pressure (STP)?

    For gas calculations R = 0.082 (L atm)/(mol K)

    The volume of 1 mol at STP is 0.082X 1 mol X 273.15 K/ 1 atm (see that all the units except for liters cancel out. I left out all of R's units and I used atmospheres because that is what real people use after they get out of school, or even basic chemistry. 1 atm =101.325 kPa)

    So 0.082 X 273.15 = 22.4 L/mol

    But you are at 10 C and STP T= 0 C

    You now know the original V,T,n and R. but you don't know P.

    PV=nRT which can morf into P=(nRT/V)

    P = (1500 mol X 0.082 atm L/mol K X 283.15 K)/1000 L = 34.8 atm

    101.325 kPa/atm X 34.8 atm =3528kPa or 3.5E2 kPa

    Source(s): Gases start to loss ideal behavior at very high or low pressures. CO2 and H2O are really not ideal gases if you want to be exact. The most important thing here to remember is that 1 mol at STP is 22.4 L/ mol. Remember this and what the conditions for STP are and you can calculate R for any units you may need.
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