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Assuming that Nitrogen will be an ideal gas at this temperature we may use the ideal gas law:

PV=nRT

To use this law we have to have a value of R that agrees with the units that we use.

1 Torr = 1/760 atmosphere so 6680 tor = 8.789 ATM

27' C = 300.15'K

347ml = .347L

so

R= 8.314472 J/'Kmol

(8.789)(.347)/(8.314472)(300.15) = n =0.00122 moles

and of course a real man has to use scientific notation. 1.22x10-3 moles of nitrogen

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