Kinetic Theory

JEE Main level, single correct, medium

121 of 168

Question 121 of 168

If 22 mole of an ideal monoatomic gas at temperature TT, is mixed with 66 mole of another ideal monoatomic gas at temperature 2T2T then the temperature of mixture is :

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Answer: (D) 74T\dfrac{7}{4}T

Using the principle of conservation of internal energy for the mixture of ideal gases:

Umix=U1+U2U_{mix} = U_1 + U_2

(n1+n2)CVTmix=n1CVT1+n2CVT2(n_1 + n_2) C_{V} T_{mix} = n_1 C_{V} T_1 + n_2 C_{V} T_2

Since both gases are monoatomic, their molar heat capacities at constant volume are equal (CV=32RC_{V} = \dfrac{3}{2}R).

(n1+n2)Tmix=n1T1+n2T2(n_1 + n_2) T_{mix} = n_1 T_1 + n_2 T_2

Substituting the given values n1=2n_1 = 2, T1=TT_1 = T, n2=6n_2 = 6, and T2=2TT_2 = 2T:

(2+6)Tmix=2(T)+6(2T)(2 + 6) T_{mix} = 2(T) + 6(2T)

8Tmix=2T+12T8 T_{mix} = 2T + 12T

8Tmix=14T8 T_{mix} = 14T

Tmix=148T=74TT_{mix} = \dfrac{14}{8}T = \dfrac{7}{4}T

Answer: 74T\dfrac{7}{4}T

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