Question

How well does argon gas al 400 K and 3 atm approximate a perfect gas? Assess the approximation by reporting the difference between the molar volumes as a percentage of the perfect gas molar volume

Answer 1

Solution:

According to perfect gas equation,

PV = nRT

V/n = RT/P

Vm = RT/P

Where, Vm = molar volume =?

R = gas constant = 0.0821 L atm K-1 mol-1

P = pressure = 3.0 atm

T = temperature = 400 K

Thus,

Vm = 0.0821 L atm K-1 mol-1 x 400 K / 3.0 atm

Vm = 10.95 L

Since, a perfect gas at 273 K temperature and 1 atm pressure,

Vm = Molar mass/ density

= 39.988 g mol-1 / 1.784 g L-1 = 22.4 L

Thus, difference in molar volume = 22.4 - 10.95 = 11.45 L

Therefore,

% difference in molar volume = (11.45 L / 22.4 L) x 100

= 51.1 %