Question

The pressure in the tyre of a car is four times the atmospheric pressure at $300K$. If this tyre suddenly brusts, its new temperature will be :
$(\gamma =1.4)$

• A. $300(4{)}^{1.4/0.4}$
• B. $300{\left(\frac{1}{4}\right)}^{-0.4/1.4}$
• C. $300(2{)}^{-4/1.4}$
• D. $300(4{)}^{-0.4/1.4}$

Answer 1

The correct option is D $300(4{)}^{-0.4/1.4}$

Burst or exploration is an adiabatic process.
For the adiabatic process:
$T^{\gamma }{P}^{1-\gamma }=$ constant or
${\left(\frac{T_2}{T_1}\right)}^{\gamma }={\left(\frac{P_1}{P_2}\right)}^{1-\gamma }$
where $T_1=$ initial temperature
$T_2=$ final temperature
$P_1=$ initial pressure = $=4\times 1atm$
$P_2=$ final pressure $=1atm$
$\gamma =1.4$
so,
$\frac{T_2}{T_1}={\left(\frac{P_1}{P_2}\right)}^{\frac{1-\gamma }{\gamma }}\frac{T_2}{300}={\left(\frac{4}{1}\right)}^{\frac{(1-1.4)}{1.4}}{T}_2=300(4{)}^{-\frac{0.4}{1.4}}$