A 0.500-mol sample of O2 gas is in a large cylinder with a movable piston on one

Question

A 0.500-mol sample of O2 gas is in a large cylinder with a movable piston on one end so it can be compressed. The initial volume is large enough that there is not a significant difference between the pressure given by the ideal gas law and that given by the van der Waals equation.

As the gas is slowly compressed at constant temperature (use 300 K), at what volume does the van der Waals equation give a pressure that is 5% higher than the ideal gas law pressure? At what volume does the van der Waals equation give a pressure that is 5% lower than the ideal gas law pressure (there are two answers for this)?Let a=0.14Nm4/mol2 and b=3.2×105m3/mol.

Explanation / Answer

By the ideal gas law:
PV = nRT.

We are given that T = 300 K and n = 0.500 mol. Taking the pressure to be in kPa, R = 8.314 L*kPa/(K*mol):
PV = nRT
     = (0.500 mol)[8.314 L*kPa/(K*mol)](300 K)
     = 1247.1 L*kPa.
PV = 1247.1.
If the pressure increases by 5%, then PV = (1247.1)(1.05) = 1309.46 .
Using the Van der Walls equation:
(P + n^2a/V^2)(V - nb) = nRT
==> (PV^2 + n^2a)(V - nb) = nRTV^2, by multiplying both sides by V^2
==> [1309.46V + (0.500)^2(0.14)][V - (0.500)(3.2 x 10^-5)] = 1309.46V^2.
(Note that PV^2 = (PV)(V) = 1309.46V and nRT = PV = 1309.46.)

Solving this

V = 4.0 x 10^-5 L

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