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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
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B
The number of air molecules increases
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C
The compression of air is adiabatic
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D
Collision between air particles increases
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Solution

## The correct option is C The compression of air is adiabaticIn a quick process or a sudden process the system does not get sufficient time to exchange heat with surroundings. Thus ΔQ=0, hence such quick/sudden process is adiabatic. Therefore, pumping of cycle tube with quick strokes is an adiabatic process. So, PVγ=constant or TVγ−1=constant ....(1) Cycle tube pumping is a compression process. Thus volume of air inside the pump barrel decreases and from Eq.(1) temperature of system increases. 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(ΔQ=0). Applying first law of thermodynamics, ⇒ΔQ=W+ΔU Or, W=−ΔU ...(i) Air is being compressed during each stroke, work is done on the system(Wsystem→−ve). Thus the work done on the system is utilised to increase its internal energy(Uf>Ui), hence temperature of system increases(Tf>Ti). Some heat from the gas gets transferred to the nozzle and thus raises its temperature.

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