Rapid Compression of Air in a Bicycle Pump: Adiabatic, Isothermal or Neither?
A student investigates rapid compression of air in a bicycle pump. The outlet is sealed and the pump barrel is insulated. Initially the air occupies 120 cm³ at 1.0 × 10⁵ Pa and 300 K. The piston is pushed in quickly in 0.8 s until the volume is 25 cm³. A fine thermocouple inside the barrel records a maximum temperature of 390 K. The student's partner says the compression is adiabatic because it is fast, and predicts a much higher temperature using γ = 1.40. The student argues that the process is isothermal because the final temperature is 'not that high'. The pump leaks slightly at the piston seal, and the thermocouple has a response time of about 0.3 s. The amount of air is small enough for the ideal-gas approximation to be reasonable. Take the gas constant as 8.31 J mol⁻¹ K⁻¹ and C_V = 20.8 J mol⁻¹ K⁻¹.
1. Calculate the final temperature predicted by (a) an adiabatic model and (b) an isothermal model. State which prediction is numerically closer to the measured 390 K, and explain why that does not settle which model is physically appropriate.
Read the worked answer
Number of moles:
[ n=\frac{pV}{RT}=\frac{(1.0\times10^5)(120\times10^{-6})}{(8.31)(300)}=4.81\times10^{-3}\ \text{mol} ]
(a) Adiabatic compression of an ideal gas with (\gamma=1.40):
[ T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} =300\left(\frac{120}{25}\right)^{0.40} ]
[ \frac{120}{25}=4.8,\qquad 4.8^{0.40}\approx1.87 ]
[ T_2\approx300(1.87)=562\ \text{K} ]
(b) Isothermal model: (T_2=300\ \text{K}).
The measured 390 K is only 90 K above the isothermal value but 172 K below the adiabatic value, so it is numerically closer to the isothermal prediction. However, that does not prove the process is isothermal. The compression takes only 0.8 s, which is fast, but the gas can still lose heat to the cooler barrel, the seal leaks, and the thermocouple has a finite response time. The real process is intermediate between the two ideal limits.
2. Explain, in terms of heat transfer, leakage and thermocouple response, why the measured maximum temperature lies between the two predictions.
Read the worked answer
The adiabatic model assumes no heat enters or leaves the gas. In reality the compressed gas becomes hotter than the barrel and loses heat to the walls during and after compression. That heat loss lowers the final temperature below 562 K.
The leak at the piston seal allows some air to escape. The escaping air carries internal energy away, and the remaining gas may also expand slightly as pressure is lost, both of which reduce the measured temperature rise.
The thermocouple does not respond instantly: its response time is about 0.3 s, comparable with part of the 0.8 s compression. It may therefore under-record the true peak temperature. Even after the piston stops, the thermocouple reading can continue to change while the gas cools.
Thus the measured 390 K is not evidence for a truly isothermal process. It is consistent with a fast but imperfectly insulated compression in which heat loss, leakage and instrument lag all reduce the peak temperature.
3. Using the adiabatic work input as an upper estimate, calculate the internal-energy increase corresponding to the measured 390 K and hence estimate the heat lost to the surroundings. Comment on the reliability of the estimate.
Read the worked answer
For the adiabatic model, temperature rise:
[ \Delta T_{\text{ad}}=562-300=262\ \text{K} ]
Work input (upper estimate):
[ W\approx nC_V\Delta T_{\text{ad}}=(4.81\times10^{-3})(20.8)(262)\approx26.2\ \text{J} ]
For the measured final temperature:
[ \Delta T_{\text{meas}}=390-300=90\ \text{K} ]
Internal-energy increase:
[ \Delta U=(4.81\times10^{-3})(20.8)(90)\approx9.0\ \text{J} ]
If the work input were about 26 J, the heat lost would be
[ Q\approx W-\Delta U\approx26.2-9.0\approx17\ \text{J} ]
This is a rough upper estimate. The actual work input is path-dependent and is probably less than the adiabatic value because heat loss lowers the pressure during compression. The leak also changes the amount of gas. Nevertheless, the calculation shows that a substantial fraction of the input energy is lost to the surroundings, which is consistent with the measured temperature being far below the adiabatic prediction.