Problem about adiabatic expansion of gases

AI Thread Summary
The discussion focuses on the calculations for the heat, internal energy, and work of 1 mol of hydrogen undergoing a reversible adiabatic expansion. Initially, the user calculated the final temperature incorrectly, leading to discrepancies in the internal energy and work values. After receiving feedback, they corrected the temperature calculation and recognized that the change in internal energy is negative, while work is positive. The final results confirmed that heat is zero, internal energy is -3873.4 J, and work is 3873.4 J. The user expressed satisfaction with the corrections and understanding of the concepts involved.
Rujano
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Hello everyone. I would like to know if I did this correctly.

Homework Statement



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



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

becuase it's an ADIABATIC expansion

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.

Thanks in advance!
 
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" (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.
 
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?
 
Much better!
Except the internal energy U is not negative; the CHANGE in U is negative.
 
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
 
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.
 
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