Expand 1.00 molof a monatomic gas initially at 5.00kPaand 600 Kfrom initial volume${\mathbf{V}}_{\mathbf{i}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$to final volume${\mathbf{V}}_{\mathbf{f}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$. At any instant during the expansion, the pressure pand volume Vof the gas are related by$\mathit{p}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{exp}\left[\left(V_i-V\right)Ia\right]$, with pin kilopascals,${\mathit{V}}_{\mathbf{i}}$ and Vin cubic meters, and$\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$. (a) What is the final pressure and (b) what is the final temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is$\mathbf{∆}\mathit{S}$for the expansion? (Hint: Use two simple reversible processes to find$\mathbf{∆}\mathit{S}$.)

Number of moles of monatomic gas, n = 1 mol

Initial pressure,${\mathrm{p}}_{\mathrm{i}}=5\mathrm{kPa}$

Initial temperature,${\mathrm{T}}_{\mathrm{i}}=600\mathrm{K}$

Initial volume,${\mathrm{V}}_{\mathrm{i}}=1.00{\mathrm{m}}^3$

Final volume,${\mathrm{V}}_{\mathrm{f}}=2.00{\mathrm{m}}^3$

Value of a,$\mathrm{a}=1.00{\mathrm{m}}^3$

Use two reversible processes to find$∆S$

We use the given condition to calculate the final pressure. Using the ratio form of the ideal gas equation, we can find the final temperature. We can use the formula of work done and entropy change of ideal gas to find the corresponding quantities.

Formulae:

The ideal gas equation,

$\mathrm{pV}=\mathrm{nRT}$ (1)

The work done for an isobaric process,

$\mathrm{W}={\int }_{\mathrm{i}}^{\mathrm{f}}\mathrm{pdV}$ (2)

The entropy change of the gas,

$∆\mathrm{S}=\mathrm{nRIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)+{\mathrm{nC}}_{\mathrm{v}}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$ (3)

The final pressure of the gas is given as:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{f}}=\left(5\mathrm{kPa}\right)\exp \left[\left({\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}}/\mathrm{a}\right)\right] \\ =\left(5\mathrm{kPa}\right)\exp \left[\left(2.00{\mathrm{m}}^3-1.00{\mathrm{m}}^3\right)/1.00{\mathrm{m}}^3\right] \\ =\left(5\mathrm{kPa}\right){\mathrm{e}}^{-1} \\ =\left(5\mathrm{kPa}\right)\times 0.3678 \\ =1.84\mathrm{kPa} \end{gathered}$$

Hence, the value of the final pressure is 1.84 kPa

Taking the ratio of two equations for the initial and final condition of the gas using equation (1), we get that the final temperature is given as:

$$\begin{gathered} \frac{{\mathrm{p}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{p}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}}=\frac{{\mathrm{nRT}}_{\mathrm{f}}}{{\mathrm{nRT}}_{\mathrm{i}}} \\ \frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}=\frac{{\mathrm{p}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{p}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}} \\ {\mathrm{T}}_{\mathrm{f}}=\left(\frac{{\mathrm{p}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{p}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}}\right){\mathrm{T}}_{\mathrm{i}} \\ {\mathrm{T}}_{\mathrm{f}}=600\mathrm{K}\frac{\left(1.84\mathrm{kPa}\right)\left(2.00{\mathrm{m}}^3\right)}{\left(5\mathrm{kPa}\right)\left(1.00{\mathrm{m}}^3\right)} \\ {\mathrm{T}}_{\mathrm{f}}=441\mathrm{K} \end{gathered}$$

Hence, the value of the final temperature is 441 K

Substituting the given value of pressure in equation (2), we get that the work done by the gas is given as:

$$\begin{gathered} \mathrm{W}=5\mathrm{kPa}{\int }_{{\mathrm{v}}_{\mathrm{i}}}^{{\mathrm{v}}_{\mathrm{f}}}{\mathrm{e}}^{\left({\mathrm{V}}_{\mathrm{i}}-\mathrm{V}\right)/\mathrm{a}}\mathrm{dV} \\ =\left(5\mathrm{kPa}\right){\mathrm{e}}^{{\mathrm{V}}_{\mathrm{i}}/\mathrm{a}}{\int }_{{\mathrm{v}}_{\mathrm{i}}}^{{\mathrm{v}}_{\mathrm{f}}}{\mathrm{e}}^{\left(-\mathrm{V}\right)/\mathrm{a}}\mathrm{dV} \\ =\left(5\mathrm{kPa}\right){\mathrm{e}}^{{\mathrm{V}}_{\mathrm{i}}/\mathrm{a}}{\left[-{\mathrm{ae}}^{\frac{\mathrm{V}}{\mathrm{a}}}\right]}_{{\mathrm{V}}_{\mathrm{i}}=1.00{\mathrm{m}}^3}^{{\mathrm{V}}_{\mathrm{f}}=2.00{\mathrm{m}}^3} \\ =\left(5\mathrm{kPa}\right){\mathrm{e}}^{\frac{1.00{\mathrm{m}}^3}{1.00{\mathrm{m}}^3}}\left(-1.00{\mathrm{m}}^3\right)\left[{\mathrm{e}}^{-2.00{\mathrm{m}}^3/1.00{\mathrm{m}}^3-{\mathrm{e}}^{-1.00{\mathrm{m}}^3/1.00{\mathrm{m}}^3}}\right] \\ =\left(5\mathrm{kPa}\right){\mathrm{e}}^1\left(-1.00{\mathrm{m}}^3\right)\left[{\mathrm{e}}^{-2}-{\mathrm{e}}^{-1}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{W}=\left(5\mathrm{kPa}\right){\mathrm{e}}^1\left(1.00{\mathrm{m}}^3\right)\left[{\mathrm{e}}^{-1}-{\mathrm{e}}^{-2}\right] \\ =\left(5\mathrm{kPa}\right)\left(2.7182\right)\left(1.00{\mathrm{m}}^3\right)\left[0.3678-0.1353\right] \\ =\left(5\mathrm{kPa}\right)\left(2.7182\right)\left(1.00{\mathrm{m}}^3\right)\left[0.2324\right] \\ =3.16\mathrm{kJ} \end{gathered}$$

Hence, the value of the work done by the gas is 3.16 kJ

Entropy change for an ideal monatomic gas using equation (3) and the given values are given as follows:

$$\begin{gathered} ∆\mathrm{S}=\mathrm{nRIn}\left(\frac{2.00{\mathrm{m}}^3}{1.00{\mathrm{m}}^3}\right)+\mathrm{n}\frac{3}{2}\mathrm{RIn}\left(2{\mathrm{e}}^{-1}\right) \\ =\frac{{\mathrm{p}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}}{{\mathrm{T}}_{\mathrm{i}}}\left[\mathrm{In}\left(2\right)+\frac{3}{2}\mathrm{In}2+\frac{3}{2}{\mathrm{Ine}}^{-1}\right] \\ =\frac{\left(5000\mathrm{Pa}\right)\left(1.00{\mathrm{m}}^3\right)}{600\mathrm{K}}\left[\mathrm{In}\left(2\right)+\frac{3}{2}\mathrm{In}2+\frac{3}{2}{\mathrm{Ine}}^{-1}\right] \\ =\frac{\left(5000\mathrm{Pa}\right)\left(1.00{\mathrm{m}}^3\right)}{600\mathrm{K}}\left[0.2328\right] \\ =1.94\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of the gas is 1.94 J/k