91Ó°ÊÓ

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:

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

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

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

Calculate the internal expression

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

Substitute values

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

Now use this in the expression:

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

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

Now substitute and solve the equation

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

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

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