Question

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.

Short answer

For 1 mol of Argon gas, Total energy=3.741KJ, Entropy=155J/K, Helmholtz energy=-42.75KJ , H =6.236 kJ and Gibbs free energy=-40.25KJ.

Step-by-step solution

Step 1. Given information

Number of moles, n=1.

Room temperature, T=300K.

Pressure P = 1atm = 1.01 x 10⁵Pa.

Step 2. Formula used

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:

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

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

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

Calculate Internal energy

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.

Calculate Entropy

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:

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

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

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

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

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

Now use these value in the original equation

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

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

Calculate Helmholtz free energy

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

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

Calculate Enthalpy

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

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

Calculate Gibb's free energy

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

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