A pistoncylinder arrangement contains a mixture composed of

A piston-cylinder arrangement contains a mixture composed of 4 kg of N_2, 3 kg of O_2, and 4 of kg CO initially at 6 bur, , 327 degree C. The mixture then expands adiabatically to 2 bar, and a temperature of 47 degree C. For this mixture, determine: a.) the analysis on a molar basis b) the initial volume of the mixture c.) the work and heat transfer for this process d.) the partial pressure of each component at the final state

Solution

For N2, we have M = 28 g/mol, Cv = 0.743 kJ/kg-K , R = 0.297 kJ/kg-K

For O2, we have M = 32 g/mol, Cv = 0.659 kJ/kg-K , R = 0.26 kJ/kg-K

For CO, we have M = 28 g/mol, Cv = 0.72 kJ/kg-K, R = 0.297 kJ/kg-K

a)

Number of moles of n_N2 = m_N2 / M_N2

= 4000 / 28

= 142.857 moles

Number of moles of n_O2 = m_O2 / M_O2

= 3000 / 32

= 93.75 moles

Number of moles of n_CO = m_CO / M_CO

= 4000 / 28

= 142.857 moles

Total moles n = 142.857 + 93.75 + 142.857 = 379.464 moles

Mole fraction N2 = 142.857 / 379.464 = 0.376

Mole fraction O2 = 93.75 / 379.464 = 0.247

Mole fraction CO = 142.857 / 379.464 = 0.376

b)

Total mass m = 4 + 3 +4 = 11 kg

Average Molar mass of mixture M_mix = 11*10³ / 379.464 = 28.99 g/mol

Gas constant for mixture R_mix = R / M_mix = 8314 / 28.99 = 286.81 J/kg-K

PV = m*R_mix*T

6*10⁵ * V = 11 * 286.81 * (327 + 273)

V = 3.155 m³

c)

Cv_mix = (4/11)*0.743 + (3/11)*0.659 + (4/11)*0.72 = 0.712 kJ/kg-K

Q - W = m*Cvmix*(T2 - T1)

Since adiabatic expansion, hence heat transfer Q = 0.

0 - W = 11 * 0.712 * (47 - 327)

W = 2192.96 kJ

d)

Partial pressure of N2 = 0.376 * 2 = 0.752 bar

Partial pressure of O2 = 0.247 * 2 = 0.494 bar

Partial pressure of CO = 0.376 * 2 = 0.752 bar