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What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow process into product streams of pure oxygen and nitrogen, also at 25°C and 1 bar, of the thermodynamic efficiency of the process is 5% and if Tσ = 300 K

User Ribsies
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Answer:

Work = 116.1 kJ/mol.

Step-by-step explanation:

Separating air into pure oxygen and nitrogen requires the removal of one component (nitrogen) from the mixture while leaving the other component (oxygen) behind. This can be accomplished using a cryogenic distillation process, which takes advantage of the different boiling points of the two components.

The thermodynamic efficiency of the process is given as 5%, which means that only 5% of the work input is converted to useful work (i.e., the separation of the components). The remaining 95% is dissipated as waste heat.

The work required for the separation of air can be calculated using the following equation:

W = ΔG / η

where W is the work required, ΔG is the Gibbs free energy change for the separation process, and η is the thermodynamic efficiency.

The Gibbs free energy change for the separation of air into pure oxygen and nitrogen can be calculated using the following equation:

ΔG = RTln(K)

where R is the gas constant, T is the temperature (in kelvin), and K is the equilibrium constant for the reaction. For the separation of air, the equilibrium constant is equal to the ratio of the vapor pressures of nitrogen and oxygen at the given temperature and pressure:

K = P_N2 / P_O2

At 25°C and 1 bar, the vapor pressures of nitrogen and oxygen are:

P_N2 = 0.79 × 1 bar = 0.79 bar
P_O2 = 0.21 × 1 bar = 0.21 bar

Therefore, the equilibrium constant is:

K = 0.79 / 0.21 = 3.76

Substituting this into the equation for ΔG gives:

ΔG = RTln(K) = (8.314 J/mol-K)(298 K)ln(3.76) = -5806 J/mol

The negative sign indicates that the separation process is thermodynamically favorable (i.e., exergonic).

Substituting ΔG and η into the equation for W gives:

W = ΔG / η = (-5806 J/mol) / 0.05 = -116,120 J/mol

The negative sign indicates that work must be done on the system to effect the separation of air. The work required is 116,120 J/mol, or 116.1 kJ/mol.

The value of Tσ = 300 K is not used in this calculation, as it represents the reference temperature for calculating the thermodynamic efficiency.

The value of 300K (or more precisely, Tσ = 298.15 K) is used as the reference temperature for calculating thermodynamic efficiency in some cases, particularly for thermodynamic cycles. However, in the problem given, we are not dealing with a thermodynamic cycle but rather a steady-flow process for the separation of air into its component gases. In this case, the temperature and pressure of the air and product streams are all specified (25°C and 1 bar), and the calculation of the work required for the separation is based on the Gibbs free energy change of the process, which depends on the actual temperature and pressure conditions. Therefore, the value of 300K (or Tσ) is not used in this calculation.
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