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The entropy substance as

$S=k\ln 鈦�\left(\mathrm{惟}\right)$

where $惟$is the number of microstates accessible substance.

For $3-d$ideal gas, the formula by

$S=Nk\left[\ln 鈦�\left(\frac{V}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]$

Where,$V$represents volume, $U$represents energy, $\text{N}$represents the number of molecules, $m$represents the mass of a single molecule, and $h$represents Planck's constant. Despite the fact that this formula appears to be a little difficult, we can see that raising either of $V,U$, or $N$increases entropy. The gas expands quasistatically in an isothermal expansion, keeping its temperature constant. This indicates that $U=3/2NkT$remains constant as well, leaving just the volume to vary. Because the gas is doing work $W$by expanding, the energy for the work must come from a source of heat $Q$introduced into the gas to keep the temperature constant. The first law of thermodynamics states:

$Q=\mathrm{螖}U+W$

But $鈭�U=0$and $W={鈭�}_{V_i}^{V_f}鈥�PdV$we get

$Q={鈭�}_{V_i}^{V_f}鈥�PdV$

From ideal-gas law,

role="math" localid="1650264837406" $\begin{array}{l}P=\frac{NkT}{V},\\ {鈭�}_{V_i}^{V_f}鈥�\frac{dV}{V}dV\\ Q=NkT\left[NkT[\ln 鈦�(V){]}_{V_i}^{V_f}\right.\\ Q=NkT\ln 鈦�\left[\frac{V_f}{V_i}\right]\end{array}$

Substituting the constant $U$and $N$we get

$\begin{array}{l}S=Nk\left[\ln 鈦�\left(\frac{V}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ S=Nk\left[\ln 鈦�\left(V\right)+\ln 鈦�\left(\frac{1}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ \ln 鈦�\left(\frac{1}{N}{\left(\frac{4m蟺U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)\end{array}$

The change in entropy process where volume only changes by

$\begin{array}{l}\mathrm{螖}S=S_f鈭�S_i=Nk\left(\ln 鈦�\left(V_f\right)鈭�\ln 鈦�\left(V_i\right)\right)\\ \mathrm{螖}S=Nk\ln 鈦�\left[\frac{V_f}{V_i}\right]\\ 鈫�\mathrm{螖}S=\frac{Q}{T}\end{array}$

The reason is valid where expanding gas is work,so the heat input provide energy for work.