Question

Calculate the entropy of mixing of 2 moles of H₂ and 3 moles of O₂ assuming that no chemical reaction occurs.

Answer 1

Let's start with the entropy of the ideal gas (note, I shall use the system where $k = 1$ so the temperature is measured in the energy units)

$$S = N\ln \frac{{eV}}{N} - Nf'(T)$$

where $f(T)$ is some function of the temperature (explicitely say $f(T) = - T\ln [{(\frac{{mT}}{{2\pi {\hbar ^2}}})^{3/2}}\sum\limits_k {{e^{ - \frac{{{\varepsilon _k}}}{T}}}} ]$ , but we don't need it).

Then the entropy before mixing is the sum of the entopies of the two gases

$${S_{{\text{before}}}} = {N_{{H_2}}}\ln \frac{{e{V_{{H_2}}}}}{{{N_{{H_2}}}}} - {N_{{H_2}}}{f_{{H_2}}}'(T) + {N_{{O_2}}}\ln \frac{{e{V_{{O_2}}}}}{{{N_{{O_2}}}}} - {N_{{O_2}}}{f_{{O_2}}}'(T)$$

The entropy before mixing again can be considered as the enropy of the two independent gases that but in volume ${V_{{H_2}}} + {V_{{O_2}}}$

$${S_{{\text{after}}}} = {N_{{H_2}}}\ln \frac{{e({V_{{H_2}}} + {V_{{O_2}}})}}{{{N_{{H_2}}}}} - {N_{{H_2}}}{f_{{H_2}}}'(T) + {N_{{O_2}}}\ln \frac{{e({V_{{H_2}}} + {V_{{O_2}}})}}{{{N_{{O_2}}}}} - {N_{{O_2}}}{f_{{O_2}}}'(T)$$

Then $\Delta S = {S_{{\text{after}}}} - {S_{{\text{before}}}}$

$$\Delta S = {N_{{H_2}}}\ln \frac{{e({V_{{H_2}}} + {V_{{O_2}}})}}{{{N_{{H_2}}}}} - {N_{{H_2}}}\ln \frac{{e{V_{{H_2}}}}}{{{N_{{H_2}}}}} + {N_{{O_2}}}\ln \frac{{e({V_{{H_2}}} + {V_{{O_2}}})}}{{{N_{{O_2}}}}} - {N_{{O_2}}}\ln \frac{{e{V_{{O_2}}}}}{{{N_{{O_2}}}}}$$

Using the properties of the logarithms

$$\Delta S = {N_{{H_2}}}\ln \frac{{{V_{{H_2}}} + {V_{{O_2}}}}}{{{V_{{H_2}}}}} + {N_{{O_2}}}\ln \frac{{{V_{{H_2}}} + {V_{{O_2}}}}}{{{V_{{O_2}}}}}$$

Now using the equation of the state $pV = NT$

$$\Delta S = {N_{{H_2}}}\ln \frac{{{N_{{H_2}}} + {N_{{O_2}}}}}{{{N_{{H_2}}}}} + {N_{{O_2}}}\ln \frac{{{N_{{H_2}}} + {N_{{O_2}}}}}{{{N_{{O_2}}}}}$$

Using that $N = \nu {N_A}$

$$\Delta S = {N_A}[{\nu _{{H_2}}}\ln \frac{{{\nu _{{H_2}}} + {\nu _{{O_2}}}}}{{{\nu _{{H_2}}}}} + {\nu _{{O_2}}}\ln \frac{{{\nu _{{H_2}}} + {\nu _{{O_2}}}}}{{{\nu _{{O_2}}}}}]$$

If you want to obtain the result in $[\frac{{\text{J}}}{{\text{K}}}]$ you need to multiply in by the Boltzmann constant, $k$ and we can use that ${N_A}k = R$ - universal gas constant. Thus in classical units

$$\Delta S = R[{\nu _{{H_2}}}\ln \frac{{{\nu _{{H_2}}} + {\nu _{{O_2}}}}}{{{\nu _{{H_2}}}}} + {\nu _{{O_2}}}\ln \frac{{{\nu _{{H_2}}} + {\nu _{{O_2}}}}}{{{\nu _{{O_2}}}}}]$$

$$\Delta S \approx 8.31[\frac{{\text{J}}}{{{\text{K}} \cdot {\text{mol}}}}] \cdot (2[{\text{mol}}] \cdot \ln \frac{{2 + 3}}{2} + 3[{\text{mol}}] \cdot \ln \frac{{2 + 3}}{3})[\frac{{\text{J}}}{{\text{K}}}] \approx 27.96[\frac{{\text{J}}}{{\text{K}}}]$$