Challenge: fast adiabatic pump

Problem

A bicycle pump compresses $200\ \text{cm}^3$ of air adiabatically from 1 atm to 4 atm (ratio $V_2/V_1\approx 0{,}3$), starting from $T_1=290\ \text{K}$, with $\gamma=1{,}4$. Calculate the final temperature and explain the “warmth” felt at the nozzle.

Solution

Adiabatic law in temperature. For a reversible adiabatic process $TV^{\gamma-1}=\text{const}$, so:

$T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=290\cdot\left(\frac{1}{0{,}3}\right)^{0{,}4}=290\cdot 3{,}33^{0{,}4}$

Computing the exponential.

$3{,}33^{0{,}4}=e^{0{,}4\cdot\ln 3{,}33}=e^{0{,}482}=1{,}619$

Final temperature.

$T_2=290\cdot 1{,}619\approx 470\ \text{K}\approx 197\ ^\circ\text{C}$

$\ev{T_2\approx 470\ \text{K}\approx 197\ ^\circ\text{C}}$

The “warmth”. The compression is too fast for the gas to lose heat to the surroundings ($Q\approx 0$): the work done on the air converts entirely into internal energy, and the temperature rises sharply. That is why, when you pump quickly, the pump nozzle heats up.

Links

Topics: Thermodynamics Concepts: Thermodynamic processes · Internal energy Objects: Ideal gas · Piston and cylinder