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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Figure 20-25, where ${{\mathbit{V}}}_{{\mathbf{23}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{{\mathbit{V}}}_{{\mathbf{1}}}$,n moles of a diatomic ideal gas are taken through the cycle with the molecules rotating but not oscillating. What are (a) ${{\mathbit{p}}}_{{\mathbf{2}}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{1}}}$, (b) ${{\mathbit{p}}}_{{\mathbf{3}}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{1}}}$, and (c) ${{\mathbit{T}}}_{{\mathbf{3}}}{\mathbf{/}}{{\mathbit{T}}}_{{\mathbf{1}}}$? For path ${\mathbf{1}}{\mathbf{\to }}{\mathbf{2}}$, what are (d) ${\mathbit{W}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$, (e) ${\mathbit{Q}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$, (f) ${\mathbf{∆}}{{\mathbit{E}}}_{\mathbf{i}\mathbf{n}\mathbf{t}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$, and (g) ${\mathbf{∆}}{\mathbit{S}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}$? For path ${\mathbf{2}}{\mathbf{\to }}{\mathbf{3}}$, what are (h)${\mathbit{W}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$ ,(i) ${\mathbit{Q}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$ (j) ${\mathbf{∆}}{{\mathbit{E}}}_{\mathbf{i}\mathbf{n}\mathbf{t}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$, and (k) ${\mathbf{∆}}{\mathbit{S}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$? For path ${\mathbf{3}}{\mathbf{\to }}{\mathbf{1}}$, what are (l) ${\mathbit{W}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$,(m) role="math" localid="1661575470031" ${\mathbf{Q}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$ (n) ${\mathbf{∆}}{{\mathbit{E}}}_{{\mathbf{int}}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}{{\mathbit{T}}}_{{\mathbf{1}}}$, and (o) ${\mathbf{∆}}{\mathbf{S}}{\mathbf{/}}{\mathbit{n}}{\mathbit{R}}$?

a) Ratio of $\frac{{\mathrm{p}}_{2}}{{\mathrm{p}}_{1}}$ is 0.333

b) Ratio of $\frac{{\mathrm{p}}_{3}}{{\mathrm{p}}_{1}}$ is 0.215

c) Ratio of $\frac{{\mathrm{T}}_{3}}{{\mathrm{T}}_{1}}$ is 0.644

d) For path $1\to 2$ ratio of $\frac{\mathrm{W}}{{\mathrm{nRT}}_{1}}$ is 1.10

e) For path $1\to 2$, the ratio of $\frac{Q}{nR{T}_{1}}$ is 1.10

f) For path $1\to 2$ ratio of $\frac{∆{E}_{int}}{nR{T}_{1}}$ is 0

g) For path $1\to 2$, the ratio of $\frac{∆S}{nR}$ is 1.10

h) For path $2\to 3$ ratio of $\frac{W}{nR{T}_{1}}$ is 0

i) For path $2\to 3$, the ratio of $\frac{\mathrm{Q}}{{\mathrm{nRT}}_{1}}$ is -0.889

j) For path $2\to 3$ ratio of $\frac{∆{E}_{int}}{nR{T}_{1}}$ is -0.889

k) For path $2\to 3$, the ratio of $\frac{∆S}{nR}$ is -1.10

l) For path $3\to 1$ ratio of $\frac{\mathrm{W}}{{\mathrm{nRT}}_{1}}$ is -0.889

m) For path $3\to 1$, the ratio of $\frac{Q}{nR{T}_{1}}$ is 0

n) For path $3\to 1$ ratio of $\frac{∆{\mathrm{E}}_{\mathrm{int}}}{{\mathrm{nRT}}_{1}}$ is 0.889

o) For path $3\to 1$, the ratio of $\frac{∆S}{nR}$ is 0

See the step by step solution

## Step 1: The given data

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.

