How much butane is present as a liquid?

given: Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 degrees celsius. A 250 mL sealed flask contains 0.55g of butane at - 22 degrees celsius.

Answer:
You get the vapor pressure of butane by the Clausius-Clapeyron relation for an ideal gas phase:
dlnpv / dT = ΔHv / (R·T²)
If integrate the relation from the normal boiling point
given with T₀= -0.4°C = 272.75K and pv₀= 1atm = 101325Pa
and solve for pv, you get:
ln (pv(T)/pv₀) = (-ΔHv/R)· (1/T - 1/T₀)
=>
pv(T) = pv₀· exp{ -(ΔHv/R)· (1/T - 1/T₀) }

at T = -22°C = 251.15K

pv = 101325Pa · exp{ -(22440J/mol / 8.314472J/molK)· (1/251.15K - 1/272.75) }
= 43263Pa

Assume liquid volume in the flask is negligible. Apply ideal gas law
p · V = m · Rb · T
(Rb = R / Mb = 8.314472J/molK / 58.08g/mol = 0.143155J/gK is the specific gas constant of butane)
If there is liquid in the flask the pressure of the gaseous butane equals the vapor pressure. Thus the mass of gaseous butane is:
m_g = (pv · V) / (Rb · T)
= (43263Pa · 0.25·10^-3m³) / ( 0.143155J/gK · 251.15K)
= 0.3g

The liquid mass in the flask is
m_l = 0.55g - m_g = 0.25g

The liquid volume is
V_l = m_l / ρ_l = 0.25g / 0.584 g/cm3 = 0.43cm³
which is much smaller than the flask volume of 250cm³. So the assumption of negligible liquid volume was correct.

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