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\n$\\mathrm{\\Delta S_{mix} = -nR\\sum_i x_i ln(x_i)}$ ,
\nwhere $\\mathrm{n}$ is the total number of moles of gas being mixed; $\\mathrm{R=8.314\\space J\\space K^{-1}\\space mol^{-1}}$ is the universal gas constant; and $\\mathrm{x_i}$ is the mole fraction of the $\\mathrm{i^{th}}$ gas in the mixture. The $\\mathrm{\\sum_i}$ indicates that we calculate its following term for each gas individually, then sum them all up.
\n\nTo use this formula for our situation, we must first calculate the mole fractions for each of our two gases. Recall that mole fraction is given by the number of moles of our specific component, divided by the total number of moles of all components:
\n$\\mathrm{x_i=\\frac{n_i}{n_{tot}}}$ .
\n\nFor our situation, we are given the number of moles of hydrogen and nitrogen, $\\mathrm{n_{_{H_2}}=\\var{nh}\\space mol}$ and $\\mathrm{n_{_{N_2}}=\\var{nn}\\space mol}$, so:
\n$\\mathrm{x_{_{H_2}}=\\frac{n_{_{H_2}}}{n_{_{H_2}}+n_{_{N_2}}}=\\frac{\\var{nh}}{\\var{nh}+\\var{nn}}=\\var{siground(xh,3)}}$ ,
\n$\\mathrm{x_{_{N_2}}=\\frac{n_{_{N_2}}}{n_{_{N_2}}+n_{_{H_2}}}=\\frac{\\var{nn}}{\\var{nn}+\\var{nh}}=\\var{siground(xn,3)}}$ .
\nAlso, we can sum our individual mole numbers to find the total number of moles in the mixture, $\\mathrm{n=\\var{n}\\space mol}$ .
\n\nWe can now use our original equation:
\n$\\mathrm{\\Delta S_{mix} = -nR\\sum_i x_i ln(x_i)=-nR[x_{_{H_2}}ln(x_{_{H_2}})+x_{_{N_2}}ln(x_{_{N_2}})]}$
\n$\\mathrm{\\Rightarrow\\space\\space\\space \\Delta S_{mix} =-\\var{n}\\times8.314\\times([\\var{siground(xh,3)}\\times ln(\\var{siground(xh,3)})]+[\\var{siground(xn,3)}\\times ln(\\var{siground(xn,3)})])=+\\var{sigformat(entro,5)}\\space J}$ .
\n\nTo quote our answer in the units asked for, we divide by one thousand to obtain:
\n$\\mathrm{\\Delta S_{mix}=+\\var{sigformat(e,3)}\\space kJ}$ .
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\nWhat is the entropy of mixing, $\\mathrm{\\Delta S_{mixing}}$?
\n[[0]] $\\mathrm{kJ\\space K^{-1}}$
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