Calculating Temperature Rise of Air in Bicycle Pump

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SUMMARY

The discussion focuses on calculating the temperature rise of air in a bicycle pump when compressed adiabatically. The initial conditions include a cylinder with a length of 20 cm and a diameter of 3.0 cm, containing air at 21.0°C and 1.0 atm. When the air is compressed to half its original volume (from 1.55 m³ to 0.775 m³), the specific heat ratio (gamma) for air, which is 1.4, is used in the isentropic process equation T₂ = T₁ × (V₁/V₂)^(k-1) to determine the final temperature.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of adiabatic processes and specific heat ratios
  • Familiarity with isentropic processes in thermodynamics
  • Basic calculus for integrating pressure and volume (W = ∫PdV)
NEXT STEPS
  • Study the Ideal Gas Law and its applications in thermodynamics
  • Learn about adiabatic processes and how they differ from isothermal processes
  • Research isentropic processes and their significance in thermodynamic cycles
  • Explore the derivation and applications of the equation T₂ = T₁ × (V₁/V₂)^(k-1)
USEFUL FOR

Students studying thermodynamics, mechanical engineers, and anyone interested in the principles of gas behavior under compression.

yossup
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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?
 
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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
 

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