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Expert-verified Found in: Page 578 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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 ${{\mathbit{V}}}_{fs}{\mathbf{=}}{\mathbf{.}}{\mathbf{30}}{{\mathbf{m}}}^{{\mathbf{3}}}$, and the scale of the horizontal axis is set by ${{\mathbit{Q}}}_{{s}}{\mathbf{=}}{\mathbf{1200}}{\mathbit{J}}$ The gas temperature is 360 K

See the step by step solution

## Step 1: Given

The graph ${V}_{f}\text{vs} Q$

${Q}_{s}=1200\text{J}\phantom{\rule{0ex}{0ex}}n=0.825\text{mol}\phantom{\rule{0ex}{0ex}}{V}_{f}=0.30{\text{m}}^{3}\phantom{\rule{0ex}{0ex}}{V}_{i}=0.20{\text{m}}^{3}$

## 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:

$W=nRT\mathrm{ln}\left(\frac{{V}_{f}}{{V}_{i}}\right)$

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,

$W=nRT\mathrm{ln}\left(\frac{{V}_{f}}{{V}_{i}}\right)$

For isothermal process,

$\therefore Q=nRT\mathrm{ln}\left(\frac{{V}_{f}}{{V}_{i}}\right)\phantom{\rule{0ex}{0ex}}\therefore T=\frac{Q}{nR\mathrm{ln}\left(\frac{{V}_{f}}{{V}_{i}}\right)}$

From the given graph,

Q = 1000 J when${V}_{fs}=0.3{\text{m}}^{2}$ ,

$T=\frac{1000}{0.825\left(8.314\right)\mathrm{ln}\left(\frac{0.30}{0.20}\right)}\phantom{\rule{0ex}{0ex}}T=359.56\text{K}\approx 360\text{K}$

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. ### Want to see more solutions like these? 