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                                    Air `(gamma = 1.4 )` is pumped at 20atm pressure in a motor tyre at `20^@C`. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmoshpheric air. | 
                            
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Answer» `gamma=1.4`. `T_1 = 20^@C=293K`, `P_1=2atm`,`P_2=1atm` We know for adiabatic process `P_1^(1-gamma) xx T_1^(1-gamma)=P^(1-gamma) xx T_2^gamma` or `(2)^(1-1.4) xx (293)^1.4 = (1)^(1-1.4) xx T_2^(1.4)` implies `(2)^(-0.4) xx (293)^1.4 = T_2^1.4` implies `2153.78 = T_2^1.4` `T_2=(2153.78)^(1//1.4)` `=240.3K`.  | 
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