Final answer:
To calculate the entropy of mixing air from pure gases using ideal gas behavior, various thermodynamic formulas are applied, such as Gibbs free energy, Boltzmann entropy, Helmholtz free energy, and noting that the enthalpy change for ideal gases during mixing is typically zero. Quantitative calculations would require details on the composition of air and the conditions of temperature and pressure at which mixing occurs. option d ids correct
Step-by-step explanation:
option d ids correct The calculation of entropy of mixing when air is prepared from the pure gases assuming ideal gas behavior involves understanding several concepts in thermodynamics:
- The Gibbs free energy (G) change for an ideal mixture can be calculated using the formula ΔG = ΔH - TΔS, where ΔH is the enthalpy change, T is the temperature in Kelvins, and ΔS is the entropy change.
- The Boltzmann entropy formula S = k ln W can be used to calculate the entropy based on the number of ways W that the particles can be arranged.
- The Helmholtz free energy (A) is given by A = U - TS, where U is the internal energy, T is the temperature, and S is the entropy of the system.
- The enthalpy change (ΔH) for mixing ideal gases is typically zero because no heat is absorbed or released during the mixing of ideal gases at constant temperature and pressure.
For a quantitative answer, you would use the properties of the individual gases making up air, their molar fractions, and the temperature at which the mixing occurs to calculate the entropy of mixing. This would typically involve applying the formula for entropy of mixing for ideal gases: ΔS_mixing = -R Σ xi ln xi, where R is the gas constant and xi is the mole fraction of each gas.
In order to determine which substance has a higher entropy, we need to consider the temperature, pressure, and the number of gas molecules present.
a) 1 mol of He(g) at 10 K and 1 atm pressure has a lower entropy compared to 1 mol of He(g) at 250°C and 0.2 atm. This is because the higher temperature and lower pressure in the second scenario lead to a greater volume and more microstates, resulting in a higher entropy.
b) A mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm has a higher entropy compared to 2 mol of NH3(g) at 25°C and 1 atm. This is because there are more molecules of gas present in the first scenario, leading to a higher entropy.