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 and Gibbs free energy=-40.25KJ.

Step-by-step solution

Step 1. Given information

Number of moles, n=1.

Room temperature, T=300K.

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

Step 2. Formula used

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

Step 3. Calculation of Total internal energy

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

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

Step 4. Calculation of Enthalpy

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,

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

Step 5. Calculation of Helmholtz free energy

Here,

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

So,

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

Step 6. Calculation of Enthalpy

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

Step 7. Calculation of Gibbs free energy

Here,

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

So,

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

Step 8. Conclusion

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