Suppose 2.00 molof a diatomic gas is taken reversibly around the cycle shown in the T-S diagram of Fig. 20-35, where${\mathbf{S}}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$and${\mathbf{S}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$. The molecules do not rotate or oscillate. What is the energy transferred as heat Qfor (a) path$\mathbf{1}\mathbf{\to }\mathbf{2}$, (b) path$\mathbf{2}\mathbf{\to }\mathbf{3}$, and (c) the full cycle? (d) What is the worxk Wfor the isothermal process? The volume V₁ in state 1 is$\mathbf{0}\mathbf{.}\mathbf{200}{\mathbf{m}}^{\mathbf{3}}$. What is the volume in (e) State 2 and (f) state 3? What is the change$\mathbf{∆}{\mathit{E}}_{\mathbf{i}\mathbf{n}\mathbf{t}}$for (g) path$\mathbf{1}\mathbf{\to }\mathbf{2}$, (h) path$\mathbf{2}\mathbf{\to }\mathbf{3}$, and (i) the full cycle? (Hint:(h) can be done with one or two lines of calculation using Module 19-7 or with a page of calculation using Module 19-9.) (j) What is the work Wfor the adiabatic process?

The number of moles is$\mathrm{n}=2.00\mathrm{mol}$

The entropies${\mathrm{S}}_1=6.00\mathrm{J}/\mathrm{K}$and${\mathrm{S}}_2=8.00\mathrm{J}/\mathrm{K}$.

The molecules do not rotate or oscillate.

The temperature,${\mathrm{T}}_2=350\mathrm{K}$.

The temperature,${\mathrm{T}}_3=300\mathrm{K}$.

The volume, $V_1=0.200{\mathrm{m}}^3$.

By using Equation 21-1 for the change in the entropy, we can find the energy transferred as heat. For an ideal gas, in an isothermal process; hence, we can find the work. By using the equation for the isothermal work from Equation 19-14, we can find the volume. By using this value in Equations 19-56, we can find the value of volume. For an isothermal process, the change in the internal energy is zero. Using this, we can find the value of the process. By using Equations 19-45, we can find the value of the process. As the cycle in the T-S diagram is a closed cycle, we can find the net change of internal energy for the entire cycle. Finally, by using the formula for the adiabatic process, we can find the work done.

Formulae:

The change in the entropy of the cycle, $∆\mathrm{S}={\int }_{\mathrm{i}}^{\mathrm{f}}\frac{\mathrm{dQ}}{\mathrm{T}}$ (1)

The area of the rectangle, role="math" localid="1661747902348" $\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{rectangle}=\mathrm{width}\times \mathrm{height}$ (2)

The area of the triangle, $\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$ (3)

The isothermal work from Equation 19-14,$\mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)$ (4)

where, n is the number of moles, R is the gas constant, T is the temperature,${\mathrm{V}}_{\mathrm{f}}$is the final volume, and${\mathrm{V}}_{\mathrm{i}}$is the initial volume.

Using adiabatic relation of volume and triangle usingEquation 19-56.

${\mathrm{T}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}^{\mathrm{y}-1}={\mathrm{T}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}^{\mathrm{y}-1}$ (5)

From Equation 19-45, the change in the internal energy,$∆{\mathrm{E}}_{\mathrm{int}}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}}\right)$ (6)

The first law of thermodynamics, $\mathrm{Q}=\mathrm{W}+∆{\mathrm{E}}_{\mathrm{int}}$ (7)

From the equation of entropy that is equation (1), we can get that energy transfer as heat is given as:

$\mathrm{Q}=\int \mathrm{TdS}$

which corresponds to the area under the curve intheT-S diagram. The area of a rectangle is given by equation (2) we find the heat as follows:

$$\begin{gathered} Q_{1\to 2}=2.00\mathrm{J}/\mathrm{K}\times 350\mathrm{K} \\ =700\mathrm{J} \end{gathered}$$

Hence, the value of energy as heat is 700 J

With no area under the curve for process,$2\to 3$, we find

$Q_{2\to 3}=0\mathrm{J}$

Hence, the value of energy as heat is 0 J .

