Question: Supposeof 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

Short Answer

Expert verified

Answer

The gas temperature is 360 K

Step by step solution

01

Step 1: Given

The graph VfvsQ

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

02

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=nRTlnVfVi

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

03

Determine the gas temperature

For isothermal process,

Q = W

Also,

W=nRTlnVfVi

For isothermal process,

Q=nRTlnVfViT=QnRlnVfVi

From the given graph,

Q = 1000 J whenVfs=0.3m2 ,

T=10000.8258.314ln0.300.20T=359.56K360K

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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