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

1. Nov 2, 2007

### Bill Foster

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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?

2. Nov 2, 2007

### 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.

3. Nov 2, 2007

Show me.

4. Nov 2, 2007

### 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 $$c_{V} + R = c_{P}$$

Last edited: Nov 2, 2007
5. Nov 2, 2007

### 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.

6. Nov 2, 2007

### 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!

7. Nov 3, 2007

### 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.