Question

Use the data below for O2 gas at 273.15 K:

P (atm)	Vm (dm3/mol)	(g/dm3)
0.750000	29.8649	1.07144
0.500000	44.8090	0.714110
0.250000	89.6384	0.356975

a. Under what pressure conditions will O₂ act most like an ideal gas and obey the ideal gas law?

b. Rearrange the ideal gas law to solve for the gas law constant, R.

Answer 1

Recall that we must avoid particle interactions

this is given when density is pretty low

from the data:

a)

Number from top to bottom, 1,2, and 3...

3 > 2 > 1

b)

Apply

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

PV = nRT

solve for R

R = PV/(nT)

R = Pv/T; where v = V/n, molar volume

now, substitute known data

i)

R = (0.75)*29.8649/273.15 = 0.082001L*atm/(mol-K)

ii)

R = (0.50)*44.8090/273.15 = 0.0820226L*atm/(mol-K)

iii)

R = (0.250)*89.6384/273.15 = 0.082041 L*atm/(mol-K)

actual R value --> 8.205733

nearest value is iii, as expected