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Found in: Page 604

### Fundamentals Of Physics

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

# (a) For 1.0mol of a monatomic ideal gas taken through the cycle in Figure, where ${{\mathbit{V}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{{\mathbit{V}}}_{{\mathbf{0}}}$ , what is ${\mathbit{W}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{0}}}{{\mathbit{V}}}_{{\mathbf{0}}}$ as the gas goes from state a to state c along path abc ?(b)What is role="math" localid="1661581522914" ${\mathbf{∆}}{{\mathbit{E}}}_{\mathbf{i}\mathbf{n}\mathbf{t}}{\mathbf{}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{0}}}{{\mathbit{V}}}_{{\mathbf{0}}}$ in going from b to c and(c)What is ${\mathbf{∆}}{{\mathbit{E}}}_{{\mathbf{int}}}{\mathbf{}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{0}}}{{\mathbit{V}}}_{{\mathbf{0}}}$ in going through one full cycle? (d)What is ${\mathbf{∆}}{\mathbit{S}}$ in going from b to c and (e) What is ${\mathbf{∆}}{\mathbit{S}}$ in going through one full cycle?

1. The value of $W/{p}_{0}{V}_{0}$ as the gas goes from state a to state c along path abc is 3 .
2. The value of $∆{E}_{\mathrm{int}}/{p}_{0}{V}_{0}$ as the gas goes from state b to state c is 6 .
3. The value of $∆{E}_{\mathrm{int}}/{p}_{0}{V}_{0}$ as the gas goes through one full cycle is zero.
4. The value of $∆S$ in going from b to c is $8.64\mathrm{J}/\mathrm{K}$.
5. The value of $∆S$ through one full cycle is zero.
See the step by step solution

## Step 1: The given data

1. Number of moles in the sample, $n=1.0\mathrm{mol}$
2. Volume of the gas, ${V}_{1}=4.00{V}_{0}$
3. P-V diagram for a monatomic ideal gas taken through the cycle.

## Step 2: Understanding the concept of thermodynamics

We can find the value of ${\mathbit{W}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{0}}}{{\mathbit{V}}}_{{\mathbf{0}}}$ as the gas goes from the state a to state c along path abc using the formula for work done by an ideal gas in terms of pressure and volume change. Then using the gas law, we can find the temperature of the gas at points b and c. Then using it in the formula for heat absorbed, we can find the amount of heat absorbed. Inserting it into the equation for the law of conservation of energy, we can find the value of ${\mathbf{∆}}{{\mathbit{E}}}_{\mathbf{i}\mathbf{n}\mathbf{t}}{\mathbf{}}{\mathbf{/}}{{\mathbit{p}}}_{{\mathbf{0}}}{{\mathbit{V}}}_{{\mathbf{0}}}$. By using the formula for entropy change, we can find its value.

Formulae:

The work done by a gas at constant pressure, $W=p∆V$ …(i)

The ideal-gas equation, $PV=nRT$ …(ii)

The molar specific heat at constant volume, ${C}_{V}=\frac{3}{2}R$ …(iii)

The internal energy of the gas from first law of thermodynamics, $∆E=Q-W$ …(iv)

The entropy change of the system, $∆S=n{C}_{V}In\frac{{T}_{f}}{{T}_{i}}$ …(v)

The heat absorbed by the body, $Q=n{C}_{V}∆T$ …(vi)

## Step 3: (a) Calculation of   from path a to c along abc

From the given p-V diagram,

Along path , the work done by the gas using equation (i) is given as:

$W={p}_{0}\left({V}_{1}-{V}_{0}\right)\phantom{\rule{0ex}{0ex}}=4{p}_{0}\left({V}_{0}-{V}_{0}\right)\phantom{\rule{0ex}{0ex}}=3{p}_{0}{V}_{0}$

The required value is given as:

$\frac{W}{{p}_{0}{V}_{0}}=3$

Therefore, the value of $W/{p}_{0}{V}_{0}$as the gas goes from state a to state c along path abc is 3.

## Step 4: (b) Calculation of  W/p0V0 as the gas moves from b to c

The temperature of gas at point b using equation (ii) is given as:

${T}_{b}=\frac{{p}_{0}{V}_{1}}{nR}\phantom{\rule{0ex}{0ex}}=\frac{4{p}_{0}{V}_{0}}{\left(1\right)R}$

The temperature of gas at point c using equation (ii) is given as:

${T}_{c}=\frac{2{p}_{0}{V}_{1}}{nR}\phantom{\rule{0ex}{0ex}}=\frac{8{p}_{0}{V}_{0}}{\left(1\right)R}$

The heat energy absorbed is given by substituting the value of equation (iii) in equation (vi) as follows:

$Q=1×\frac{3}{2}R\left({T}_{c}-{T}_{b}\right)\phantom{\rule{0ex}{0ex}}=1×\frac{3}{2}R\left(\frac{8{p}_{0}{V}_{0}}{\left(1\right)R}-\frac{4{p}_{0}{V}_{0}}{\left(1\right)R}\right)\phantom{\rule{0ex}{0ex}}=6{p}_{0}{V}_{0}$

Since along path bc , the gas undergoes an isochoric process. So, $W=0$

The change in internal energy using equation (iv) can be given as:

$∆E=Q\phantom{\rule{0ex}{0ex}}=6{p}_{0}{V}_{0}\phantom{\rule{0ex}{0ex}}\frac{∆E}{{p}_{0}{V}_{0}}=6$

Therefore, the value of $∆E/{p}_{0}{V}_{0}$ as the gas goes from state b to state c along path abc is 6.

## Step 5: (c) Calculation of  ∆Eint/p0V0 as the gas goes full cycle

Since energy is a state function, $∆{E}_{int}=0$

Therefore, the value of $∆E/{p}_{0}{V}_{0}$ as the gas is going through one full cycle is zero.

## Step 6: (d) Calculation of entropy change from path b to c

The change in entropy along path bc using equation (v) and the given values is given as:

Therefore, the value of $∆S$ in going from b to c is $8.64\mathrm{J}/\mathrm{K}$ .

## Step 7: (e) Calculation of entropy change for complete cycle

For a complete cycle, ${T}_{i}={T}_{f}$

Substituting it in the above formula of equation (v) for entropy change gives $∆S=0$

Therefore, the value of $∆S$ through one full cycle is zero.