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Q18P

Expert-verifiedFound in: Page 605

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${{\mathit{T}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{400}}{\mathbf{.}}{\mathbf{0}}{\mathbf{K}}$ ****and the scale of the horizontal axis is set by ${{\mathit{S}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{J}}{\mathbf{/}}{\mathbf{k}}$****. (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 $4.5\times {10}^{3}\mathrm{J}$.

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

c) Work done by the gas is 9.5 kJ.

Number of moles of a monatomic gas n=2 mol

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

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

**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,

${\mathrm{Q}}_{\mathrm{straight}}=\frac{\left({\mathrm{T}}_{\mathrm{f}}+{\mathrm{T}}_{\mathrm{i}}\right)}{2}\u2206\mathrm{S}$ (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,

$Q=\mathrm{W}+\u2206{\mathrm{E}}_{\mathrm{int}}$ (3)

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 ${\mathrm{T}}_{\mathrm{i}}={\mathrm{T}}_{\mathrm{s}}=400\hspace{0.33em}\mathrm{K}$ and ${\mathrm{T}}_{\mathrm{f}}=200\hspace{0.33em}\mathrm{K}$, 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:

$\u2206\mathrm{S}={\mathrm{S}}_{\mathrm{f}}-{\mathrm{S}}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}=20\mathrm{J}/\mathrm{K}-5\mathrm{J}/\mathrm{K}\phantom{\rule{0ex}{0ex}}=15\mathrm{J}/\mathrm{K}$,

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

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

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

When a confined ideal gas undergoes a temperature change $\u2206\mathrm{T}$ the change in internal energy is given as:

For a monatomic ideal gas,${\mathrm{C}}_{\mathrm{V}}=\frac{3}{2}\mathrm{R}$ ,$\mathrm{n}=2,\hspace{0.33em}\mathrm{R}=8.31\hspace{0.33em}\mathrm{J}/\mathrm{mol}\cdot \mathrm{K}$ and,

$\u2206\mathrm{T}={\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{I}}\phantom{\rule{0ex}{0ex}}=200\mathrm{K}-400\mathrm{K}\phantom{\rule{0ex}{0ex}}=-200\mathrm{K}$)

$\u2206{\mathrm{E}}_{\mathrm{int}}=2\left(\frac{3}{2}\times 8.31\mathrm{J}/\mathrm{mol}.\mathrm{K}\right)\left(-200\mathrm{K}\right)\phantom{\rule{0ex}{0ex}}=-5\times {10}^{3}\mathrm{J}$

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

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

$\mathrm{W}=\mathrm{Q}-\u2206{\mathrm{E}}_{\mathrm{int}}\phantom{\rule{0ex}{0ex}}=4.5\mathrm{kJ}-\left(-5\mathrm{kJ}\right)\phantom{\rule{0ex}{0ex}}=9.5\mathrm{kJ}$

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

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