TERMODYNAMICS - gas leaking from container at T=const

[Niksha]

Homework Statement

Tank contain an ideal gas. Known is type of gas, temperature of gas T, volume of tank V=const, pressure of gas in tank p1, and the temperature of surrounding air T (same as temperature of gas in the tank). That is state 1.

Tank has small valve that was not closed completely. Gas was leaking out very slowly, so that temperature in the tank was all of the time equal to T. After valve was closed, we noticed that pressure in tank is now p2<p1. So, new state of gas i T, V, and p2<p1.

So this is isotermic (T=const) change of gas state, but with mass m that is not constant.

We need to calculate amounth of heat (Q) exchanged in this process (heat given to the gas in the tank while leaking), change of inside energy (U2-U1) of gas in the tank, and work (W).

Homework Equations

The Attempt at a Solution

I have solved this example, but I'm not sure that it is correct. I have used formula for enthropy chanage and the formula for the second law:

dS = - R (dp/p)
dQ = T dS

From that we have:

dQ = - T R (dp/p)

Mass m is depending of pressure (T,V and R are constant for this problem) so I used state equatation:

m = (p V)/(R T)

Now, I got:

dQ = - V dp

Integration gives:

Q = - V (p2 - p1) = V (p1 - p2)

And that's it. I don't know is that correct. Also I don't know how to calculate (U2- U1) and W.



If anybody has stumbled to similar example , or knows solution to this one let me know.

Thanks.

Niksha.

 

[quietrain]

erm, i think its like this, i am probably wrong? i don't know

since number of moles n is changing, assuming volume of gas is constant,

p₁V / n₁ = p₂V / n₂

where n₁ is initial number of moles of gas
n₂ is number of moles of gas left after leak

so p₁ / p₂ = n₁ / n₂

heat in, for reversible isothermal, is given by q_rev = n₁RTln {p₁ / p₂}

so q = nRT ln {n₁ / n₂} = nRT ln {m₁ / m₂}


for part 2, although it is isothermal, the number of moles is changing

U = 3/2 nRT , so change in U = (3/2)RT [n₂-n₁]

lastly, the work done on the gas can be obtained by 1st law of thermodynamics

change in U = q_in + w_on


i am not sure whether the above is correct, someone might want to double check

 

[rude man]

This is not a reversible process. This falls under "free expansion":

dW = pdV so work = 0 since ∆V = 0.

So by 1st law,
dU = dQ.

For ideal gas, dU = Cv*dT, Cv = n*cv. cv is specific heat at constant volume and depends on the gas in question ( ~ 3R/2 for monatomic gas, 5R/2 for diatomic gas except H2).

So U2 - U1 = cv*(n2 - n1)T.

But n = pV/RT for ideal gas, so we can write
U2 - U1 = cv*(V/RT)*(n2 - n1)T or
U2 - U1 = cv*(V/R)*(p2 - p1) < 0

Then, Q2 - Q1 = U2 - U1 < 0

 

[quietrain]

why is this considered free expansion?

if the gas is leaking out, it has to do work against the external pressure?

 

[quietrain]

oh isee,

is it because we don't care about the volume of the gas that has leaked, and only take the volume in the tank which is constant, so work done is 0.

so for reversible isothermal process, the change in internal energy is 0. but since this question changes the number of moles of the gas, so its internal energy will also change in relation to the number of moles?

with regards to dU = Cv*dT, if T is constant, then wouldn't dU be 0?

 

[rude man]

Confusing, I admit. With me adding to the confusion. dU = 0 isothermally only if the amount of the gas (n) stays the same. Instead of writing dU = CvdT I should have written it as

dU/dT = Cv = n*cv. If n changes, so does n*cv.

dU/dT = n*cv is a differential equation. When integrated, assuming cv independent of temperature, U = n*cv*T + constant. If we talk about differences in U, then the constant does not matter & can be = 0 (it actually can always be = 0 unless a chemical reaction or "state of aggregation" takes place). So,

U1 = n1*cv*T
U2 = n2*cv*T
U2 - U1 = cv(n1 - n2)T
etc.

 

[quietrain]

ah i see thank you!

 

[Niksha]

Thanks guys. I'm puting this on paper so I can check solutions. I have managed to do something on my own and I have found similar example in one book (this one is with p=const). I will check everything than post back.

See ya!

 

[rude man]

Big 10-4, Niksha!

 

[Niksha]

Ok, I'm pretty sure now that

Q = V (p1 - p2)

and I agree with

U2 - U1 = cv T (m2 - m1) = ...which gives... = - (p1 - p2)(cv V)/R.

Folloving the same approach, I have calculated the work

----
dw = p dv multiplying with mass m

dW = m p dv = m p d( (RT)/p) = ( (pV)/(RT) ) p d( (RT)/p ) = p*2 ( V/(RT) ) R T d(1/p) =
= p*2 V d(1/p) = p*2 V (- (dp)/(p*2) ) = - V dp

W = - V (p2 - p1) = V (p1 - p2)
----

But, if we check this with first law, it is not good!

But, if I make calculation with same oproach for 1kg, that is, without mass m , results fit in equatation of the 1. law.

So, do I making mistake, or the 1. law of termodynamics cannot be used only for container because mass is changing?

 

[rude man]

niksha - my post of sept 17 does obey the 1st law!
Q = ΔU + W with
W = 0