91Ó°ÊÓ

Number of moles, n=1.

Room temperature, T=300K.

Pressure P = 1atm = 1.01 x 10⁵Pa.

Internal energy is calculated by

U = 3/2 (nRT)

Where

n= Numbers of moles of gas

R = Gas constant

T = Temperature

Sackur-Tetrode equation is:

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

Where,

N = number of molecules,

k= Boltzmann constant,

T= temperature,

P= pressure,

m= mass and

h= Planck's constant.

And the enthalpy is calculated as :

$\mathrm{H}=\mathrm{U}+\mathrm{PV}$

Where,

U= internal energy,

P= pressure and

V= 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 Internal energy using

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

Substitute the values and calculate:

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

So internal energy is 3.741 kJ for one mole of gas.

To calculate entropy use the expression of Sackur-Tetrode equation which is related to the internal energy of the molecules to the volume and mass of the gas moelcules:

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

For simplified calculation first calculate the value of expression inside the log.

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

Substitute the values

k = 1.38 x 10^-23 J/K
T= 300 K
P= 1.03 x10⁵N/m²
m=66.8 x 10^-27kg
h=6.63 x 10^-34 J.s

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

Now use these value in the original equation

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

N = number of molecules and k= boltzmann constant.

The product of N and k is gas constant.
So

N k=R
The value of Boltzmann constant is: 1.38 x 10^-23 J/K
The value of gas constant is: 8.314 J / K.mol

Now find entropy using these value in original equation

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

Helmholtz free energy can be calculated using below equation

F = U -TS

Use our calculated values

U = 3.741KJ

S = 155J/K,

T = 300 K

Substitute in equation, we get

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

Enthalpy can be calculated by using ideal gas equation

From the thermodynamics:
H = U + PV

As know from the ideal gas equation:
PV=nRT
Substitute this, we get

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

Gibb's free energy from therodynamics can be calculated as

G = F + PV

Substitute from ideal gas equation PV= nRT

So

G = F + nRT

Substitute the values

n = 1 mol, R = 8.314 J/K . mol and forT= 300K ,

F = - 42.75kJ

we get

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