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.

PART A)

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? Let a=0.14N⋅m4/mol2 and b=3.2×10−5m3/mol.

Express your answer numerically using two significant figures. If there is more than one answer, enter each answer separated by a comma.

PART B)

At what volume does the van der Waals equation give a pressure that is 5% lower than the ideal gas law pressure?

Express your answer numerically using two significant figures. If there is more than one answer, enter each answer separated by a comma.

Answer 1

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.

I'll just drop the units at this point, so:

PV = 1247.1.

If the pressure increases by 5%, then PV = (1247.1)(1.05) = 1309.46 (to 2 d.p.).

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 for V gives V = 4.0 x 10^-5 L = 4.0*10^-8 m^3

You can re-do this using PV = (1247.1)(0.95) to answer the second part.