91Ó°ÊÓ

Volume${\mathrm{V}}_{23}=3{\mathrm{V}}_1$

n is the moles of an ideal diatomic gas taken through the cycle with the molecules rotating but not oscillating.

We use various adiabatic relations and the ideal gas equation to find the ratios of temperature and pressure. Using the formula of work done, entropy change, and the first law of thermodynamics, we find the different ratios.

Formulae:

The ideal gas equation

$pV=nRT$ (1)

The adiabatic relations of pressure, volume, temperature,

$$\begin{gathered} {\mathrm{pV}}^{\mathrm{y}}=\mathrm{constant} \\ {\mathrm{TV}}^{\mathrm{y}-1}=\mathrm{constant} \end{gathered}$$ (2)

The work done in an adiabatic process,

$\mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$ (3)

The first law of thermodynamics,

$Q=\mathrm{Δ}{E}_{int}+W$ (4)

The entropy change of gas,

$∆S=\frac{Q}{T}$ (5)

The change in internal energy of the gas,

$\mathrm{Δ}{E}_{int}=n{C}_V\mathrm{Δ}T$ (6)

The entropy change of gas,

$∆\mathrm{S}={\mathrm{nC}}_{\mathrm{V}}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)+\mathrm{nRIn}\left(\frac{{\mathrm{V}}_{\mathrm{final}}}{{\mathrm{V}}_{\mathrm{initial}}}\right)$ (7)

For path$1→2$the process is isothermal.

So${\mathrm{T}}_2={\mathrm{T}}_1$

Using equation (1) for paths$1→2$, at a constant temperature, we can get that

$$\begin{gathered} \frac{p_2}{p_1}=\frac{V_1}{V_2} \\ =\frac{1}{3} \\ =0.333 \end{gathered}$$

Hence, the ratio is$p_2/p_1=0.333$

The path $3→1$is an adiabatic process

Using the adiabatic relation of pressure-volume of equation (2) for paths$3→1$, we can get that

$$\begin{gathered} {\mathrm{p}}_3{\mathrm{V}}_3^{\mathrm{y}}={\mathrm{p}}_1{\mathrm{V}}_1^{\mathrm{y}} \\ \frac{{\mathrm{p}}_3}{{\mathrm{p}}_1}=\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_3}\right) \\ \frac{{\mathrm{p}}_3}{{\mathrm{p}}_1}={\left(\frac{1}{3}\right)}^{\mathrm{y}} \end{gathered}$$

For diatomic gas,$y=\frac{7}{5}$or 1.4

$$\begin{gathered} \frac{p_3}{p_1}={\left(\frac{1}{3}\right)}^{1.4} \\ =0.215 \end{gathered}$$

Hence, the value of the ratio is$p_3/p_1=0.215$

Using the adiabatic relation of temperature and volume of equation (2)for path$3→1$, we can get that

$$\begin{gathered} T_3{\mathrm{V}}_3^{\mathrm{y}-1}=T_1{\mathrm{V}}_1^{\mathrm{y}-1} \\ \frac{{\mathrm{T}}_3}{T_1}={\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_3}\right)}^{y-1} \\ \frac{{\mathrm{T}}_3}{{\mathrm{T}}_1}={\left(\frac{1}{3}\right)}^{1.4-1} \\ \frac{{\mathrm{T}}_3}{{\mathrm{T}}_1}={\left(\frac{1}{3}\right)}^{0.4} \\ \frac{{\mathrm{T}}_3}{{\mathrm{T}}_1}=0.644 \end{gathered}$$

Hence, the value of the ratio is$T_3/T_1=0.664$

Work done for an isothermal process for the path$1→2$is given using equation (3) as:

$$\begin{gathered} \mathrm{W}={\mathrm{nRT}}_1\mathrm{In}\left(3\right) \\ \mathrm{W}=1.10{\mathrm{nRT}}_1 \\ \frac{\mathrm{W}}{{\mathrm{nRT}}_1}=1.10 \end{gathered}$$

Hence, the value of the ratio is$\mathrm{W}/{\mathrm{nRT}}_1=1.10$

The change in internal energy is$∆E_{int}=0$ since this is an ideal gas process without a temperature change.

Using the relation of equation (4), we get that

$$\begin{gathered} \mathrm{Q}=1.10{\mathrm{nRT}}_1 \\ \frac{\mathrm{Q}}{{\mathrm{nRT}}_1}=1.10 \end{gathered}$$

Hence, the value of the ratio is$Q/nR{T}_1=1.10$

Since,$∆E_{int}=0$

The value of the ratio,

$\frac{∆E_{int}}{nR{T}_1}=0$

Hence, the value of the ratio is $∆E_{int}/nR{T}_1=0$.

The entropy change of the gas from path$1→2$using equation (5) is given as:

$$\begin{gathered} ∆S=\frac{1.10nR{T}_1}{T_1} \\ \frac{∆S}{nR}=1.10 \end{gathered}$$

Hence, the value of the ratio is$∆S/nR=1.10$

There is no change in volume for the process$2→3$, therefore, work done is zero.

