Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.

Number of moles, n=1.

Room temperature, T=300K.

Pressure,$P=1atm=1.01\times {10}^{5}Pa$.

Internal Energy,

$$\begin{gathered} \text{U=}\frac{3}{2}\text{nRT} \\ where \\ \text{n=Numberofmolesofgas} \\ R=\text{Gasconstant} \\ \text{T=temperature} \end{gathered}$$

Sackur-Tetrode equation:

$$\begin{gathered} \text{S=Nk}\left[\text{ln}\left(\frac{\text{kT}}{\text{P}}{\left(\frac{2\mathrm{\pi }\text{nKT}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right)\text{+}\frac{5}{2}\right] \\ \text{where} \\ \text{N=numberofmolecules} \\ \text{k=Boltzmannconstant} \\ \text{T=temperature} \\ \text{P=Pressure} \\ \text{m=mass} \\ \text{h=Plancksconstant} \end{gathered}$$

Enthalpy is

$$\begin{gathered} \text{H=U+PV} \\ where \\ \text{U=internalenergy} \\ \text{P=pressure} \\ \text{V=Volume} \end{gathered}$$

Ideal gas equation is

$\text{PV=nRT}$.

So,$$\begin{gathered} \text{H=}\frac{3}{2}\text{nRT+nRT} \\ =\frac{5}{2}\text{nRT} \end{gathered}$$

$$\begin{gathered} \text{U=}\frac{3}{2}\text{nRT} \\ =\frac{3}{2}\left(1mol\right)\left(8.314J/K.mol\right)\left(300K\right) \\ =3.741KJ \end{gathered}$$

So, internal energy for 1 mole of gas,$\text{U=3.741KJ}$.

First calculate the value of $\left(\frac{\text{kT}}{P}{\left(\frac{2\mathrm{\pi nKT}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right)$.

Here,

$$\begin{gathered} {\text{k=1.38×10}}^{-23}{\text{JK}}^{-1} \\ \text{T=300K} \\ {\text{P=1.013×10}}^5{\text{N/m}}^2 \\ {\text{m=66.8×10}}^{-27}\text{kg} \\ {\text{h=6.63×10}}^{-34}\text{J·s} \end{gathered}$$

So,

$$\begin{gathered} \left(\frac{\text{kT}}{P}{\left(\frac{2\mathrm{\pi nKT}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right) \\ =\frac{\left(\left(1.38\times {10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\right){\left(2\mathrm{\pi }\left(66.8\times {10}^{-27}\text{kg}\right)\left(1.38\times {10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\right)}^{{\displaystyle \frac{3}{2}}}}{\left(1.013\times {10}^5{\text{N/m}}^2\right){\left(6.63\times {10}^{-34}\text{J·s}\right)}^3} \\ =1.0149\times {10}^7 \end{gathered}$$

The product of N and k results in the formation of a new constant known as gas constant, $\text{Nk=R}$.

Here $\text{R=8.314J/K·mol}$.

So,

$$\begin{gathered} \text{S=8.314}\left[\ln \left(1.0149\times {10}^7\right)\text{+}\frac{5}{2}\right] \\ \text{=}\left(8.314\right)\left(16.13+2.5\right) \\ \text{=134.1+20.78} \\ =155\text{J/K} \end{gathered}$$

Here,

$$\begin{gathered} \text{U=3.741KJ} \\ \text{S=155J/K} \\ \text{T=300K} \end{gathered}$$

So,

role="math" localid="1647277895317" $$\begin{gathered} \text{F=U-TS} \\ =3741\text{J-}\left(300\text{K}\right)\left(155\text{J/K}\right) \\ \text{=-42759J} \\ =-42.75\text{KJ} \end{gathered}$$

$$\begin{gathered} \text{H=}\frac{5}{2}\text{nRT} \\ =\frac{5}{2}\left(1\right)\left(8.314\right)\left(300\right) \\ \text{=6.236KJ} \end{gathered}$$

Here,

$$\begin{gathered} \text{F=-42.75KJ} \\ \text{n=1mol} \\ \text{R=8.314J/K} \\ \text{T=300K} \end{gathered}$$

So,

$$\begin{gathered} \text{G=F+nRT} \\ =-42.75\text{KJ+}\left(1\text{mol}\right)\left(8.314\text{J/mol}\right)\left(300\text{K}\right) \\ \text{=-40.25KJ} \end{gathered}$$

For 1 mol of Argon gas, Total energy=3.741KJ, Entropy=155J/K, Helmholtz energy=-42.75KJ and Gibbs free energy=-40.25KJ.