Question

The pressure inside a tyre is $4$ times that of atmosphere. If the tyre burst suddenly at temperature $300K$, what will be the new temperature?

• A. $300(4{)}^{7/2}$
• B. $300(4{)}^{2/7}$
• C. $300(2{)}^{7/2}$
• D. $300(4{)}^{-2/7}$

Answer 1

The correct option is D $300(4{)}^{-2/7}$

Under adiabatic change
$\frac{T_2}{T_1}={\left(\frac{P_1}{P_2}\right)}^{\frac{1-\gamma }{\gamma }}$
or $T_2=T_1(\frac{P_1}{P_2}{)}^{\left(\frac{1-\gamma }{\gamma }\right)}$
$\therefore$$T_2=300(4{)}^{\frac{1-\left(7/5\right)}{7/5}}$; $\gamma =1.4=7/5$ for air
or $T_2=300(4{)}^{-2/7}$