Constant Volume Heat Addition

A 28 liter rigid enclosure contains air at 140 kPa and 20^oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.

Formulae:

For an ideal gas, PV = nRT

From first law of thermodynamics,

Q + W = ΔU

For an ideal gas ΔU is a function of temperature only, and work done at constant volume = 0

Therefore, Q = ΔU.

And, ΔU = mC_V(T₂ - T₁)

m = n x Molecular weight

n = PV / (RT) = 140 x 1000 x 0.028 / (8314 x 293) = 1.609 x 10^-3 kmol

m = 1.609 x 10^-3 x 29 = 0.0467 kg.

For constant volume system, P ∝ T.

Therefore, P₁/T₁ = P₂/T₂

140/293 = 345/T₂

T₂ = 722 K.

C_V = 0.718 kJ/kg.^oC (data, for air at 20^oC)

Q = ΔU = 0.0467 x 0.718 x (722 - 293) = 14.385 kJ

i.e., Heat added = 14.385 kJ