Calculating Temperature Rise of Air in Bicycle Pump

AI Thread Summary
The discussion focuses on calculating the temperature rise of air in a bicycle pump when compressed adiabatically. The initial conditions include a volume of 1.55 m³ at 21.0°C and 1.0 atm, with the final volume being 0.775 m³. Participants confirm that air behaves as a diatomic ideal gas with a specific heat ratio (gamma) of 1.4. The relevant equation for temperature change during adiabatic compression is T2 = T1 * (V1/V2)^(k-1). The conversation emphasizes the need to apply the correct thermodynamic principles to solve the problem.
yossup
Messages
28
Reaction score
0

Homework Statement



A bicycle pump is a cylinder 20cm long and 3.0cm in diameter. The pump contains air at 21.0C and 1.0atm. 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?

Homework Equations



PV=nRT
PV^gamma = PV^gamma

The Attempt at a Solution



So the intial volume is 1.55m^3. The final volume is .775m^3. And because the handle is pushed very quickly, the process is adiabatic. So I want to know what gamma is but do I use the one for di or triatomic? Then i just use the W = integral P dv?
 
Physics news on Phys.org
yossup said:

Homework Statement



A bicycle pump is a cylinder 20cm long and 3.0cm in diameter. The pump contains air at 21.0C and 1.0atm. 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?

Homework Equations



PV=nRT
PV^gamma = PV^gamma

The Attempt at a Solution



So the intial volume is 1.55m^3. The final volume is .775m^3. And because the handle is pushed very quickly, the process is adiabatic. So I want to know what gamma is but do I use the one for di or triatomic? Then i just use the W = integral P dv?

Air is composed mostly of Nitrogen and Oxygen (which are diatomic molecules). Thus, the specific heat ratio of air is assigned the value of 1.4.

Since the air is compressed quickly, you may assume it is adiabatic. At relatively low pressures air is also considered to behave as an ideal gas. If you further assume it is a reversible process then you end up with an isentropic process.

Do you know of any relationships that contain the volume, temperature, and specific heat ratio for an isentropic process?

CS
 
Um...we haven't even gone over isentropic processes yet but is the equation your're talking about this? PdV + VdP = nRdT
 
yossup said:
Um...we haven't even gone over isentropic processes yet but is the equation your're talking about this? PdV + VdP = nRdT

T_2 = T_1 \cdot \left(\frac{V_1}{V_2}\right)^{k-1}

where k is the specific heat ratio.

CS
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top