1. Sep 20, 2011

### Rujano

Hello everyone. I would like to know if I did this correctly.

1. The problem statement, all variables and given/known data

Calculate the heat, internal energy and work of 1 mol of hydrogen, which undergoes a reversible adiabatic expansion from a volume of 5.25 m^3 at 300 K to a volume of 25.5 m^3

3. The attempt at a solution

The first thing that I have to do is to find the final temperature using the following equation:

(Vf/Vi)^1-y * Ti = Tf

y = gamma

y = Cp/Cv

Because it's HYDROGEN (diatomic gas):

y = 7/2/5/2 = 1.4

(25.5 m^3/5.25 m^3)^1.4 * 300 K = Tf

Tf = 2741.9 K

so...

a) Heat

Q = 0 J

b) Internal energy:

E = n Cv (Tf - Ti)

Cv = 5/2 * R

E = 1 mol * (5/2) * 8.31 J/mol K * ( 2741.9 K - 300 K)

E= 50730.47 J

c) Work

I've seen the equation in two different ways:

E = Q + W

AND...

E = Q - W

Anyway, I'll use the second one (because is the one I have to use according to the worksheet)...

W = Q - E

W = 0 J - 50730.47 J

W = - 50730.47 J

Is it ok?

Please check it out, because the worksheet doesn't have the answers and I want to know if I did it correctly.

2. Sep 21, 2011

### rude man

" (25.5 m^3/5.25 m^3)^1.4 * 300 K = Tf "

Do you see an error here? You should be suspicious since you have computed an increase in E while at the same time having done positive work W ( = integral pdV). The 1st law doesn't like that if, as you correctly state, Q = 0 ... in the same vein, computing W as negative can't be right either ...

Tip - I would always use U = Q - W rather than U = Q + W, this defines W as the work done BY the system. This goes back to the early days of thermodynamics when steam engines were the fad and W was of course always positive. (Most people use U rather than E).

I would also add that the formulas for Cv are approximate - you can use the ideal gas law to get a precise answer without reverting to Cv.

3. Sep 28, 2011

### Rujano

Oh thank you. I see it now.

It was 1 - y and I forgot about doing this:

1 - 1.4 = - 0.4

So it plug into the equation the -0.4 instead of the 1.4 It get 113.6 K (final temperature)

So dT would be: 113.6 K - 300 K = - 186.4 K

That means that the Internal Energy is NEGATIVE

Therefore

U = q - w

q = 0

U = - w

w = - U

w = - (- 3873.4 J )

w = 3873.4 J

Is it ok now?

4. Sep 28, 2011

### rude man

Much better!
Except the internal energy U is not negative; the CHANGE in U is negative.

5. Sep 28, 2011

### Rujano

Thank you again! I can't believe that I skipped that.

So, the final correct answer would be that:

* q = 0

* U = -3873.4 J

* w = 3873.4 J

U negative and w positive

6. Sep 28, 2011

### rude man

It certainly would be. Except I haven't checked your algebra. But all your equations look right now. And Q = 0, ΔU < 0 and W = - ΔU are all most definitely correct.