vela said:
I have to admit I'm having a great deal of difficulty understanding what you're thinking. I'm not sure if you're just not expressing yourself clearly, but to me, some of your questions and statements seem to reveal really fundamental misconceptions about thermodynamics.
Let's start over. Here's what you said was the original problem:
Then later you say this:
In the first problem, you have an isobaric expansion from A to B. In the second problem, you have an isothermal expansion from A to B. Why would you expect applying the exact same steps of the second problem would give you the correct answer for the first problem? They are different problems.
In the second problem, you got the correct answer because you applied a formula for an isothermal process and the process connecting A to B was an isothermal expansion. In the first problem, however, you tried to use the same formula even though the expansion from A to B was isobaric. The formula won't work here because the expansion is not isothermal. This is what I was getting at in my reply, yet you seemed to completely miss that.
As you noted later, for an isobaric process, the work is given by W=P \Delta V. Unfortunately, you're not given the pressure or volumes, so you might want to look for a more appropriate formula for the work done by an isobaric expansion (or derive it using the ideal gas law).
Once you get this part, we can move on to part (ii).
I'm terribly sorry, I think I've delieved my question wrong which confused you. What I meant was apart from the question I posted at the beinging of the post I was also given another set of question to work with, which is:
one mole of an ideal monoatomic gas doubles its initial volume in an isothermal expansio from state A to state B. The gas is then compressed isobarically to state C and finally heated isochorically until it returns to state A.
a) If state C corresponds to a pressure p=4 atm, and temperature T = 1069K, determine the temperature of the gas in state B.
I worked out temerature of B by:
P. V = nRT
SO (P.V)/T = n . R Where R is a constant
SO (P(B)*V(B))/T(B) = (P(C)*V(C))/T(C)
SO T(C) * (P(B)*V(B))/T(B) = P(C)*V(C)
SO T(C) * V(B))/T(B) = (P(C)*V(C)) * T(B)
SO T(B) = T(C) * P*(B)/P(C) * V(B)/V(C)
Therefore TB = 1069 X 1 X 2 (2 because volume doubles)
= 2138 K
b)
i) A > B
For this part I use the following equation:
I know W = delta U, I also know U = P*V
Therefore W = P/V, BUT I know P is calculated from n.R.T
SO W= (intergration)n.R.T/V.dv
SO n.R.T(B).(intergration) 1/V. dv
SO n.R.T(B).ln(V(B)/V(A))
n = 1 (one mole of ideal monoatomic gas)
R = 8.314
T(B) = 2138
ln(V(B)/V(A)) = ln(2)
Therefore n.R.T(B).ln(V(B)/V(A)) = 1*8.314*2138*ln(2) = 12321J
ii) B > C
W=n.R.(T(B)-T(C))
I use (T(B)-T(C)) because it is compress to state C from state B, when there is compression made temperature decrease (whearse if it is expand to state C from state B, I would use (T(B)+T(C))
=1*8.314*(2138-1069)
=8887.7J
iii) C > A
0
These are the steps I took for this particular question, which somehow I got it all right. Therefore I used these steps again for the question I posted originally in this forum. WHICH gives me a wrong answer... I don't know why...
