Question

A cycle pump becomes hot near the nozzle after a few quick stokes even if they are smooth because

• A. The volume of air decreases
• B. The number of air molecules increases
• C. The compression of air is adiabatic
• D. Collision between air particles increases

Answer 1

The correct option is C The compression of air is adiabatic

In a quick process or a sudden process the system does not get sufficient time to exchange heat with surroundings.
Thus $\Delta Q=0,$ hence such quick/sudden process is adiabatic.

Therefore, pumping of cycle tube with quick strokes is an adiabatic process.
So, $P{V}^{\gamma }=\text{constant}$
or $T{V}^{\gamma -1}=\text{constant}....\left(1\right)$
Cycle tube pumping is a compression process. Thus volume of air inside the pump barrel decreases and from Eq.$\left(1\right)$ temperature of system increases.

$\underset{–}{\text{Alternative:}}$
During pumping of cycle tube with quick strokes the system doesnot get sufficient time to exchange heat with surroundings. Thus it is an adiabatic process$(\Delta Q=0)$.
Applying first law of thermodynamics,
$\Rightarrow \Delta Q=W+\Delta U$
Or, $W=-\Delta U...\left(i\right)$
Air is being compressed during each stroke, work is done on the system$(W_{\text{system}}\to -ve)$.
Thus the work done on the system is utilised to increase its internal energy$(U_f>U_i)$, hence temperature of system increases$(T_f>T_i)$.

Some heat from the gas gets transferred to the nozzle and thus raises its temperature.