A sample of 1.00 moles of oxygen at 50°C and 98.6 kPA occupies what volume?

27.3L

Explanation:

Looking at the parameters given;

Number of moles (n) = 1.00 moles

Temperature (T) = 50° + 273 = 323K

Pressure (P) = 98.6KPa or 0.97 atm (to obtain this, divide 98.6*10^3/101325)

R = 0.082 L. atmK-1mol-1

Then we apply the ideal gas equation

PV=nRT

V = nRT/P

V = 1.00*0.082*323/0.97

V = 27.3L