Answer to Question #106015 in Physical Chemistry for Bharti

Question

a stream of nitrogen flowing at the rate of 2 kg/sec and the stream of hydrogen is flowing at the rate 0.5 kg/sec mixed adiabatically in a steady state flow process. if the gases are assumed ideal what is the rate of entropy change as a result of a process. in chemical engineering thermodynamics

Answer 1

The entropy of adiabatic mixing n₁ moles of an ideal gas 1 and n₂ moles of an ideal gas 2 is given by the formula:

$\Delta S = R*(n1*ln((n1+n2)/n1)+$

$+n2*ln((n1+n2)/n2))$

the moles of nitrogen N₂ are: n₁ = 2000 g/(28 g/mole) = 71.4286 mole

the moles of hydrogen H₂ are: n₂ = 500 g/(2 g/mole) = 250 mole

Putting n₁ and n₂ in the formula and multiplying by R = 8.31 J/(K*mole) gives that the rate of entropy increase per 1 second is:

$\Delta$S = 1414.883 J/(K*sec)