The temperature of a gas is -68^(@)C.To what temperature should it be heated so that (a)the average translational KE of the molecules be double (b)the root mean square velocity of the molecules be doubled?

The temperature of a gas is -68^(@)C.To what temperature should it be

1.	The temperature of a gas is -68^(@)C.To what temperature should it be heated so that (a)the average translational KE of the molecules be double (b)the root mean square velocity of the molecules be doubled?
Answer» Solution :(a)According to kinetic theory of gases,average TRANSLATIONAL KE ofa molecule is given by `E=(3)/(2)kT` so that `(E_(2))/(E_(1))=(T_(2))/(T_(1)),2=(T_(2))/(T_(1))` so `T_(2)=2T_(1)`=2[273+(-68)]=410K (b)According to K inetic theory of gases,TMS speed of gas molecules is given by `v_(rms)=sqrt((3RT)/(M))` so that `((V_(rms))_(2))/((V_(rms))_(1))=sqrt((T_(2))/(T_(1)))implies2=sqrt((T_(2))/(T_(1)))` `therefore T_(2)=4T_(1)`=4[273+(-68)]=820K,`t_(2)=547^(@)C`