One mole of an ideal gas is heated slowly so that it goes from the PV state (Pi,

Question

One mole of an ideal gas is heated slowly so that it goes from the PV state (Pi, V) to (6Pi, 6V) in such a way that the pressure is directly proportional to the volume. (a) How much work is done on the gas in the process? (Use any variable or symbol stated above along with the following as necessary: n and R.) w= (b) How is the temperature of the gas related to its volume during this process? (Do not substitute numerical values; use variables only. Use the following as necessary: n, R, V, Vi and P.)

Explanation / Answer

The work done can easily be determined if you first draw a diagram of this process; that is, a pVdiagram. You can see that the diagram will just be a diagonal linefrom (V, p) to (6V, 6p)

Work done is the area between this line and the V axis.This consists of a rectangle of height p and width 6V - V = 5V      or     area = 5 pV

and a triangle of height 6p - p = 5p and width 6V - V = 5V

or    area   = (1/2) x 5p x 5V = (25/2) pV

So the work done BY the gas is 5 pV + (25/2) pV = 17.5 pV

The work done ON thegas is = - 17.5 pV

Using the ideal gas law, pV = nRT , we can see that temp is proportional to the product pV. But Since p is proportional to V, then

T = (Pi / nRVi) V2

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