Air (γ = 1.4) is pumped at 2 atm pressure in a motor tyre at 20C. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmospheric air.

Sol. γ = 1.4, T base 1 = 20°C = 293 k, P base 1 = 2 atm, p base 2 = 1 atm We know for adiabatic process, P base 1^1-γ * T base 1^γ = P base 2^1-γ * T base 2^γ or (2)^1-1.4 * (293)^1.4 = (1)^1-1.4 * T base 2^1.4 ⇒ (2)^0.4 * (293)^1.4 = T base 2^1.4 ⇒ 2153.78 = T base^1.4 ⇒ T base 2 = (2153.78)^1/1.4 = 240.3 K

Sol. Sol. γ = 1.4, T base 1 = 20°C = 293 k, P base 1 = 2 atm, p base 2 = 1 atm We know for adiabatic process, P base 1^1-γ * T base 1^γ = P base 2^1-γ * T base 2^γ or (2)^1-1.4 * (293)^1.4 = (1)^1-1.4 * T base 2^1.4 ⇒ (2)^0.4 * (293)^1.4 = T base 2^1.4 ⇒ 2153.78 = T base^1.4 ⇒ T base 2 = (2153.78)^1/1.4 = 240.3 K