IIT JEE , AIEEE Forum - Physics , Thermal Physics

joyfrancis

demonstrate that the process in which the work performed by an ideal gas is proportional to the corresponding increment of its internal energy is described by the equation pvⁿ= const., where n is a constant.

Asmita

Given that w/

U=k where k is a constant.

Now

U=q+w

Dividing by

U ,1=q/

U +k

Since

U varies so the only possibility is q=0

So the system is adiabatic.

Hence PV^

is a constant.

umang

pls explain why q has to be zero ??

1=q/dU + k . So , why q has to be zero ?

Pls explain -

Since

U varies so the only possibility is q=0

joyfrancis

according to ashmita k = 1

but in an adiabatic process

w +

U = 0

w/del.U = -1 .....not 1

The first law is violated!!!!

Whats going on???

kiran

Hi

Proof that PV^g = constant for an ideal gas undergoing an adiabatic process where g is the ration of the molar or specific heat capacities. We use molar heat capacities g = c _p/c_v since R = c_p - c_v.

First start with the differential form of first law of thermodynamics applied to an adiabatic process,

In general we showed that for any type of process (for an ideal gas)

Combining the last two equations,

Replacing the pressure using the ideal gas law, p = nRT/V

Substituting R = c _p - c _v and integrating both sides,

Raising both sides to the e ^th power we obtain,

This is a useful relation between the temperature and volume for an adiabatic process all by itself.

Using the ideal gas law we can eliminate the temperature we can convert this last equation into the form for which we are looking.

Thus we see that an adiabatic process is a polytropic process with n = g.

Cheers!

joyfrancis

adiabatic process??...where is it given that the process is adiabatic?...

priyesh

yes joyfrancis u're correct it's not given that the process is adiabatic.

by the way i read this question in i.e irodov's book.

joyfrancis

solve it dude

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