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=300 K
Pressure, P=1 atm =1.01 x 10⁵Pa

$$\begin{gathered} \mathrm{G}=-42.75\mathrm{kJ}+(1\mathrm{mol})(8.315\mathrm{J}/\mathrm{mol}·\mathrm{K})(300\mathrm{K}) \\ \mathrm{G}=-40.25\mathrm{kJ} \end{gathered}$$Internal energy can be given by:

$\mathrm{U}=\frac{3}{2}\mathrm{nRT}$
n = Number of moles of gas
R = Gas constant
T = Temperature

From Sackur-Tetrode equation:

$\mathrm{S}=\mathrm{Nk}\left[\ln \left(\frac{\mathrm{kT}}{\mathrm{P}}{\left(\frac{2\pi \mathrm{mkT}}{{\mathrm{h}}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]$

Where, N is the number of molecules, k is the Boltzmann constant, T is the temperature, P is the pressure, m is the mass and h is the Planck's constant.

The enthalpy: $\mathrm{H}=\mathrm{U}+\mathrm{PV}$

Where, U is the internal energy, P is pressure and V is the volume.

Ideal gas equation is:

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{H}=\frac{3}{2}\mathrm{nRT}+\mathrm{nRT} \\ \mathrm{H}=\frac{5}{2}\mathrm{nRT} \end{gathered}$$

First, calculate the total internal energy using $\mathrm{U}=\frac{3}{2}\mathrm{nRT}$

Substitute the values and solve:

$$\begin{gathered} \mathrm{U}=\frac{3}{2}(1\mathrm{mol})(8.314\mathrm{J}/\mathrm{K}·\mathrm{mol})(300\mathrm{K}) \\ \mathrm{U}=3.741\mathrm{kJ} \end{gathered}$$

Now, use the Sackur-Tetrode equation, we get

$\mathrm{S}=\mathrm{Nk}\left[\ln \left(\frac{\mathrm{kT}}{\mathrm{P}}{\left(\frac{2\pi \mathrm{mkT}}{{\mathrm{h}}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]$

Calculate the internal expression

$\left(\frac{\mathrm{kT}}{\mathrm{P}}{\left(\frac{2\pi \mathrm{mkT}}{{\mathrm{h}}^2}\right)}^{\frac{3}{2}}\right)$

Substitute values

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

Now use this in the expression:

$S=Nk\left[\ln \left(\frac{kT}{P}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\pm \frac{5}{2}\right]$

The value of Boltzmann constant is: $1.38\times {10}^{-23}\mathrm{J}{\mathrm{K}}^{-1}$
The value of gas constant is: $8.314\mathrm{J}/\mathrm{K}·\mathrm{mol}$

Now substitute and solve the equation

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

Use the expression of Helmholtz free energy (F)
$\mathrm{F}=\mathrm{U}-\mathrm{TS}$
Using calculated values in this expression

$\mathrm{U}=3.741\mathrm{kJ},\mathrm{S}=155\mathrm{J}/\mathrm{K}and\mathrm{T}=300\mathrm{K}$

$$\begin{gathered} \mathrm{F}=3741\mathrm{J}-\left(\right(300\mathrm{K}\left)\right(155\mathrm{J}/\mathrm{K}\left)\right) \\ \mathrm{F}=-42759\mathrm{J}\left(\frac{1\mathrm{kJ}}{1\mathrm{kJ}}\right) \\ \mathrm{F}=-42.75\mathrm{kJ} \end{gathered}$$

Now, from the thermodynamics:
$\mathrm{H}=\mathrm{U}+\mathrm{PV}$

From the ideal gas equation:
$\mathrm{PV}=\mathrm{nRT}$

From above two we get

$$\begin{gathered} \mathrm{H}=\frac{3}{2}\mathrm{nRT}+\mathrm{nRT} \\ \mathrm{H}=\frac{5}{2}\mathrm{nRT} \\ \mathrm{H}=\frac{5}{2}\left(1\right)(8.314)\left(300\right) \\ \mathrm{H}=6.236\mathrm{kJ} \end{gathered}$$

Gibbs free energy is calculated as

$\mathrm{G}=\mathrm{F}+\mathrm{PV}$

Susbtitute nRT for ideal gas, we get

$\mathrm{G}=\mathrm{F}+\mathrm{nRT}$

Calculate by substituting given values the given values

n =1 mol

Gas constant = 8.314 J/K mol

T = 300 K

F= -42.75 kJ

$$\begin{gathered} \mathrm{G}=-42.75\mathrm{kJ}+(1\mathrm{mol})(8.315\mathrm{J}/\mathrm{mol}·\mathrm{K})(300\mathrm{K}) \\ \mathrm{G}=-40.25\mathrm{kJ} \end{gathered}$$