What is the pressure in a 10.6-L cylinder filled with 11.2 g of oxygen gas at a temperature of 345 K?

1 Answer
May 6, 2016

Approx. "1 atm"1 atm

Explanation:

We use the Ideal gas equation with appropriate units. R =0.0821*L*atm*K^-1*mol^-1R=0.0821LatmK1mol1, is probably most useful for chemists.

P=(nRT)/VP=nRTV == ((11.2*cancelg)/(32.0*cancelg*cancel(mol^-1))*0.0821*cancelL*atm*cancel(K^-1)*cancel(mol^-1)xx345*cancel(K))/(10.6*cancelL) ~= 1*atm

The most difficult part of this question is choosing the most appropriate gas constant, R. For chemists, R=0.0821*L*atm*K^-1*mol^-1 is probably most useful, because it uses the units that with which chemists typically deal. In an examination, several values of the gas constant should be quoted as supplementary material.