Question

Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at –108.0 °C. What is the vapor pressure of xenon at -148.0 °C? (R = 8.314 J/K⋅mol)

Answer 1

Answer:

P₁ = 0.0562 atm

Explanation:

Using the Clausius-Clapeyon equation

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁)       ------ (eqn 1)

Step 1: From the question given, we state out the parameters given

P₁ = ?                T₁ = -148.0⁰C

P₂ = 1atm          T₂ = -108.0⁰C

ΔHvap = 12.6kJ/mol      R = 8.314J/K.mol

Step 2: Do conversions where necessary for unit consistency since our R value is in J/K.mol

a) convert ⁰C to K

1K = ⁰C + 273.15

T₁ = -148.0⁰C => -148.0⁰C + 273.15

T₁ = 125.15K

T₂ = -108.0⁰C => -108.0⁰C + 273.15

T₂ = 166.15K

b) convert kJ/mol to Joules

ΔHvap = 12.6kJ/mol = 12600Joules

substituting parameters into eqn 1

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁)

ln (P₁/1atm) = 12600J / 8.314 (1/166.15 - 1/125.15)

                  = 1515.51 (0.0060 - 0.0079)

                  = 1515.51(-0.0019)

ln (P₁/1atm) = -2.8794

taking exponential of both sides to get rid of the natural log

P₁ = e^ -2.8794

P₁ = 0.05616 atm

P₁ = 0.0562atm

Key Words

1) Clausius-Clapeyon: shows the relationship between pressure and temperature and it is used to estimate the vapour of a solution at a different temperature