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Fundamentals Of Physics
Found in: Page 578

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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


The gas temperature is 360 K

See the step by step solution

Step by Step Solution

Step 1: Given

The graph VfvsQ

Qs=1200Jn=0.825 molVf=0.30m3Vi=0.20m3

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 lnVfVi

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


W=nRT lnVfVi

For isothermal process,

Q=nRT lnVfViT=QnR lnVfVi

From the given graph,

Q = 1000 J whenVfs=0.3 m2 ,

T=10000.8258.314ln0.300.20T=359.56 K360 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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