A motor car tyre is pumped up to pressure of two atmosheres at `15^(@)C` when it suddenly bursts. Calculate the resulting drop in temperature of the escaping air `(gamma =14)`.

Answer

2.3734

Explanation

As it bursts suddenly, the changes is adiabatic.
We have `T^(gamma)p^(1-gamma) =` a constant
`T_(1)^(gamma)P_(1)^(1-gamma) = T_(2)^(gamma)p_(2)^(1-gamma)`
or `(273+15)^(gamma)(2P_(0))^(1-gamma) = T_(2)^(gamma)(2p_(0))^(1-gamma) Where p_(0) = 1 atmoshpere`
or `288^(1.4) 2^(1-1.4)p_(0)^(1-gamma) = T_(2)^(1.4)p_(0)^(1-gamma)`
`288^(1.4) 2^(-0.4) = T_(2)^(1.4)`
or ` T_(2)^(1.4) = (288^(1.4))/(2^(0.4))`
or `log T_(2) = log288 - 0.4log 2//1.41 =2.3734`