Consider 2kg mixture of H2 and N2 ideal gases inside a closed rigid tank at 140k

Question

Consider 2kg mixture of H2 and N2 ideal gases inside a closed rigid tank at 140kPa and 100 degree C. This mixture has 0.4 mass fraction of H2. This mixture undergoes a heating process to final temperature 190 degree C. For this mixture, determine R, M, Cv and initial volume of the mixture. Heat energy added to the mixture. Use these Given data: Cv for H2 = 10.183kJ/(kg k) Cv for N2 = 0.743 kJ/(kg K) Universal Gas Constant = 8.314 kJ/(kmol k), Molar mass for H2 = 2 kg/kmol Molar mass for N2 = 28.013 kg/mol

Explanation / Answer

Total Mass =2 kg

Mass of H2 = 0.4*2 =0.8 kg

Mass of N2 =1.2 kg

kmol of H2 =0.8/2 =0.4

kmol of N2 =1.2/28.013 =0.043

a)Now M(mixture) =(0.4*2 +28.013*0.043)/(0.4+0.043) =2/0.443 = 4.515 kg/kmol [ANS]

b) R = Summation of {(Mole fraction)i * Ri}

= (0.4/0.443) *(8.314/2) +(0.043/0.443) * (8.314/28.013)

= 3.782 KJ/kg-K

=3.782*4.515 KJ/Kmol -K

=17.077 KJ/Kmol-K

c) Cv= Summation of {(Mole fraction)i * Cvi}

=(0.4/0.443) *(10.183) +(0.043/0.443) * (0.743)

=9.267 KJ/kg-K

d)Initial Volume V =(n*R*T)/P = 20.156 Litre

e) Heat added =m*Cv*dT =1668.06 KJ [ANS]

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