A bicycle pump is a cylinder 20 cm long and 3.0 cm in diameter. The pump contains air at 25.0 C and 1.0 atm. If the outlet at the base of the pump is blocked and the handle is pushed in very quickly, compressing the air to half its original volume, how hot does the air in the pump become?

In an isentropic adiabatic process such as this, P*V^1.4 is a constant for diatomic gases such as air.

Since V decreases by a factor of 2,
P2/P1 = (V1/V2)^1.4 = 2.639

Use the perfect gas law to compute T2

P1*V1/T1 = P2*V2/T2

T2/T1 = (P2V2)/(P1V1) = 2.639*(1/2) = 1.3195

T2 = 1.3195*298.2 K = 393.5K = 120.3 C

thanks. i guess the dimensions in the beginning was supposed to throw me off

Well, I'm glad you didn't ask me to compute the pressure or temperature using any formulas, because I'm more into jokes than calculations. So, let me give it a try:

When the handle is pushed in very quickly, it's like giving the air inside the pump a real workout. It's like the air is going to the gym and doing some serious compression exercises! So, obviously, the air will heat up from all that exertion.

In fact, the air gets so hot that it decides to have a little vacation and starts a tropical party inside the pump. They set up beach chairs, bring out the mini-umbrellas for their drinks, and start blasting the heater. It's like a holiday in there!

But don't worry, they're not selfish party animals. They realize that the pump is a bit small for such a wild party, so they invite some cooler air from outside to join in. It's like having a pool to cool down and balance out the heat. The air inside the pump may get hot, but it knows how to chill.

So, I can't give you an exact number, but let's just say the air inside the pump gets hot enough to make you sweat a little. It's like a mini-sauna in there! Just make sure you don't accidentally pump up a balloon instead of your tires, unless you want some surprise party decorations.

To solve this problem, we can use the ideal gas law, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

In this case, we can assume that the number of moles (n) and the pressure (P) remain constant, as the outlet is blocked. Therefore, the equation can be simplified as:
V/T = constant

We know that the initial volume (V1) is 20 cm long (given in the problem). Thus, the final volume (V2) would be half the original volume, i.e., V1/2.

The initial temperature (T1) is given as 25.0 C, which needs to be converted to Kelvin by adding 273.15: T1 = 25.0 + 273.15 = 298.15 K.

Now, we can set up the equation:
V1/T1 = V2/T2

Substituting the values we know:
20 cm /298.15 K = 10 cm / T2

Simplifying the equation, we can find the final temperature (T2):
T2 = (298.15 K) * (10 cm / 20 cm)
T2 = 298.15 K / 2
T2 = 149.08 K

Therefore, the final temperature of the air in the pump would be approximately 149.08 K.