448 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th9-3.pm5(i) The temperature of the equilibrium mixture,
(ii) The pressure of the mixture, and
(iii) The change in entropy for each component and total value.
cv(N ) 2 = 0.744 kJ/kg K, cv(H ) 2 = 10.352 kJ/kg K
cp(N ) 2 = 1.041 kJ/kg K, cp(N ) 2 = 14.476 kJ/kg K.
[Ans. (i) 342.6 K ; (ii) 2.661 bar ; (iii) ()∆SH 2 = 0.00235 kJ/K ;
()∆SN 2 = 0.1908 kJ/K ; ∆S = 0.19315 kJ/K]- 3 kg of N 2 and 5 kg of CO 2 at a pressure of 3 bar and a temperature of 20°C comprise a perfect gas
mixture. Calculate cv and cp of the mixture.
If the mixture is heated at constant volume to 40°C, find the change in internal energy, enthalpy and
entropy of the mixture.
Take : cv(N ) 2 = 0.7448 and cp(N ) 2 = 1.0416 kJ/kg K
cv(CO ) 2 = 0.6529 and cp(CO ) 2 = 0.8418 kJ/kg K.
[Ans. 0.6873 kJ/kg K, 0.9167 kJ/kg K ; 109.96 kJ, 146.67 kJ, 0.363 kJ/K]