Question

Show that during the quasistatic isothermal expansion of a monatomic ideal gas, the change in entropy is related to the heat input $Q$by the simple formula

$∆s=\frac{Q}{T}$

In the following chapter I'll prove that this formula is valid for any quasistatic process. Show, however, that it is not valid for the free expansion process described above.

Short answer

The Formula of monatomic ideal gas of during quasistatic isothermal expansion is proved as change in entropy,$∆s=\frac{Q}{T}$and the reason is valid where expanding gas is work,so the heat input provide energy for work.

Step-by-step solution

Step: 1 Sackur-Tetrode equation:

The entropy substance as

$S=k\ln \left(\mathrm{\Omega }\right)$

where $\Omega$is the number of microstates accessible substance.

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

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

Step; 2 Equating amount of energy:

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:

$Q=\mathrm{\Delta }U+W$

But $∆U=0$and $W={\int }_{V_i}^{V_f} PdV$we get

$Q={\int }_{V_i}^{V_f} PdV$

From ideal-gas law,

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

Step: 3 Change in entropy:

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

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

The change in entropy process where volume only changes by

$\begin{array}{l}\mathrm{\Delta }S=S_f-S_i=Nk\left(\ln \left(V_f\right)-\ln \left(V_i\right)\right)\\ \mathrm{\Delta }S=Nk\ln \left[\frac{V_f}{V_i}\right]\\ \to \mathrm{\Delta }S=\frac{Q}{T}\end{array}$

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