Solving for final temp of adiabatic ideal air expansion

AI Thread Summary
To solve for the final temperature of air expanding adiabatically, the relationship between pressure and temperature can be utilized, specifically the correlation involving the ratio of specific heats, gamma. The equations discussed include the isentropic relation, which suggests that TP^((1-γ)/γ) remains constant during the process. It was clarified that the number of moles (n) can be treated as a constant, simplifying calculations. The key takeaway is to use the appropriate relationships that do not require volume to derive the final temperature. Understanding these principles allows for effective problem-solving in adiabatic processes.
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.
 
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takeachance said:

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

takeachance said:
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##.
 
stockzahn said:
##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.
 
stockzahn said:
##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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What you do is substitute ##V=nRT/P## into your original equation. Then you have an equation involving only P and T.
 
  • #10
Chestermiller said:
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
takeachance said:
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
Chestermiller said:
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
Yes. That's exactly what I mean. It will be absorbed into the constant to form a new constant C'
 
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