For the complete cycle, the net energy transferred as heat should be the area inside the figure, which is the area of a triangle that is given as equation (3) as:

$$\begin{gathered} Q_{net}=\frac{1}{2}\times 2.00\mathrm{J}/\mathrm{K}\times 50\mathrm{K} \\ =50\mathrm{K} \end{gathered}$$

Hence, the value of the energy for full cycle is$50\mathrm{K}$

We know that for an ideal gas, the change in internal energy is$∆{\mathrm{E}}_{\mathrm{int}}=0$in an isothermal process. Thus, the work done during the isothermal process is given as:

$$\begin{gathered} {\mathrm{W}}_{1\to 2}={\mathrm{Q}}_{1\to 2} \\ =700\mathrm{J} \end{gathered}$$

Here, the positive sign indicates expansion. Hence, the value of work is 700 J

Using equation (4), we can get the volume of stage 2 as given:

$$\begin{gathered} {\mathrm{V}}_2={\mathrm{V}}_1\exp \left(\frac{{\mathrm{W}}_{1\to 2}}{\mathrm{nRT}}\right) \\ =0.200{\mathrm{m}}^3\times \exp \left(\frac{700\mathrm{J}}{2.00\mathrm{mol}\times 8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 350\mathrm{K}}\right) \\ =0.200{\mathrm{m}}^3\times {\mathrm{e}}^{\left(0.12\right)} \\ =0.226{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of volume for stage 2 is$0.226{\mathrm{m}}^3$

Processis adiabatic. Using the given values in equation (5),$\lambda =\frac{5}{3}$(Only translational degree of freedom is relevant here), we have the volume of stage 3 as given:

$$\begin{gathered} {\mathrm{V}}_3^{\mathrm{y}-1}=\frac{{\mathrm{T}}_2{\mathrm{V}}_2^{\mathrm{y}-1}}{{\mathrm{T}}_3} \\ {\mathrm{V}}_3^{\left(\frac{5}{3}-1\right)}=\frac{350\mathrm{K}\times {\left(0.226{\mathrm{m}}^3\right)}^{\left({\displaystyle \frac{5}{3}}-1\right)}}{300\mathrm{K}} \\ {\mathrm{V}}_3={\left(0.43286\right)}^{\left(\frac{3}{2}\right)}{\mathrm{m}}^3 \\ {\mathrm{V}}_3=0.2848{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the volume of stage 3 is$0.2848{\mathrm{m}}^3$

As stated in part d), since we are dealing with an ideal gas, the change in the internal energy $∆{\mathrm{E}}_{1\to 2}$for process$1\to 2$is 0

By using the given values in equation (6), we get the change in the internal energy$∆{\mathrm{E}}_{2\to 3}$for process$2\to 3$ as given: (where,${\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\mathrm{R}$is molar specific heat at constant volume.)

$$\begin{gathered} ∆{\mathrm{E}}_{2\to 3}=2.00\mathrm{mol}\times \frac{3}{2}\times 8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}\times \left(300\mathrm{K}-350\mathrm{K}\right) \\ =-1.25\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the internal energy is$-1.25\times {10}^3\mathrm{J}$

Since the cycle in the T-S diagram is a closed cycle, the net change of internal energy for the entire cycle is zero, that is,$∆{\mathrm{E}}_{\mathrm{int}}=0\mathrm{J}$

Hence, the value of change in internal energy is zero ( 0 J ).

For adiabatic process $2\to 3$, the energy transferred as heat is zero. Thus, using equation (7), the work done in the adiabatic process is given as:

$$\begin{gathered} {\mathrm{W}}_{2\to 3}=-∆{\mathrm{E}}_{2\to 3} \\ =-\left(-1.25\times {10}^3\mathrm{J}\right) \\ =1.25\times {10}^3\mathrm{J} \end{gathered}$$

Here, the positive sign indicates expansion. Hence, the value of the work is$1.25\times {10}^3\mathrm{J}$