Thus,$\frac{\mathrm{W}}{{\mathrm{nRT}}_1}=0$

Hence, the value of the ratio is $W/nR{T}_1=0$.

The change in internal energy using equation (6) from the path$2→3$is given by:

$$\begin{gathered} ∆E_{int}=n\left(\frac{5}{2}R\right)\left(T_3-T_2\right)\left(fordiatomicgas,C_V=\frac{5}{2}R\right) \\ =n\left(\frac{5}{2}R\right)\left(\frac{T_1}{3^{0.4}}-T_1\right) \\ =n\left(\frac{5}{2}R\right)\left(0.6444T_1-T_1\right) \\ =n\left(\frac{5}{2}R\right)\left(0.6444-1\right)T_1 \\ =0.889nR{T}_1 \\ \frac{∆E_{int}}{nR{T}_1}=0.889 \end{gathered}$$

Using the first law of thermodynamics for path$2→3$:

Since there is no change in volume for this path, the work done is zero.

$$\begin{gathered} Q=∆E_{int} \\ \frac{Q}{nR{T}_1}=-0.889 \end{gathered}$$

Hence, the value of the ratio is$Q/nR{T}_1=-0.889$

Since,$\mathrm{Δ}{E}_{int}=-0.889nR{T}_1$from the above calculations, we get that,

$\frac{∆E_{int}}{nR{T}_1}=0.889$

Hence, the value of the ratio is$∆{\mathrm{E}}_{\mathrm{int}}/{\mathrm{nRT}}_1=0.889$

Entropy change for an ideal gas is given using equation (7) as follows:

Dividing both sides by R,

$$\begin{gathered} \frac{∆\mathrm{S}}{\mathrm{R}}=\frac{{\mathrm{nC}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_{\mathrm{final}}}{{\mathrm{V}}_{\mathrm{initial}}}\right) \\ =\frac{{\mathrm{nC}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_3}{{\mathrm{T}}_1}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_1}\right) \end{gathered}$$.

For this path, volume is constant i.e.$V_3=V_1$and the second term in the above equation vanishes because In(1)=0.

$$\begin{gathered} \frac{∆\mathrm{S}}{\mathrm{R}}=\frac{{\mathrm{nC}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{3^{{\displaystyle \frac{{\mathrm{T}}_1}{0.4}}}}{{\mathrm{T}}_1}\right) \\ \frac{∆\mathrm{S}}{\mathrm{nR}}=\frac{1}{\mathrm{R}}×\frac{5}{2}\mathrm{R}×\mathrm{In}\left(\frac{3^{{\displaystyle \frac{{\mathrm{T}}_1}{0.4}}}}{{\mathrm{T}}_1}\right) \\ \frac{∆\mathrm{S}}{\mathrm{nR}}=\frac{5}{2}\mathrm{In}\left(3^{-0.4}\right) \\ \frac{∆\mathrm{S}}{\mathrm{nR}}=-1.10 \end{gathered}$$

Hence, the value of the ratio is$∆S/nR=-1.10$

The process is adiabatic and for an adiabatic process Q=0

Hence, there is no change in entropy.

For a complete cycle, the change in internal energy must be zero, so all internal energies must add up to zero.

$$\begin{gathered} {\mathrm{Δ·¡}}_{(\mathrm{int},1→2}+{\mathrm{Δ·¡}}_{\mathrm{int},2→3}+{\mathrm{Δ·¡}}_{(\mathrm{int},3→1}=0 \\ 0+(-0.889{\mathrm{nRT}}_1)+{\mathrm{Δ·¡}}_{\mathrm{int},3→1}=0 \\ {\mathrm{Δ·¡}}_{\mathrm{int},3→1}=0.889{\mathrm{nRT}}_1 \\ \frac{∆{\mathrm{E}}_{\mathrm{int},3→1}}{{\mathrm{nRT}}_1}=0.889 \end{gathered}$$

Now, using equation (4), we can get the work done by the gas from the path$3→1$is given as:

$$\begin{gathered} W=0-0.889nR{T}_1 \\ \frac{W}{nR{T}_1}=-0.889 \end{gathered}$$

Hence, the value of the ratio is$W/nR{T}_1=-0.889$

This process is adiabatic, so Q=0

Hence, the ratio is zero.

$\frac{\mathrm{Q}}{{\mathrm{nRT}}_1}=0$

Hence, the value of the ratio is zero.

This is the same as calculated in part(l)but positive.

$\frac{∆E_{int}}{nR{T}_1}=0.889$

Hence, the value of the ratio is$∆{\mathrm{E}}_{\mathrm{int}}/{\mathrm{nRT}}_1=0.889$

Because we have completed one complete cycle, so its initial temperature and volume are the same as the final temperature and volume and the ratio is zero since, In 1=0.

$$\begin{gathered} \frac{∆\mathrm{S}}{\mathrm{nR}}=\frac{{\mathrm{C}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_1}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_1}\right) \\ \frac{∆\mathrm{S}}{\mathrm{nR}}=0 \end{gathered}$$

Hence, the value of the ratio is zero.