Question

A perfect gas mixture consists of 4 kg of N₂ and 6 kg of CO₂ at a pressure of 4 bar and a temperature of 25°C. Calculate c_v and c_p of the mixture. 

If the mixture is heated at constant volume to 50°C, find the change in internal energy, enthalpy and entropy of the mixture. 

Take : c_{v(N2 )} = 0.745 kJ/kg K, 

c_v(CO2) = 0.653 kJ/kg K 

c_p(N2) = 1.041 kJ/kg K, 

c_p(CO2) = 0.842 kJ/kg K.

Answer 1

m_N2 = 4 kg, m_CO2 = 6 kg, p_mix = 4 bar

T₁ = 25 + 273 = 298 K, T₂ = 50 + 273 = 323 K

c_v(mix) = ?, c_p(mix) = ?

Using the relation,

Change in internal energy, ∆U :

∆U = [mc_v(T₂ – T₁)]_mix 

= (4 + 6) × 0.6898(323 – 298) = 172.45 kJ.

Change in enthalpy, ∆H :

∆H = [mc_p(T₂ – T₁)]_mix 

= (4 + 6) × 0.9216(323 – 298) = 230.4 kJ.

Change in entropy, ∆S :

= 4 × 0.745 log_e \(\cfrac{323}{298}\) + 6 × 0.653 log_e \(\cfrac{323}{298}\)

= 0.5557 kJ/K.

Note. ∆S may also be found out as follows :

∆S = (mN₂ + m_CO) c_v(mix) loge \(\cfrac{T_2}{T_1}\)

= (4 + 6) × 0.6898 log_e \(\cfrac{323}{298}\) = 0.5557 kJ/K