# Calculate the entropy of 1 mole of helium at standard temperature pressure, assume that all the atoms are indistinguishable.

Problem | Second Law of Thermodynamics

## Problem – Calculate the entropy of 1 mole of helium at standard temperature pressure, assume that all the atoms are indistinguishable.

Solution

Using Sackur-Tetrode Equation,

The total entropy of an ideal gas,

S=N$\kappa\left[ln(\frac{V}{N}(\frac{4\pi mU}{3Nh^{2}})^{3/2})+\frac{5}{2}\right]$ — Eq. (1)

For standard temperature and pressure,

T=298 K, P=1.01*10^{5} Pa

And,

For 1 mole of helium,

mass of 1 atom of helium (m) = $\frac{0.004kg}{6.022*10^{23}}$

Number of particles, N=6.022 * $10^{23}$

U = $\frac{3}{2}nRT$

= $\frac{3}{2}(1)RT$

Now,

Using ideal gas equation

PV = nRT

PV=(1)RT

V=$\frac{RT}{P}$

Substituting all values in Eq. (1),

S=N$\kappa\left[ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right]$

We get,

S=N$\kappa\left[ln\left(\frac{RT}{PN}\left(\frac{4\pi m(\frac{3}{2}RT)}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right]$

Or,

S=$\left(6.022*10^{23}\right)\left(1.38*10^{23}\right)\left[ln\left(\frac{(8.314)(298)}{(1.01*10^{5})\left(6.022*10^{23}\right)}\left(\frac{4\pi\left(\frac{0.004}{6.022*10^{23}}\right)(\frac{3}{2}(8.314)(298))}{3\left(6.022*10^{23}\right)\left(6.6*10^{-34}\right)^{2}}\right)^{3/2}\right)+\frac{5}{2}\right]$

By using,

We get,

S=$\left(6.022*10^{23}\right)\left(1.38*10^{23}\right)\left[ln(318391.823 +\frac{5}{2})\right]$

S=$\left(6.022*10^{23}\right)\left(1.38*10^{23}\right)\left[15.171\right]$

S=126.076 J/K