Calculate the density of oxygen gas at 25deg.C and 763 mmHg?

Question

I can get as far as changing from C to K... so 298.15 K =T and P=763mmhg =763 torr(1atm/760torr)=1.004

But now I don't know the steps to get the density. Please help! Thank you!

Answer 1

Ok assuming O2 is ideal then at 100 kPa and 298 K 1 mole occupies 24.8 dm3. Your pressure is slightly more than 100 kPa (1 bar) so the volume will be reduced to 24.38 dm3.

1 mole for oxygen is 32 g so the density (mass/volume) = 1.31 g dm-3.

Answer 2

Take the ideal gas law and rewrite n (moles) as m/Mm (mass in grams / molar mass).

PV = (m / Mm)RT . . .multiply both sides by Mm

PVMm = mRT . . . . . divide both sides by V

PMm = (m/V)RT . . . but m/V = density (d)

PMm = dRT

d = PMm / RT = (1.004)(32) / (0.0821)(298.15) = 1.31 g / L