Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A sample of an ideal gas is taken through the cyclic process abca as shown in figure. The scale of the vertical axis is set by $p_{b} = 7.5 kPa$ and $p_{ac} = 2.5 kPa$ . At point a, $T = 200 K$ .
aHow many moles of gas are in the sample?
What are:
b.The temperature of gas at point b
c.Temperature of gas at point c
d.The net energy added to the gas as heat during the cycle?

The number of moles of gas present in the sample are $1.5 mol$ .
The temperature of the gas at point b is . $1.8 \times 10^{3} K$
The temperature of the gas at point c is $6.0 \times 10^{2} K$ .
The net energy added to the gas as heat during the cycle is $5.0 \times 10^{3} J .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

$P_{b} = 7.5 kPa = 7500 Pa P_{a c} = 2.5 kPa = 2500 Pa$

At point a, $T = 200 K$
At point a, $V = 1 m^{3}$
At point b, $V = 3 m^{3}$

Step 2: Determine the concept

Find the number of moles and the temperature of the gas at points b and c present in the sample using the gas law. Net energy added can be found from the net work done by the gas using the given graph.

Formula is as follow:

$p_{i} v_{i} = n R T_{i}$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

Step 3: (a) Determine the number of moles of gas present in the sample

According to the gas law,

$P V = n R T n = \frac{P V}{R T}$

From the given graph, at point a,

$n = \frac{2500 (1)}{8.314 (200)} n = 1.5 mol$

Hence, the number of moles of gas present in the sample are . $1.5 mol$

Step 4: (b) Determine the temperature of the gas at point b

By gas law,

$\frac{P_{b} V_{b}}{P_{a} V_{a}} = \frac{T_{b}}{T_{a}}$

$T_{b} = \frac{P_{b} V_{b}}{P_{a} V_{a}} T_{a} T_{b} = \frac{7500 (3)}{2500 (1)} (200)$

$T_{b} = 1.8 \times 10^{3} K$

Hence, the temperature of the gas at point b is $1.8 \times 10^{3} K$ .

Step 5: (c) Determine the temperature of the gas at point c

Similarly,

$T_{c} = \frac{P_{c} V_{c}}{P_{a} V_{a}} T_{a} T_{c} = \frac{2500 (3)}{2500 (1)} (200)$

$T_{c} = 600 = 6.0 \times 10^{2} K$

Hence, the temperature of the gas at point c is $6.0 \times 10^{2} K$ .

Step 6: (d) Determine the net energy added to the gas as heat during the cycle

First law of thermodynamics gives,

$△ E = Q - W$

For isothermal process,

$Q = W Q = Area under the curve Q = \frac{1}{2} (ac) (bc)$

Substitute the values as:

$Q = \frac{1}{2} (2) (7500 - 2500)$

$Q = 5000 = 5.0 \times 10^{3} J$

Hence, the net energy added to the gas as heat during the cycle is $5.0 \times 10^{3} J .$

Therefore, the number of moles of gas present in the sample,temperature at a point, and heat added can be foundusing gas law.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determine the concept

Step 3: (a) Determine the number of moles of gas present in the sample

Step 4: (b) Determine the temperature of the gas at point b

Step 5: (c) Determine the temperature of the gas at point c

Step 6: (d) Determine the net energy added to the gas as heat during the cycle

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determine the concept

Step 3: (a) Determine the number of moles of gas present in the sample

Step 4: (b) Determine the temperature of the gas at point b

Step 5: (c) Determine the temperature of the gas at point c

Step 6: (d) Determine the net energy added to the gas as heat during the cycle

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