Question

Calculate the molar entropy of Krypton at 298.15K and 1 bar. Assume it behaves as a
monatomic ideal gas with m = 0.08308 kg mol-1 and the degeneracy of the ground
electronic state equal to one.
Hint:
Q(N, V, T)= 1/N![(2πmkBT)/h^2] V^N*gel

Answer 1

Total differential of the enthalpy is given by
dH = T dS + V dP
when you divide by the amount it can be expressed in terms of the molar properties
h = H/n , s = S/n and V/n.
=>
dh = T ds + v dp
So total differential of the entropy is,
ds = (1/T) dh - (v/T) dp

For an ideal gas
dh = cp dT
and
v/T = V/(n∙T) = R/p
Therefore
ds = (cp/T) dT - (R/p) dp

When you integrate this from some reference state with T₀, p₀ and the known entropy s₀ = s(T₀,p₀) to another state at T and p you get
s....... T............... p
∫ ds = ∫ (cp/T) dT - ∫ (R/p) dp
s₀....... T₀............. p₀
<=>
s - s₀ = cp∙ln(T/T₀) - R∙ln(p/p₀)

Hence, when you konw the entropy for a reference state you can compute the entropy at any other ideal gas state from:
s(T,p) = s(T₀,p₀) + cp∙ln(T/T₀) - R∙ln(p/p₀)


So the molar entropy of Krypton at 298K and 1 bar is
s(T,p) = 164.1 J∙K⁻¹∙mol⁻¹ + 20.8∙ J∙K⁻¹∙mol⁻¹∙ln( 298 / 298) - 8.3145 J∙K⁻¹∙mol⁻¹∙ln(1/1)

= 164.1 J∙K⁻¹∙mol⁻¹