Question

Why does a cycle pump become hot near the nozzle after a few quick strokes even if they are smooth?
A. the volume of air decreases
B. the number of molecules increase
C. the compression is adiabatic
D. collision between air particles increases

Answer 1

The adiabatic process is the process which occurs without transferring heat to its surroundings but transfers energy only as work to the surroundings. There is always a variation in temperature. The internal energy for the cycle pump must be increased, so there is a change in temperature.

Complete step-by-step solution:
The adiabatic process can be defined as the process in which there is no transfer of heat or mass from the thermodynamic system to its surroundings. It transfers energy to its surroundings only as work. In adiabatic processes, there is variation of temperature. The conditions for this process to occur are:
-> There should be an insulated thermodynamics system from surroundings.
-> The time taken for the process should be quick for heat transfer to take place.
The equation for adiabatic process of an ideal gas is –
$P{V^\gamma } = C$
where, $P$ is the pressure
$V$ is the volume
$\gamma$ is the adiabatic index
$C$ is the constant
The nozzle of the cycle pump becomes hot due to the adiabatic process. Due to adiabatic compression, the internal energy of the system is increased to compensate for the compression of gas, hence there is an increase in temperature. Therefore, the nozzle gets hot due to adiabatic compression.
Hence, option (C) is the correct answer.

Note:- Adiabatic compression is the compression in which there is no transfer of heat from surroundings but the internal energy increases which is equal to the external work done on surroundings. During compression, when temperature is increased the pressure of air is more than the volume.