Find the entropy of mixing of 80% N2 and 20% O2 in air
To find the entropy of mixing of two gases, you can use the formula:
ΔS_mix = -R * ∑(xi * ln(xi))
Where:
- ΔS_mix is the entropy of mixing
- R is the gas constant (8.314 J/(mol·K))
- xi is the mole fraction of each component (N2 and O2)
In this case, we have a mixture of 80% N2 (0.8) and 20% O2 (0.2). So the calculation would be:
ΔS_mix = -8.314 * [ (0.8 * ln(0.8)) + (0.2 * ln(0.2)) ]
To simplify the calculation, you can use the fact that ln(x) = -ln(1/x). Therefore:
ΔS_mix = 8.314 * [ (0.8 * -ln(1/0.8)) + (0.2 * -ln(1/0.2)) ]
Now you can solve this equation using a scientific calculator:
ΔS_mix ≈ 3.96 J/(mol·K)
So the entropy of mixing of 80% N2 and 20% O2 in air is approximately 3.96 J/(mol·K).