Two separateexperiments were carried out involving adiabaticreversible expansion, taking monoatomic gas in one case and a diatomic gas in the second case.In each case, the initialtemperature wasT_(i) and expansion was from volume V_(1)to volume V_(2). The finaltemperaturesattained were T_(M) formonoatomic gasand T_9D) for diatomicgas. Then

Two separateexperiments were carried out involving adiabaticreversible

1.	Two separateexperiments were carried out involving adiabaticreversible expansion, taking monoatomic gas in one case and a diatomic gas in the second case.In each case, the initialtemperature wasT_(i) and expansion was from volume V_(1)to volume V_(2). The finaltemperaturesattained were T_(M) formonoatomic gasand T_9D) for diatomicgas. Then
Answer» `T_(M) =T_(D) ltT_(i)` `T_(M) ltT_(D) ltT_(i)` `T_(D) gtT_(M) gtT_(i)` `T_(D) ltT_(M) ltT_(i)` Solution :For ADIABATIC expansion, `(T_(2))/(T_(1))=((V_(1))/(V_(2)))^(gamma-1),gamma=1.66` formonoatomic gas and1.40 for diatomic gas. HENCE, for monoatomicgas, `(T_(M))/( T_(i)) =((V_(1))/(V_(2)))^(0.66)` and for diatomcgas, `(T_(D))/(T_(i))=((V_(1))/(V_(2)))^(0.40)` As`(V_(1))/(V_(2)) lt 1( :' V_(2) gt V_(1)) `. Hence,`(T_(M))/(T_(i)) LT1` i.e.,`T_(M) lt T_(i)` and`(T_(D))/(T_(i))lt 1` , i.e., `T_(D) lt T_(i)` Suppose `V_(2)= 10V_(1) `. Then `(T_(M))/(T_(D)) =((1)/(10))^(0.66-0.40) = ((1)/(10))^(0.26)` i.e. `T_(M)lt T_(D)`. Hence, `T_(M) ltT_(D) lt T_(i)`