## Step 2: Understanding the concept of adiabatic relations

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,

${\mathrm{pV}}^{\mathrm{y}}=\mathrm{constant}\phantom{\rule{0ex}{0ex}}{\mathrm{TV}}^{\mathrm{y}-1}=\mathrm{constant}$ (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,

$Q=\Delta {E}_{int}+W$ (4)

The entropy change of gas,

$∆S=\frac{Q}{T}$ (5)

The change in internal energy of the gas,

$\Delta {E}_{int}=n{C}_{V}\Delta T$ (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)

## Step 3: (a) Calculation of p2/p1

For path$1\to 2$ the process is isothermal.

So ${\mathrm{T}}_{2}={\mathrm{T}}_{1}$

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

$\frac{{p}_{2}}{{p}_{1}}=\frac{{V}_{1}}{{V}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}=0.333$

Hence, the ratio is ${p}_{2}/{p}_{1}=0.333$

## Step 4: (b) Calculation of p3/p1

The path $3\to 1$ is an adiabatic process

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

${\mathrm{p}}_{3}{\mathrm{V}}_{3}^{\mathrm{y}}={\mathrm{p}}_{1}{\mathrm{V}}_{1}^{\mathrm{y}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{p}}_{3}}{{\mathrm{p}}_{1}}=\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{3}}\right)\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{p}}_{3}}{{\mathrm{p}}_{1}}={\left(\frac{1}{3}\right)}^{\mathrm{y}}$

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

$\frac{{p}_{3}}{{p}_{1}}={\left(\frac{1}{3}\right)}^{1.4}\phantom{\rule{0ex}{0ex}}=0.215$

Hence, the value of the ratio is ${p}_{3}/{p}_{1}=0.215$

## Step 5: (c) Calculation of T3/T1

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

${T}_{3}{\mathrm{V}}_{3}^{\mathrm{y}-1}={T}_{1}{\mathrm{V}}_{1}^{\mathrm{y}-1}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{T}}_{3}}{{T}_{1}}={\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{3}}\right)}^{y-1}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{T}}_{3}}{{\mathrm{T}}_{1}}={\left(\frac{1}{3}\right)}^{1.4-1}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{T}}_{3}}{{\mathrm{T}}_{1}}={\left(\frac{1}{3}\right)}^{0.4}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{T}}_{3}}{{\mathrm{T}}_{1}}=0.644$

Hence, the value of the ratio is ${T}_{3}/{T}_{1}=0.664$

## Step 6: (d) Calculation of W/nRT1 for path 1→2

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

$\mathrm{W}={\mathrm{nRT}}_{1}\mathrm{In}\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{W}=1.10{\mathrm{nRT}}_{1}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{W}}{{\mathrm{nRT}}_{1}}=1.10$

Hence, the value of the ratio is $\mathrm{W}/{\mathrm{nRT}}_{1}=1.10$

## Step 7: (e) Calculation of Q/nRT1 for paths 1→2

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

$\mathrm{Q}=1.10{\mathrm{nRT}}_{1}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{Q}}{{\mathrm{nRT}}_{1}}=1.10$

Hence, the value of the ratio is$Q/nR{T}_{1}=1.10$

## Step 8: (f) Calculation of ∆Eint=nRT1 for paths 1→2

Since,$∆{E}_{int}=0$

The value of the ratio,

$\frac{∆{E}_{int}}{nR{T}_{1}}=0$

Hence, the value of the ratio is $∆{E}_{int}/nR{T}_{1}=0$.

## Step 9: (g) Calculation of ∆S/nR for path 1→2

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

$∆S=\frac{1.10nR{T}_{1}}{{T}_{1}}\phantom{\rule{0ex}{0ex}}\frac{∆S}{nR}=1.10$

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

## Step 10: (h) Calculation of W/nRT1 for path 2→3

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

Thus,$\frac{\mathrm{W}}{{\mathrm{nRT}}_{1}}=0$

Hence, the value of the ratio is $W/nR{T}_{1}=0$.

