A tyre pumped to a pressure of 2 atmosphere, suddenly bursts. If the temperature of air before expansion is T then the air temperature after the tyre…

Question

A tyre pumped to a pressure of 2 atmosphere, suddenly bursts. If the temperature of air before expansion is T then the air temperature after the tyre busts is (Assume the expansion is adiabatic and adiabatic constant γ=3/2

MARKS Content Team · Accepted Answer

Correct answer: (B) 12^(1/3)T. The above process belongs to the thermodynamic process known as the adiabatic process and the relation between pressure and volume in the adiabatic process is, PV^(γ)=K, now use ideal gas equation, ⇒PnRTP^(γ)=K⇒T^(γ)P^(γ-1)=K ⇒T1^(γ)P1^(γ-1)=T2^(γ)P2^(γ-1)⇒T^(γ)2^(γ-1)=T2^(γ)1^(γ-1)⇒T^322^12=T2^321⇒T2^3=1/2×T^3⇒T_2=12^(1/3)T

A tyre pumped to a pressure of $2$ atmosphere, suddenly bursts. If the temperature of air before expansion is $T$ then the air temperature after the tyre busts is
(Assume the expansion is adiabatic and adiabatic constant $γ = \frac{3}{2})$

Solution

Relevant Questions

A tyre pumped to a pressure of 2 atmosphere, suddenly bursts. If the temperature of air before expansion is T then the air temperature after the tyre busts is(Assume the expansion is adiabatic and adiabatic constant γ=32

Solution

Relevant Questions

A tyre pumped to a pressure of $2$ atmosphere, suddenly bursts. If the temperature of air before expansion is $T$ then the air temperature after the tyre busts is
(Assume the expansion is adiabatic and adiabatic constant $γ = \frac{3}{2})$