Finding Final temp for Adiabatic process

[speny83]

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=nC_vmdt
dU=dw...as q=0 adiabaitc
dw=-P_exdv
using these above equations my notes derive the following
V_iT_i^c=V_fT_f^c where c=Cvm/R
P_iV_i^$\gamma$=P_fV_f^$\gamma$ 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!^$\gamma$

 

[speny83]

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?

 

[DrClaude]

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.

 

[Andrew Mason]

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=nC_vmdt
dU=dw...as q=0 adiabaitc
dw=-P_exdv
using these above equations my notes derive the following
V_iT_i^c=V_fT_f^c where c=Cvm/R
P_iV_i^$\gamma$=P_fV_f^$\gamma$ 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!^$\gamma$

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

 

[DeeNos]

= \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