Solving for final temp of adiabatic ideal air expansion

In summary: 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...What you do is substitute ##V=nRT/P## into your original equation. Then you have an equation involving only P and...
  • #1
takeachance
6
0

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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
Last edited:
  • #4
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.
 
  • #5
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.
 
  • #6
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##.
 
  • #7
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.
 
  • #8
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...
 
  • Like
Likes takeachance
  • #9
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'
 
  • Like
Likes takeachance

Related to Solving for final temp of adiabatic ideal air expansion

1. How is the final temperature of adiabatic ideal air expansion calculated?

The final temperature of adiabatic ideal air expansion can be calculated using the equation T2 = T1 * (P2/P1)^((γ-1)/γ), where T1 is the initial temperature, T2 is the final temperature, P1 is the initial pressure, P2 is the final pressure, and γ is the specific heat ratio of the gas.

2. What is the difference between adiabatic and isothermal ideal air expansion?

Adiabatic ideal air expansion is a process in which no heat is exchanged between the system and its surroundings, while isothermal ideal air expansion is a process in which the temperature remains constant. In adiabatic expansion, the final temperature will be lower than the initial temperature, while in isothermal expansion, the final temperature will be the same as the initial temperature.

3. Can the final temperature of adiabatic ideal air expansion be lower than 0 Kelvin?

No, the final temperature of adiabatic ideal air expansion cannot be lower than 0 Kelvin. This is because temperature is a measure of the average kinetic energy of particles, and 0 Kelvin represents the absence of any kinetic energy. Therefore, it is not physically possible to have a temperature lower than 0 Kelvin.

4. How does the specific heat ratio affect the final temperature of adiabatic ideal air expansion?

The specific heat ratio, represented by the symbol γ, is a constant value that depends on the type of gas being expanded. It is a measure of how much a gas can be compressed or expanded without changing its temperature. A higher value of γ means that the gas is less compressible, and therefore, the final temperature of adiabatic ideal air expansion will be higher.

5. What are some real-world applications of solving for the final temperature of adiabatic ideal air expansion?

Calculating the final temperature of adiabatic ideal air expansion is important in understanding the behavior of gases in various systems, such as internal combustion engines, gas turbines, and compressors. It is also crucial in designing and optimizing these systems for maximum efficiency and performance.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
915
  • Introductory Physics Homework Help
Replies
4
Views
843
  • Introductory Physics Homework Help
Replies
1
Views
962
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
954
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
962
  • Introductory Physics Homework Help
2
Replies
54
Views
3K
Back
Top