Question

The temperature inside a pressure cooker is 117°C. Calculate the vapor pressure of water inside the pressure cooker. The enthalpy of vaporization of water is 40.7 kJ/mol.

What would be the temperature inside the pressure cooker if the vapor pressure of water was 2.50 atm?

The enthalpy of vaporization of water is 40.7 kJ/mol.

Answer 1

Answer – We are given, Vapor pressure of water, P₁ = ? atm, T₁ = 117^oC + 273.15 = 390 K, P 2 = 1.0 atm

T₂ = 100 + 273 = 373 K , ΔH_vap = 40.7 kJ/mol = 4.07*10⁴ J/mole

We need to use the Clausius Clapeyron equation –

ln P₁/ P₂ = ΔH_vap / R *( 1/T₂ -1/T₁)

ln P1/ 1 atm = 4.07*10⁴ J.mol^-1 / 8.3145 J.mol^-1. K^-1 * ( 1/373 K – 1/390 K)

ln P1 / 1.00 atm = 0.572

Taking antiln from both side

P1 / 1.00 atm = e ^0.572

P1 / 1.00 atm = 1.77

P1 = 1.77 *1.00 atm

     = 1.77 atm

The vapor pressure of water inside the pressure cooker is 1.77 atm

B) vapor pressure P₁ = 1.00 atm, T₁ = 373 K

T₂ = ? K , ΔH_vap = 4.07*10⁴ J/mole, P₂ = 2.50atm

We need to use the Clausius Clapeyron equation –

ln P₁/ P₂ = ΔH_vap / R *( 1/T₂ -1/T₁)

ln 1.00 atm /2.50 atm = 4.07*10⁴ J.mol^-1 / 8.3145 J.mol^-1. K^-1 *( 1/T₂ – 1/373 K)

-0.916   = 4895.6 * 1/T₂ – 13.12

-0.916 +13.12 = 4895.6 * 1/T₂

12.20 = 4895.6 * 1/T₂

1/T₂ = 11.5 / 4936

        = 0.002469

   T₂ = 401 KT

The temperature inside the pressure cooker if the vapor pressure of water was 2.50 atm is 401 K.