Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T_(0), while box B contains one mole of helium at temperature (7//3)T_(0). The boxes are then put into thermal contact with each other , and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then , the final temperature of gases, T_(f) , in terms of T_(0) is -

Two rigid boxes containing different ideal gases are placed on a table

1.	Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T_(0), while box B contains one mole of helium at temperature (7//3)T_(0). The boxes are then put into thermal contact with each other , and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then , the final temperature of gases, T_(f) , in terms of T_(0) is -
Answer» `T_f=5/2T_0` `T_f=3/7T_0` `T_f=7/3T_0` `T_f=3/2T_0` SOLUTION :`DeltaE_"int"=(DeltaE_"int")_(N_2)+(DeltaE_"int")_(He)`…(1) but `DeltaE_"int"=0` `(DeltaE_"int")_(N_2)=(MU C_V DELTAT)_(N_2)(C_V)_(N_2)=5/2R` `(DeltaE_"int")_(He)=(mu C_V DeltaT)_(He) (C_V)_(He)=3/2R` From equation (1), `0=1xx5/2(T_f-T_0)+1xx3/2xx(T_f-7/3T_0)` `therefore 0=5 T_f-5T_0+3T_f - 7T_0` `therefore 12T_0 =8T_f` `therefore T_f=3/2T_0`