Thermodynamics, gas expansion

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fluidistic
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Homework Statement


Two moles of a monoatomic ideal gas expand adiabatically from a temperature [itex]T_i=0°C[/itex] and a pressure of 1 atm until the temperature reaches [itex]T=-50°C[/itex].
1)What are the initial and final volumes and the final pressure of the system?
2)Calculate the work done by the gas. What are the initial and final internal energies of the gas?

Homework Equations


I)PV=NRT
II)[itex]U=\frac{3RT}{2}[/itex].
III)[itex]PV^{3/2}=constant[/itex]

The Attempt at a Solution


1)Using I), I get that [itex]P_f V_f=3707.81J[/itex].
Using III), I get that [itex]P_f V_f ^{3/2}=P_i V_i ^{3/2}=45400J[/itex].
Then I isolated [itex]P_f[/itex] from I), plugged into III) and isolated [itex]V_f[/itex] to get [itex]V_f=3.5 m^3[/itex].
This gives me [itex]P_f \approx 1059.61 Pa[/itex] or 0.01 atm. The answer of another student is 0.6 atm. What did I do wrong?!
2)Since the process is adiabatic, there's no heat exchenge of the system with the environment, the change in internal energy is worth the work done on the gas. Using II), I get that the work done by the system is worth 297.9 cal (multiply by 4.186 to get joules). Initial internal energy[itex]=6808.67J[/itex] and final internal energy=[itex]5561.67J[/itex]. Again, the difference is the work done.
Are these numbers correct for part 2)?
 

Answers and Replies

  • #2
I like Serena
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Hi fluidistic!

Your formula (III) is off.
The power should be 5/3 (γ=Cp/Cv).

And let's see.
If I understand your problem statement correctly, you have:
$$\begin{array}{|c|c|c|}\hline
& initial & final \\
\hline
T & 273.15\textrm{ K} & 223.15\textrm{ K}\\
P & 1\textrm{ atm} & ?\\
V & ? & ?\\
n & 2\textrm{ mol} & 2\textrm{ mol} \\
\hline \end{array}$$

You seem to start from the final state.
But how did you get that first result from (III)? :confused:
 
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  • #3
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using the data table that 'i like serena' created, you can solve for the volume in the initial state by doing v = nRT/P where R=.08205 (the pressure constant for atms)

then you can use PiVi = PfVf because you know the values for all except Vf
i got very different numbers than you did in the first part


for part two the formula W = -PΔT, which gives you an answer in joules
from this method, since P = 1 and T = -50K, I would think the work done would be equal to fifty joules
 
  • #4
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then you can use PiVi = PfVf because you know the values for all except Vf

I'm afraid this is not true, since the expansion is adiabatic.
 
  • #5
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I'm afraid this is not true, since the expansion is adiabatic.

but if its adiabatic doesn't that mean that there is no heat lost to the environment? and wouldn't that make the conditions of the system ideal, meaning that the standard equations are applicable?

I haven't been formally taught thermodynamics yet if there is a fallacy in my understanding I appreciate you helping me out! thank you
 
  • #6
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but if its adiabatic doesn't that mean that there is no heat lost to the environment? and wouldn't that make the conditions of the system ideal, meaning that the standard equations are applicable?

I haven't been formally taught thermodynamics yet if there is a fallacy in my understanding I appreciate you helping me out! thank you

Yep. No heat lost. Heat being different from temperature.
This means ##P_i V_i^\gamma=P_f V_f^\gamma##, where ##\gamma=\frac {C_p} {C_v}##.

See for instance:
http://en.wikipedia.org/wiki/Table_of_thermodynamic_equations#Equation_Table_for_an_Ideal_Gas
 
  • #7
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Ok that makes sense since Q is the quantity of heat in Joules, I can see how that differs from a temperature in Kelvin

But for the very first part of the problem, I did:
PV = nRT
V = (2*.o8205*273)/1
Vi = 44.8 L , or , .0448cubic meters
is this correct?
 
  • #8
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Well, I'd like to give the OP, fluidistic, a chance to solve the problem, before I try to help you solve the problem...
 
  • #9
fluidistic
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Ok thanks a lot for the 5/3 correction.
I've redone everything.
First I calculate [itex]V_i=0.0454 m^3[/itex]. I reached this from I).
So now I calculate [itex]P_iV_i^{5/3}=577.59=V_fP_f^{5/3}[/itex] (*).
From I), [itex]V_fP_f=NRT_f=3707.775[/itex]. I isolate [itex]P_f[/itex] and plug this into (*).
I reach [itex]V_f=0.1558m^3[/itex]. So that [itex]P_f=23801.6 Pa \approx 0.238 atm[/itex].
Still not the answer from the other student. Did I make another mistake?
 
