One mole of an ideal monoatomic gas at an initial temperature T expands adiabatically until its root mean square (rms) speed becomes 1 3 times its initial value. The work done by the gas in this process is:

Step-by-step solution

The root mean square speed of gas molecules is given by v_ rms = 3RT M . This implies that v_ rms T . Given that the final rms speed is 1 3 of the initial rms speed: v_ rms , f v_ rms , i = 1 3 = T_f T_i Squaring both sides gives the ratio of temperatures: T_f T = 1 3 T_f = T 3 For a monoatomic gas, the adiabatic index = 5 3 . The work done by the gas during an adiabatic expansion is: W = nR(T_i - T_f) - 1 Substitute n = 1 , T_i = T , T_f = T 3 , and = 5 3 : W = 1 R (T - T 3 ) 5 3 - 1 W = R ( 2T 3 ) 2 3 W = RT Answ