Question

At normal temperature, when the wheel of the cycle gets punctured the compressed air inside the tube of the wheel starts to come out suddenly. Find out what happens to the air inside.
A) The air starts getting hotter.
B) The air starts getting colder.
C) The air remains at the same temperature.
D) The air may get hotter or colder depending on the amount of water vapour present.

Answer 1

When the tube gets punctured, the compressed air inside the tube starts coming out suddenly, i.e., the air inside the tube expands leading to a drop in the pressure of the gas. The corresponding change in the temperature is explained based on the ideal gas equation $PV = nRT$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles present in the gas, $R$ is the universal gas constant and $T$ is the temperature of the gas.

Complete step by step answer:
Step 1: Describe what happens to the air inside the tube when the tube gets punctured.
At normal temperature, the tube punctures. As a result, the air inside the tube which initially existed in a compressed state now expands and comes out of the puncture. The compressed state of the air suggests that the pressure was high inside the tube. Now, as the air expands the pressure drops.
Step 2: Based on the ideal gas equation explain what happens to the temperature of the air.
The ideal gas equation is given by, $PV = nRT$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles present in the gas, $R$ is the universal gas constant and $T$ is the temperature of the gas.
The volume of the tube remains constant and neglect $n$ and $R$.
Then from the ideal gas equation, we see that the pressure is directly proportional to the temperature.
i.e., $P \propto T$ .
Thus, as the pressure drops on the expansion of air, the temperature of the air will also decrease. Therefore, the air inside the tube starts to get colder.

Note:
Compressed air coming out suddenly from a puncture at normal temperature is an adiabatic process. Hence the decrease in temperature of the air can also be explained based on adiabatic expansion.
For an adiabatic process, $\dfrac{{{P^{\gamma - 1}}}}{{{T^\gamma }}}$ is constant. If ${T_1}$ and ${P_1}$ are the initial temperature and pressure of the air while ${T_2}$ and ${P_2}$ are its final temperature and pressure, then ${P_1}^{\gamma - 1}{T_2}^\gamma = {P_2}^{\gamma - 1}{T_1}^\gamma$ ----- (1).
As the air expands, pressure drops, i.e., ${P_2} < {P_1}$ . So for equation (1) to be true, ${T_2} < {T_1}$ .
Therefore, the air inside the tube becomes colder.