Finding Final temp for Adiabatic process

In summary: I ON THE RIGHT TRACK HERE?In summary, the air in the tire becomes significantly warmer after it has been pumped up.
  • #1
speny83
17
0

Homework Statement



A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Tf 287K . The final pressure in the tire is Pf = 3.75 bar.

Calculate the final temperature of the air in the tire. Assume that CV,m = 5R/2.

Homework Equations


dU=nCvmdt
dU=dw...as q=0 adiabaitc
dw=-Pexdv
using these above equations my notes derive the following
ViTic=VfTfc where c=Cvm/R
PiVi[itex]\gamma[/itex]=PfVf[itex]\gamma[/itex] where gamma=Cpm/Cvm

The Attempt at a Solution



I am really stuck on where to even start this problem. It seems as if i am missing info, that i need a number of moles or a volume or something. I can't seem to find a way to rearange any of the above. In other problems it boiled down to using the boring old ideal gas pv=nrt to solve for one variable but again this leaves me with 2 unknows so I am not sure what to do.

Any tips, hints, explanations anything would be a great help![itex]\gamma[/itex]
 
Last edited:
Physics news on Phys.org
  • #2
can i say that So I am thinking can i say that
(Tf/Ti)^c=Vi/Vf and (Pf/Vf)^(1/gamma)=ViVf

so that (Tf/Ti)^c=(Pf/Vf)^(1/gamma)


is this legit?
 
  • #3
speny83 said:
can i say that So I am thinking can i say that
(Tf/Ti)^c=Vi/Vf and (Pf/Vf)^(1/gamma)=ViVf

so that (Tf/Ti)^c=(Pf/Vf)^(1/gamma)


is this legit?
You have a couple of typos in there. Rewrite it more carefully.
 
  • #4
speny83 said:

Homework Statement



A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Tf 287K . The final pressure in the tire is Pf = 3.75 bar.

Calculate the final temperature of the air in the tire. Assume that CV,m = 5R/2.

Homework Equations


dU=nCvmdt
dU=dw...as q=0 adiabaitc
dw=-Pexdv
using these above equations my notes derive the following
ViTic=VfTfc where c=Cvm/R
PiVi[itex]\gamma[/itex]=PfVf[itex]\gamma[/itex] where gamma=Cpm/Cvm

The Attempt at a Solution



I am really stuck on where to even start this problem. It seems as if i am missing info, that i need a number of moles or a volume or something. I can't seem to find a way to rearange any of the above. In other problems it boiled down to using the boring old ideal gas pv=nrt to solve for one variable but again this leaves me with 2 unknows so I am not sure what to do.

Any tips, hints, explanations anything would be a great help![itex]\gamma[/itex]
Try expressing the adiabatic condition in terms of T and P. Think of the air in the tire at the end as a compression of a much greater volume of air at atmospheric pressure into the final volume of the tire.

AM
 
  • #5
= \frac{C_p}{C_v}

To solve this problem, we need to use the ideal gas law and the equations for adiabatic processes.

First, we can use the ideal gas law to find the initial volume of air in the tire. We know the initial pressure (Pi) and temperature (Ti), so we can rearrange the ideal gas law to solve for volume (Vi).

Vi = \frac{nRTi}{Pi}

Next, we can use the equation for adiabatic processes to find the final temperature (Tf). We know the initial and final pressures, as well as the specific heat ratio (gamma), so we can substitute these values into the equation and solve for Tf.

Tf = Ti * \left(\frac{Pf}{Pi}\right)^{\frac{\gamma-1}{\gamma}}

Finally, we can use the ideal gas law again to find the final volume (Vf). We know the final pressure (Pf) and final temperature (Tf), so we can rearrange the ideal gas law to solve for volume (Vf).

Vf = \frac{nRTf}{Pf}

Now, we can plug in the values we have found for Vi and Vf into the equation ViTic=VfTfc, where c=Cvm/R, to solve for the number of moles (n).

n = \frac{PiVi}{RTi} * \frac{Ti}{Tf} * \frac{Pf}{Vf}

Once we have found the number of moles, we can use the equation dU=nCvmdt to find the change in internal energy (dU). Since this process is adiabatic, we know that q=0 and dw=-Pexdv, so we can substitute these values into the equation and solve for dU.

dU = nCvm(Tf-Ti)

Finally, we can use the equation dU=dw to find the final temperature (Tf). We know that dw=-Pexdv, and we can solve for the change in volume (dv) by rearranging the equation for adiabatic processes.

dv = \frac{Vf-Vi}{\gamma}

Substituting this into the equation for work, we get:

dw = -PiVi\left(\frac{Vf}{Vi}\right)^{\gamma-1}

Now we can equate d
 

1. What is an adiabatic process?

An adiabatic process is a type of thermodynamic process in which no heat enters or leaves the system. This means that the process is carried out without any transfer of thermal energy between the system and its surroundings.

2. How do you calculate the final temperature for an adiabatic process?

The final temperature for an adiabatic process can be calculated using the adiabatic equation: T2 = T1 * (P2/P1)^((γ-1)/γ), where T1 and T2 are the initial and final temperatures, P1 and P2 are the initial and final pressures, and γ is the ratio of specific heats for the gas.

3. What is the significance of an adiabatic process in thermodynamics?

Adiabatic processes are important in thermodynamics as they can help us understand the behavior of gases and how they exchange energy with their surroundings. They also have practical applications in engineering and the study of atmospheric processes.

4. How does the final temperature change in an adiabatic process?

In an adiabatic process, the final temperature depends on the initial temperature, pressure, and the properties of the gas. The final temperature can either increase or decrease, depending on the specific conditions of the process.

5. What are some real-life examples of adiabatic processes?

Some examples of adiabatic processes include the compression or expansion of a gas in a piston, the heating or cooling of gas in a closed container, and the expansion of air in a bicycle pump.

Similar threads

Replies
1
Views
498
Replies
22
Views
2K
  • Classical Physics
Replies
1
Views
861
  • Advanced Physics Homework Help
Replies
5
Views
934
  • Biology and Chemistry Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
25
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Replies
8
Views
24K
  • Biology and Chemistry Homework Help
Replies
4
Views
6K
  • Classical Physics
2
Replies
61
Views
5K
Back
Top