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

Concept :

When a tyre bursts suddenly, the process is considered adiabatic, meaning no heat is transferred. We use the adiabatic process formula for gases.

For an adiabatic process, P₁V₁^γ = P₂V₂^γ and T₁P₁^(1-γ)/γ = T₂P₂^(1-γ)/γ

Where γ (gamma) is the adiabatic index (for air, γ = 7/5)

Calculation:

Given P₁ = 4P_atm and P₂ = P_atm

⇒ T₂ = T₁ × (P₂/P₁)^(γ-1)/γ

⇒ T₂ = 300K × (P_atm/4P_atm)^{(7/5-1)/(7/5)}

⇒ T₂ = 300K × (1/4)^(2/5)

⇒ T₂ = 300K × (4)^-2/5

⇒ T₂ = 300K × (4)^-0.4

⇒ T₂ = 300K × (4)^-2/5

⇒ T₂ = 300 × (4)^-2/5

⇒ T₂ = 300 × 4^-2/5

∴ T₂ = 300(4)^-2/5

The correct answer is option 4.