A 19.0 g sample of liquid methane is heated at a constant pressure of 1 atm from a temperature of 109.1 K to a temperature of 185.3 K. How much energy in kJ is required?

The energy required is 12.887KJ

Explanation:

There are two separate heat inputs involved in this problem:

q₁ = heat added to vaporize the methane at 109.1 K q₂ = energy added to heat the vapor from 109.1 K to 185.3 K

q = q₁ + q₂

q = nΔH + mCΔT

where;

n is number of moles of methane

ΔH is the molar enthalpy of vaporization of methane = 8.17 kJ/mol

m is the mass of methane = 19g

C is the specific heat capacity of gaseous methane = 2.20 J/g. K

ΔT = T₂ - T₁ = 185.3 - 109.1 = 76.2 K

n = Reacting mass/Molar mass

molar mass of methane (CH₄) = 16g/mol

n = 19/16 = 1.1875 mol

⇒q₁ = nΔH = 1.1875 X 8.17 = 9.702 kJ

⇒q₂ = mCΔT, = 19 X 2.2 X 76.2 = 3185.16 J = 3.18516KJ

q = q₁ + q₂, ⇒ 9.702 kJ + 3.18516KJ = 12.887KJ

Therefore, the energy required is 12.887KJ