Question:

Consider the van der Waals constants, a and b, for the following gases.
GasArNeKrXe
a/ (atm dm6 mol-2)1.30.25.14.1
b/ (10-2 dm3 mol-1)3.21.71.05.0

Which gas is expected to have the highest critical temperature?

Show Hint

Critical temperature is \(T_c = \dfrac{8a}{27Rb}\), so it is proportional to \(a/b\).
Updated On: Nov 5, 2026
  • Ne
  • Kr
  • Xe
  • Ar
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The Correct Option is B

Solution and Explanation

For a van der Waals gas, \(T_c = \dfrac{8a}{27Rb}\), i.e. \(T_c \propto \dfrac{a}{b}\).
Compute \(a/b\) (with b in units of \(10^{-2}\) dm3/mol):
Ar: \(1.3/3.2 = 0.41\)
Ne: \(0.2/1.7 = 0.12\)
Kr: \(5.1/1.0 = 5.1\)
Xe: \(4.1/5.0 = 0.82\)
Kr has the largest ratio, hence the highest critical temperature.
Answer: (2) Kr
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