## Step 11: (i) Calculation of Q/nRT1 for path 2→3

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

$∆{E}_{int}=n\left(\frac{5}{2}R\right)\left({T}_{3}-{T}_{2}\right)\left(fordiatomicgas,{C}_{V}=\frac{5}{2}R\right)\phantom{\rule{0ex}{0ex}}=n\left(\frac{5}{2}R\right)\left(\frac{{T}_{1}}{{3}^{0.4}}-{T}_{1}\right)\phantom{\rule{0ex}{0ex}}=n\left(\frac{5}{2}R\right)\left(0.6444{T}_{1}-{T}_{1}\right)\phantom{\rule{0ex}{0ex}}=n\left(\frac{5}{2}R\right)\left(0.6444-1\right){T}_{1}\phantom{\rule{0ex}{0ex}}=0.889nR{T}_{1}\phantom{\rule{0ex}{0ex}}\frac{∆{E}_{int}}{nR{T}_{1}}=0.889$

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

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

$Q=∆{E}_{int}\phantom{\rule{0ex}{0ex}}\frac{Q}{nR{T}_{1}}=-0.889$

Hence, the value of the ratio is $Q/nR{T}_{1}=-0.889$

## Step 12: (j) Calculation of ∆Eint/nRT1 for path 2→3

Since, $\Delta {E}_{int}=-0.889nR{T}_{1}$ from the above calculations, we get that,

$\frac{∆{E}_{int}}{nR{T}_{1}}=0.889$

Hence, the value of the ratio is $∆{\mathrm{E}}_{\mathrm{int}}/{\mathrm{nRT}}_{1}=0.889$

## Step 13: (k) Calculation of ∆S/nR for path 2→3

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

Dividing both sides by R,

$\frac{∆\mathrm{S}}{\mathrm{R}}=\frac{{\mathrm{nC}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_{\mathrm{final}}}{{\mathrm{V}}_{\mathrm{initial}}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{nC}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_{3}}{{\mathrm{T}}_{1}}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_{3}}{{\mathrm{V}}_{1}}\right)$.

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.

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

## Step 14: (l) Calculation of W/nRT1 for path 3→1

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.

${\mathrm{\Delta E}}_{\left(\mathrm{int},1\to 2}+{\mathrm{\Delta E}}_{\mathrm{int},2\to 3}+{\mathrm{\Delta E}}_{\left(\mathrm{int},3\to 1}=0\phantom{\rule{0ex}{0ex}}0+\left(-0.889{\mathrm{nRT}}_{1}\right)+{\mathrm{\Delta E}}_{\mathrm{int},3\to 1}=0\phantom{\rule{0ex}{0ex}}{\mathrm{\Delta E}}_{\mathrm{int},3\to 1}=0.889{\mathrm{nRT}}_{1}\phantom{\rule{0ex}{0ex}}\frac{∆{\mathrm{E}}_{\mathrm{int},3\to 1}}{{\mathrm{nRT}}_{1}}=0.889$

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

$W=0-0.889nR{T}_{1}\phantom{\rule{0ex}{0ex}}\frac{W}{nR{T}_{1}}=-0.889$

Hence, the value of the ratio is $W/nR{T}_{1}=-0.889$

## Step 15: (m) Calculation of Q/nRT1 for path 3→1

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.

## Step 16: (n) Calculation of ∆Eint/nRT1 for path 3→1

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 $∆{\mathrm{E}}_{\mathrm{int}}/{\mathrm{nRT}}_{1}=0.889$

## Step 17: (o) Calculation of ∆S/nR for path 3→1

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.

$\frac{∆\mathrm{S}}{\mathrm{nR}}=\frac{{\mathrm{C}}_{\mathrm{V}}}{\mathrm{R}}\mathrm{In}\left(\frac{{\mathrm{T}}_{1}}{{\mathrm{T}}_{1}}\right)+\mathrm{nIn}\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{1}}\right)\phantom{\rule{0ex}{0ex}}\frac{∆\mathrm{S}}{\mathrm{nR}}=0$

Hence, the value of the ratio is zero.