A tyre pumped to a pressure 3.375 atm at 27° C suddenly bursts. What is the final temperature (γ=1.5) ?

Question

A tyre pumped to a pressure 3.375 atm at 27° C suddenly bursts. What is the final temperature (&gamma;=1.5) ? (A) 27° C (B) -27° C (C) 0° C (D) -73° C

Tardigrade Admin · Accepted Answer

T11-&gamma; P11-&gamma;=T2&gamma; P21-&gamma; or ((T1/T2))&gamma;=((P1/P2))&gamma;-1 =((300/T2))3 / 2=((3.375/1))3 / 2-1 or T2=(300/(3.375)1 / 3) =200 K =-73° C

Q. A tyre pumped to a pressure $3.375 a t m$ at $2 7^{\circ} C$ suddenly bursts. What is the final temperature $(γ = 1.5)$ ?

$2 7^{\circ} C$

$- 2 7^{\circ} C$

$0^{\circ} C$

$- 7 3^{\circ} C$

$T_{1}^{1 - γ} P_{1}^{1 - γ} = T_{2}^{γ} P_{2}^{1 - γ}$
or $(\frac{T _{1}}{T _{2}})^{γ} = (\frac{P _{1}}{P _{2}})^{γ - 1}$
$= (\frac{300}{T _{2}})^{3/2} = (\frac{3.375}{1})^{3/2 - 1}$
or $T_{2} = \frac{300}{( 3.375 ) ^{1/3}}$
$= 200 K = - 7 3^{\circ} C$

Q. A tyre pumped to a pressure 3.375atm at 27∘C suddenly bursts. What is the final temperature (γ=1.5) ?

27∘C

−27∘C

0∘C

−73∘C