A 28 liter rigid enclosure contains air at 140 kPa and 20^{o}C. 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 = DU

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

Therefore, Q = DU.

And, DU = mC_{V}(T_{2} - T_{1})

m = n x Molecular weight

Calculations:

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 a T.

Therefore, P_{1}/T_{1} = P_{2}/T_{2}

140/293 = 345/T_{2}

T_{2} = 722 K.

C_{V} = 0.718 kJ/kg.^{o}C (data,
for air at 20^{o}C)

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

i.e., Heat added = **14.385 kJ**