  • #10
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Looks good, except you seem to have swapped Vf and Pf around in (*).
 
  • #11
fluidistic
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for part two the formula W = -PΔT, which gives you an answer in joules
from this method, since P = 1 and T = -50K, I would think the work done would be equal to fifty joules
The problem with the formula you give is that P isn't a constant. So what would you put in "W = -PΔT"?
I think I got that part right in my first post, just waiting for someone to confirm. :)
Basically the work done by the gas is worth the loss of internal energy of the gas. (The gas had to spend its internal energy to do some work).
Edit: To IlikeSerena: Haha, thanks. Yeah indeed, a typo mistake here while posting, on my draft it's correct.
 
  • #12
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I think I got that part right in my first post, just waiting for someone to confirm. :)
Basically the work done by the gas is worth the loss of internal energy of the gas. (The gas had to spend its internal energy to do some work).

Yep. That part is correct. :)
 
  • #13
fluidistic
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Ok thank you very much I Like Serena!!!
If the other guy wants some help, welcome.
 
  • #14
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Edit: To IlikeSerena: Haha, thanks. Yeah indeed, a typo mistake here while posting, on my draft it's correct.

So did you find the correct answer?
Because it seems you have somehow plugged in your results incorrectly, since I also get 0.6 atm.
 
  • #15
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Ok that makes sense since Q is the quantity of heat in Joules, I can see how that differs from a temperature in Kelvin

But for the very first part of the problem, I did:
PV = nRT
V = (2*.o8205*273)/1
Vi = 44.8 L , or , .0448cubic meters
is this correct?

So if you hadn't already noticed, this is correct. :)
 
  • #16
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yes, I comprehend that much but its the process after that which is causing me troubles

I've never seen the formula with the 5/3 in it, I'll have to do some more research on the topic, and I'm not sure to what depth AP Physics B gets into thermodynamics, considering there's only three formulas given on the official formula sheet

thanks for your help anyhow!
 
  • #17
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Oh well, if you know how to solve a differential equation, you can derive it yourself from formulas (I) and (II), combined with the fact that the change in internal energy equals the work done (in an adiabatic process of an ideal gas).
 
  • #18
fluidistic
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Yes thanks ILS, I get the 0.6 atm answer and P final about 0.0615 m³.
I wonder why we don't usually use the fundamental relation for ideal gas instead of the 2 typical state functions.
 
  • #19
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Yes thanks ILS, I get the 0.6 atm answer and P final about 0.0615 m³.
I wonder why we don't usually use the fundamental relation for ideal gas instead of the 2 typical state functions.

Which state functions do you mean?

In thermodynamics you typically have a long list of relations and formulas.
They can be derived from a very small set, but you have to be careful when to apply a formula and check if all conditions apply.

As for the ideal gas relation, most real substances are not ideal gasses.
And for real substances, you usually have to consult tables.
 
  • #20
fluidistic
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Which state functions do you mean?

In thermodynamics you typically have a long list of relations and formulas.
They can be derived from a very small set, but you have to be careful when to apply a formula and check if all conditions apply.

As for the ideal gas relation, most real substances are not ideal gasses.
And for real substances, you usually have to consult tables.

The fundamental equation of a monoatomic ideal gas is given by [itex]S=\underbrace { \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0}_{S_0}+NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ][/itex].
From my understanding this equation is equivalent to the Lagrangian in classical mechanics. In other words, with this equation alone, if one is given [itex]S_0[/itex] (initial conditions), one has all possible information (equation of motion in CM) about the monoatomic ideal gas.
So from that equation, one has more or at least equally -not clear to me yet... :( - information than just the 2 equations of state: [itex]PV=NRT[/itex] and [itex]U=\frac{3NRT}{2}[/itex]. In theory, one needs 3 equations of states to get all the information of a thermodynamics system, whereas here 2 were enough to solve the problem.
 
  • #21
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I guess that would be the fundamental equation of a monoatomic gas "in the entropy representation".
There are others for the other representations.

To be honest, I've haven't learned them, so I'm not aware of their importance.

I usually start from the equations that are always true:
dU=dQ+dW
dU=TdS-PdV
(Conservation of energy)

And for an ideal gas:
PV=NRT
U=NCvT
Cv=(k/2)R, where k is the number of freedoms (equipartition theorem: k=3, 5, or 6).
 
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