Gujarati
Hindi
11.Thermodynamics
normal

Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside

A

Starts becoming hotter

B

Remains at the same temperature

C

Starts becoming cooler

D

May become hotter or cooler depending upon the amount of water vapour present

Solution

Since tube is punctured suddenly, the process is adiabatic.

In adiabatic process : $TV ^{\gamma-1}=$ constant

$\frac{ T _2}{ T _1}=\left(\frac{ V _1}{ V _2}\right)^{\gamma-1}$

since air is expanding,

$V _2 > V _1$

$\therefore T _1 > T _2$

Standard 11
Physics

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