Thermodynamics: Amount of work required to increase pressure at constant volume

[Bill Foster]

Homework Statement

By means of a hand pump, you inflate a tire from 0 psi to 35 psi (overpressure). The volume of the tire remains constant at 3.5 ft^3. How much work must you do on the air pump? Assume each stroke of the pump is an adiabatic process and that the air is initially at STP.

Homework Equations

pV^\gamma=constant
dW=pdV
\Delta{Q}=nC_v\Delta{T}
pV=nRT

The Attempt at a Solution

I can't use dW=pdV because the change in volume is 0.

p_0V^\gamma=p_1V^\gamma

Since V is constant, that implies p_0=p_1

So I'm left with \Delta{Q}=nC_v\Delta{T} and pV=nRT

Combining them, I get \Delta{Q}=\frac{C_v\Delta{p}V}{R}

The problem is I don't know what C_v is.

Since the value isn't given in my book, I assume that I'm suppose to calculate it.

I know that C_p=\frac{5}{2}R for monotomic gas and C_p=\frac{7}{2}R for diatomic. I assume air is a mixture of both.

But when I plug in the numbers (R=8.314) it just doesn't come out correctly.

The correct answer is 8.1\times{10^3} ft\cdot{lb}=10982 J. Working backwards from that answer I can see that C_v=3.82.

So what am I missing?

 

[Shooting Star]

You have written that change of volume is zero. Change of which volume? The gas in the cylinder of the pump is getting compressed.

The air which is inside the tyre is definitely compressed, so there must have been a change in volume.

Now think which formula you can apply.

 

[Bill Foster]

Show me.

 

[dynamicsolo]

Here's the thing about this problem: you want to track the amount of gas being pumped into the tire. The air in the tire starts at 1 atm absolute (the pressures given in the problem are gauge pressures) and ends at 1 atm + 35 psi . The volume remains at 3.5 ft^3 . I think we may have to assume that while the process is adiabatic, the temperature of the air in the tire reaches equilibrium at STP. What you want to find is how many moles of air are injected into the tire and then find what volume that air originally occupied. That would give you a way of assessing the adiabatic work done in getting the air into the tire.

Air is treated as entirely diatomic, BTW (it's predominately nitrogen and oxygen). Also c_{V} + R = c_{P}

 

[Bill Foster]

Trying to calculate the volume of the air before it was put in the tire:

p_1V_1^\gamma=p_2V_2^\gamma

V_1=V_2(\frac{p_2}{p_1})^\frac{1}{\gamma}

W=\int{pdV}

p=p_1\frac{V_1^\gamma}{V^\gamma}

W=\int{pdV}=\int{\frac{p_1V_1^\gamma}{V^\gamma}dV}=\frac{1}{\gamma-1}p_1V_1(1-(\frac{V_1}{V_2})^{\gamma -1})

I'm still getting the wrong answer.

 

[Bill Foster]

Maybe I'm using the wrong value for \gamma.

All the problems in the book as well as the examples suggest \gamma = 1.4. So that's what I've been using.

Well, even though it's driving me crazy, I've got to forget about this. I'm taking the GRE subject test in physics tomorrow and I still have to go over electromagnetics and oscillations. It's been 20 years since I've had undergrad physics!

 

[Shooting Star]

Hi Bill,

Your calculation is correct, even though it's better to eliminate V1 from the final expression since that’s the unknown but that’s immaterial, since you are putting the value of V1 later. I converted everything to MKS and got the result as 437213 Joules. The arithmetic is very messy, and there’s no guarantee that the number is correct.

And gamma is 7/2 for air.