Solving for final temp of adiabatic ideal air expansion

  • #1
takeachance
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Homework Statement


Given a problem where air expands adiabatically with a given initial temperature, initial pressure, final pressure, and gamma (ratio of specific heats) , how would one go about solving for the final temperature? I'm just eager to get a general sense of direction.

Homework Equations


##PV^\gamma = constant##
(but it doesn't feel particularly useful since volume isn't provided)
##\frac{P_i V_i}{T_i}=\frac{P_f V_f}{T_f}##
(but again, I'm not sure what to do in the absence of volume)

The Attempt at a Solution


Just looking for a sense of direction. I guess that either the volume can be ignored/cancelled or there is an equation that doesn't require volume.
 

Answers and Replies

  • #2
stockzahn
Homework Helper
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Homework Statement


Given a problem where air expands adiabatically with a given initial temperature, initial pressure, final pressure, and gamma (ratio of specific heats) , how would one go about solving for the final temperature? I'm just eager to get a general sense of direction.

Homework Equations


##PV^\gamma = constant##
(but it doesn't feel particularly useful since volume isn't provided)
##\frac{P_i V_i}{T_i}=\frac{P_f V_f}{T_f}##
(but again, I'm not sure what to do in the absence of volume)

The Attempt at a Solution


Just looking for a sense of direction. I guess that either the volume can be ignored/cancelled or there is an equation that doesn't require volume.

For an isentropic change of state there is a correlation between ##p## and ##T## depending on ##\gamma## in the form of

$$\left(\frac{p}{T}\right)^{f\left(\gamma\right)} = const.$$

You can find this exponent ##f\left(\gamma\right)## very easily in the internet.
 
  • #3
takeachance
6
0
For an isentropic change of state there is a correlation between ##p## and ##T## depending on ##\gamma## in the form of

$$\left(\frac{p}{T}\right)^{f\left(\gamma\right)} = const.$$

You can find this exponent ##f\left(\gamma\right)## very easily in the internet.
Just to double check, ##f(γ)## is perfectly okay as a constant, yes? I guess when I was searching for this I wasn't using the right keywords, because this is my first time seeing P and T related in such a manner in recent memory. Thanks for the help.
 
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  • #4
stockzahn
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Just to double check, ##f(γ)## is perfectly okay as a constant, yes? I guess when I was searching for this I wasn't using the right keywords, because this is my first time seeing it in recent memory. Thanks for the help.

First of all, I just saw that I made a mistake, sorry. Of course the correlation should be ## Tp^{f\left(\gamma\right)} = const## similar to the correlation you posted

And yes, the function of ##\gamma## is constant for an ideal gas. If you are searching for "isentropic change of state" or "isentropic process" you should find what you are looking for.
 
  • #5
takeachance
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First of all, I just saw that I made a mistake, sorry. Of course the correlation should be ## Tp^{f\left(\gamma\right)} = const## similar to the correlation you posted

And yes, the function of ##\gamma## is constant for an ideal gas. If you are searching for "isentropic change of state" or "isentropic process" you should find what you are looking for.
Just a quick search reveals this website that suggests that ##TP^{(1-\gamma)/\gamma} = constant## and not just ##TP^{\gamma} = constant##. Any thoughts?

Messing around with some quick numbers seems to suggest that ##TP^{\gamma} = constant## gives an absurdly high constant.
 
  • #6
stockzahn
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Just a quick search reveals this website that suggests that ##TP^{(1-\gamma)/\gamma} = constant## and not just ##TP^{\gamma} = constant##. Any thoughts?

Just a quick search reveals this website that suggests that ##TP^{(1-\gamma)/\gamma} = constant## and not just ##TP^{\gamma} = constant##. Any thoughts?

Messing around with some quick numbers seems to suggest that ##TP^{\gamma} = constant## gives an absurdly high constant.

##f\left(\gamma\right)## can stand for each function containing ##\gamma##.
 
  • #7
takeachance
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##f\left(\gamma\right)## can stand for each function containing ##\gamma##. I use the correlation is: ##TP^{\gamma/(\gamma-1)} = constant##. Maybe those are equivalent, I didn't check.
Well, I think that get's me to where I need to be. Regardless, all of the online sources I've consulted suggest using ##TP^{((γ−1)/γ)}## or some variation, but this compared to your preferred one seem to yield different exponents.
 
  • #8
stockzahn
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##f\left(\gamma\right)## can stand for each function containing ##\gamma##.

Yes, sorry again, your first proposal, according to the website you quoted was right. It's ##Tp^{(1-\gamma)/\gamma}##, that's the correlation you can use to solve the problem. You can find these correlation for all "special" changes of states (isochor, isothermal, ...).

Edit: Really sorry for my mistakes, no excuse for that. Next time I concentrate, when responding...
 
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  • #9
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What you do is substitute ##V=nRT/P## into your original equation. Then you have an equation involving only P and T.
 
  • #10
takeachance
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What you do is substitute ##V=nRT/P## into your original equation. Then you have an equation involving only P and T.
Thanks, I appreciate an alternate approach, but I don't know how much of the gas I have. I don't think I would be able to deal with n. Does your suggestion still work without knowing or having a way to solve for n?
 
  • #11
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Thanks, I appreciate an alternate approach, but I don't know how much of the gas I have. I don't think I would be able to deal with n. Does your suggestion still work without knowing or having a way to solve for n?
n is a constant.
 
  • #12
takeachance
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n is a constant.
I thought n was the number of moles of gas you have and R was a gas constant?

Edit: Do you mean that the n is constant in the situation (unchanging during expansion) so the n will eventually not be an issue?
 
  • #13
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Yes. That's exactly what I mean. It will be absorbed into the constant to form a new constant C'
 
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