Choose any gas: 

        1 mole of Argon  =  39.948 g Ar  =  22.4 L Ar @ STP

        STP = Standard temperature (0^oC or 273 K)  & Standard Pressure  =  1 atmosphere

        Assume the gas acts ideally.  Then  PV =  nRT

        Solve for R:   

        Substitute into equation: 

        Solve:  R  =  0.0821 atm L / mol K

        If R is needed in units of pressure (kPa):

                =  8.314 kPa L / mol K

        If R is needed in units of pressure (mm Hg)

                 =  62.396 mm Hg L / mol K

    I would suggest always using the value of 0.0821 atm L / mol K for R.

   You must be sure the pressure you calculate with in the problem is ALWAYS in atmosphere if using the value                                                 R  =  0.0821 atm L / mol K.

R  =  0.0821 atm L / mol K

8.314 kPa L / mol K

62.396 mm Hg L / mol K

Last modified