Question

( mathrm{PV}^{wedge} mathrm{gamma}= ) constant.
( mathrm{PV}=n mathrm{RT} )
( mathrm{P}^{*}(mathrm{nRT} / mathrm{P})^{wedge} mathrm{g} ) amma ( =mathrm{constant} )
( left(mathrm{P}^{wedge} 1-mathrm{g} mathrm{amma}right)^{*}(mathrm{T})^{wedge} mathrm{g} ) amma ( = ) constant.
( mathrm{T}=3 mathrm{OO} mathrm{K} ) and ( mathrm{P}=3 mathrm{atm} )
for ( mathrm{P}=1 ) atm as initial and final pressure of tyre is the same! Get T
( mathrm{T}=219.19 mathrm{K} mathrm{so} mathrm{dt}=80.2 mathrm{K} ) or ( mathrm{C} )
Thus the temperature is decreased

# 14. A motor tyre pumped to a pressure of 3 atm. suddenly bursts. Calculate the fall in temperature due to adiabatic expansion. The temperature of air before expansion is 27°C. Given y=1.4. Initial pressure, P, = 3 atm ; Final pressure, P, = 1 atm Initial temperatura T 797

Solution