91Ó°ÊÓ

Number of moles, n=1.

Room temperature, T=300K.

Pressure,$P=1atm=1.01×{10}^{5}Pa$.

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:

$$\begin{gathered} \text{S=Nk}\left[\text{ln}\left(\frac{\text{kT}}{\text{P}}{\left(\frac{2\mathrm{Ï€}\text{nKT}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right)\text{+}\frac{5}{2}\right] \\ \text{where} \\ \text{N=numberofmolecules} \\ \text{k=Boltzmannconstant} \\ \text{T=temperature} \\ \text{P=Pressure} \\ \text{m=mass} \\ \text{h=Plancksconstant} \end{gathered}$$

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

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

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

First calculate the value of $\left(\frac{\text{kT}}{P}{\left(\frac{2\mathrm{Ï€²Ô°­°Õ}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right)$.

Here,

$$\begin{gathered} {\text{°ì=1.38×10}}^{-23}{\text{JK}}^{-1} \\ \text{T=300K} \\ {\text{±Ê=1.013×10}}^5{\text{N/m}}^2 \\ {\text{³¾=66.8×10}}^{-27}\text{kg} \\ {\text{³ó=6.63×10}}^{-34}\text{´³Â·²õ} \end{gathered}$$

So,

$$\begin{gathered} \left(\frac{\text{kT}}{P}{\left(\frac{2\mathrm{Ï€²Ô°­°Õ}}{{\text{h}}^2}\right)}^{\frac{3}{2}}\right) \\ =\frac{\left(\left(1.38×{10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\right){\left(2\mathrm{Ï€}\left(66.8×{10}^{-27}\text{kg}\right)\left(1.38×{10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\right)}^{{\displaystyle \frac{3}{2}}}}{\left(1.013×{10}^5{\text{N/m}}^2\right){\left(6.63×{10}^{-34}\text{´³Â·²õ}\right)}^3} \\ =1.0149×{10}^7 \end{gathered}$$

The product of N and k results in the formation of a new constant known as gas constant, $\text{Nk=R}$.

Here $\text{¸é=8.314´³/°­Â·³¾´Ç±ô}$.

So,

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

Here,

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

So,

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

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

Here,

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

So,

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

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