problem 5utaphdqualifving exam problemfebruary2010 the helmholtz function f of a non ideal gas i

(a) The fundamental relation dF=-SdT-PdV gives us S(V,T) =mKV1/2T2 and P(V,T)=nKV1/2T2.

Problem 5utaphdqualifving exam problemfebruary2010 the...

The Helmholtz function F of a non-ideal gas is given by F(V,T) = -KV^(1/2)T^3 where K is a positive constant and the other symbols have their usual meaning. (a) Use the fundamental relation dF = -SdT - PdV to show that the entropy S and pressure P are given by S(V,T) = mKV^(1/2)T^2 and P(V,T) = nKV^(1/2)T^2. (b) Calculate the work done by the gas W, the heat transferred Q, and the change in internal energy U for a reversible isothermal expansion from initial volume Vo to final volume 9Vo at temperature To. (c) Obtain the final temperature Tr in a reversible adiabatic expansion from initial temperature T and initial volume Vo to final volume 9Vo. Express your answer in terms of only To.

Transcript

00:01 So in this question the initial volume the initial volume of the monotomic ideal gas is given as v0 equals to 0 .15 meter cube the final volume is given as 0 .0 .8 meter cube the initial value of pressure is equal to 55 kios so cp will be equal to 5 r by 2 and cv will be equal to 3 r by 2 now the t0 t0 is the temperature at temperature at state 0 similarly t1 is the temperature at state 1 and t2 is the temperature at state 2 now the heat that the heat that is absorbed for the first subpart for the first subpart we have to identify the pv diagram so this will be the pv diagram this is the p axis this is the v axis this is the first process then the second process and this will be the third process now for subpart v we have to find the work done by the gas so work done by the gas work done by the gas is equal to p delta v that is equal to pressure is equal to 55 kilo pascal so 55 into 10 to the part 3 and will take the difference of the volumes so 0 .1 minus 0 .8, 0 .15 minus 0 .8.

02:09 This will be in jules.

02:10 Let's calculate the value.

02:12 55 into 10 to the power 3.

02:16 55 into 10 to the power 3 into 0 .15 minus 0 .8 gives us the value that is equal to minus 35 .75 kilojoules.

02:34 I'm sorry, this will be positive.

02:36 This will be positive because this is 0 .1 .0 .8 minus 0 .15.

02:44 This is 0 .8 minus 0 .15.

02:47 So the work done by the gas will be positive.

02:52 Now the heat absorbed.

02:55 The heat absorbed by the gas will be equal to ncp t1 minus t0.

03:03 And this is equal to n into 5 r by 2 into t1 minus t0.

03:11 This will be equal to 5 by 2 into nr t1 minus nr t0.