Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: Suppose of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. If figure shows the final volume V_f versus Q, what is the gas temperature? The scale of the vertical axis is set by $V_{f s} = . 30 m^{3}$ , and the scale of the horizontal axis is set by $Q_{s} = 1200 J$

Answer

The gas temperature is 360 K

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The graph $V_{f} vs Q$

$Q_{s} = 1200 J n = 0.825 mol V_{f} = 0.30 m^{3} V_{i} = 0.20 m^{3}$

Step 2: Determine the concept

Using that formula for work done by the gas at a constant temperature and substituting the values from the given graph, find the gas temperature.

Formula is as follow:

$W = n R T \ln (\frac{V_{f}}{V_{i}})$

Here, W is work done, V is volume, T is temperature, R is universal gas constant and n is number of moles.

Step 3: Determine the gas temperature

For isothermal process,

Q = W

Also,

$W = n R T \ln (\frac{V_{f}}{V_{i}})$

For isothermal process,

$∴ Q = n R T \ln (\frac{V_{f}}{V_{i}}) ∴ T = \frac{Q}{n R \ln (\frac{V_{f}}{V_{i}})}$

From the given graph,

Q = 1000 J when $V_{f s} = 0.3 m^{2}$ ,

$T = \frac{1000}{0.825 (8.314) \ln (\frac{0.30}{0.20})} T = 359.56 K \approx 360 K$

Hence, the gas temperature is 360 K

Therefore, the temperature of the gas for isothermal process can be found by using the formula for work done by the gas at a constant temperature.

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Step by Step Solution

Step 1: Given

Step 2: Determine the concept

Step 3: Determine the gas temperature

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Short Answer

Question: Suppose of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. If figure shows the final volume Vf versus Q, what is the gas temperature? The scale of the vertical axis is set by Vfs=.30m3, and the scale of the horizontal axis is set by Qs=1200J

Step by Step Solution

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Step 2: Determine the concept

Step 3: Determine the gas temperature

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