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Found in: Page 605

### Fundamentals Of Physics

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

# A 2.0 mol sample of an ideal monatomic gas undergoes the reversible process shown in Figure. The scale of the vertical axis is set by${{\mathbit{T}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{400}}{\mathbf{.}}{\mathbf{0}}{\mathbf{K}}$ and the scale of the horizontal axis is set by ${{\mathbit{S}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{J}}{\mathbf{/}}{\mathbf{k}}$. (a) How much energy is absorbed as heat by the gas? (b) What is the change in the internal energy of the gas? (c) How much work is done by the gas?

a) Energy absorbed as heat by the gas is $4.5×{10}^{3}\mathrm{J}$.

b) Change in internal energy of the gas is $-5×{10}^{3}\mathrm{J}$.

c) Work done by the gas is 9.5 kJ.

See the step by step solution

## Step 1: The given data

Number of moles of a monatomic gas n=2 mol

Temperature ${T}_{s}=400\mathrm{K}$

Entropy ${\mathrm{S}}_{\mathrm{s}}=20\mathrm{J}/\mathrm{K}$

## Step 2: Understanding the concept of thermodynamic cycle

We construct the formula for heat by using the area under the curve. Using the equation of change in internal energy and the first law of thermodynamics, we can find the change in internal energy and work done by the gas.

Formulae:

From the graph, the heat transferred by the gas in terms of entropy,

${\mathrm{Q}}_{\mathrm{straight}}=\frac{\left({\mathrm{T}}_{\mathrm{f}}+{\mathrm{T}}_{\mathrm{i}}\right)}{2}∆\mathrm{S}$ (1)

The change in internal energy of the gas,

${\mathrm{\Delta E}}_{\mathrm{int}}=n{C}_{V}\Delta T$ (2)

The first law of thermodynamics,

$Q=\mathrm{W}+∆{\mathrm{E}}_{\mathrm{int}}$ (3)

## Step 3: (a) Calculation of the energy absorbed as heat by the gas

Heat can be found by the area under the curve in the TS diagram, so we have to find the area under the straight line path in the given diagram. The diagram looks like a rectangular trapezoid and its area is given by,

From the graph, we see that ${\mathrm{T}}_{\mathrm{i}}={\mathrm{T}}_{\mathrm{s}}=400 \mathrm{K}$ and ${\mathrm{T}}_{\mathrm{f}}=200 \mathrm{K}$, and ${\mathrm{S}}_{\mathrm{s}}={\mathrm{S}}_{\mathrm{f}}=20\mathrm{J}/\mathrm{K}$and

${\mathrm{S}}_{\mathrm{i}}=5\mathrm{J}/\mathrm{K}$

Thus, the net entropy change of the gas is given as:

$∆\mathrm{S}={\mathrm{S}}_{\mathrm{f}}-{\mathrm{S}}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}=20\mathrm{J}/\mathrm{K}-5\mathrm{J}/\mathrm{K}\phantom{\rule{0ex}{0ex}}=15\mathrm{J}/\mathrm{K}$,

Substituting all values in the equation (1), we get that the energy transferred by the gas as heat is given as follows:

${\mathrm{Q}}_{\mathrm{straight}}=\left(\frac{200+400}{2}\right)\left(15\mathrm{J}/\mathrm{K}\right)\phantom{\rule{0ex}{0ex}}=300\mathrm{K}×15\mathrm{J}/\mathrm{K}\phantom{\rule{0ex}{0ex}}=4.5×{10}^{3}\mathrm{J}$

Hence, the value of the energy transferred as heat is $4.5×{10}^{3}J$

## Step 4: (b) Calculation of the change in internal energy of the gas

When a confined ideal gas undergoes a temperature change $∆\mathrm{T}$ the change in internal energy is given as:

For a monatomic ideal gas,${\mathrm{C}}_{\mathrm{V}}=\frac{3}{2}\mathrm{R}$ ,$\mathrm{n}=2, \mathrm{R}=8.31 \mathrm{J}/\mathrm{mol}\cdot \mathrm{K}$ and,

$∆\mathrm{T}={\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{I}}\phantom{\rule{0ex}{0ex}}=200\mathrm{K}-400\mathrm{K}\phantom{\rule{0ex}{0ex}}=-200\mathrm{K}$)

$∆{\mathrm{E}}_{\mathrm{int}}=2\left(\frac{3}{2}×8.31\mathrm{J}/\mathrm{mol}.\mathrm{K}\right)\left(-200\mathrm{K}\right)\phantom{\rule{0ex}{0ex}}=-5×{10}^{3}\mathrm{J}$

Hence, the value of the change of internal energy of the gas is $-5×{10}^{3}\mathrm{J}$

## Step 5: (c) Calculation of the work done by the gas

Using equation (3) and the given values, we get that the work done by the gas is given as:

$\mathrm{W}=\mathrm{Q}-∆{\mathrm{E}}_{\mathrm{int}}\phantom{\rule{0ex}{0ex}}=4.5\mathrm{kJ}-\left(-5\mathrm{kJ}\right)\phantom{\rule{0ex}{0ex}}=9.5\mathrm{kJ}$

Hence, the value of the work done by the gas is 9.5 kJ