Question

When comparing two gases to determine which one is closest to an ideal gas, which strategy is best?

Answer 1

This problem can be solved by approching this by Van der waals problem strategy .

consider Van Der Waal's equation

(P + a n² / V²) x (V - nb) = nRT

certainly if a = 0 and b = 0, the equation reduces to
P x V = nRT (the ideal gas law)

now let's look at your choices.

(1).. the higher the "a", the more ideal. This is false. it's backwards. The lower the "a" the more ideal the gas. More importantly, it's the combination of both lower "a" and lower "b" that makes a gas more ideal than another.
(2).. calculate R.... R = (P + an²/V²)x(V-nb) / (nT). the closer the R is to 0.08206, the more ideal the gas
(3).. generally speaking the lower the molar mass, the smaller the volume, the more ideal the gas. So this is false. But it's not always the case.
(4).. this one's tricky. the lower the "n" value, the more ideal the gas. This is sort of true but.. it's actually the combination of "a" and "b" that makes the gas more ideal. If BOTH are lower, then you can say the gas behaves more ideally at the same P, V, n, and T.

The point is this. Because of the terms in that Van Der waal's equation, You must consider both "a" and "b" and P,V,n, and T all at once to determine which gas is more ideal.

the best way to do that, is calculate R .