Some amount of gas at 27^(@)C is suddenlycompressed to 8 times its initial pressure. If gamma = 1.5, find out the rise in temperature.

Some amount of gas at 27^(@)C is suddenlycompressed to 8 times its ini

1.	Some amount of gas at 27^(@)C is suddenlycompressed to 8 times its initial pressure. If gamma = 1.5, find out the rise in temperature.
Answer» <P> Solution :As the gas is suddenly compressed, the PROCESS is adiabatic. So, `T_(1)^(gamma) p_(1)^(1-gamma) = T_(2)^(gamma)p_(2)^(1-y) or, ((T_1)/(T_2))^(gamma) = ((p_2)/(p_1))^(1 - gamma)` or, `T_(2) = T_(1) ((p_1)/(p_2))^((1-gamma)/gamma)` Here, `T_(1) = 27^(@)C = 300 K , (p_2)/(p_1) = 8`, `gamma = 1.5 = 3/2 or, 1 - gamma = 1 - 3/2 = -1/2` or, `(1-gamma)/(gamma) = -1/3` `:. (1-gamma)/(gamma) = -1/3` `:.T_(2) = 300 XX (1/8)^(-1//3)` `= 300 xx (8)^(1//3)` ` = 300 xx 2 = 600 K` `:.` Rise in temperature = 600 - 300 = 300 K = 300^(@)C`.