If 2 mole of an ideal monoatomic gas at temperature T , is mixed with 6 mole of another ideal monoatomic gas at temperature 2T then the temperature of mixture is :

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Quantrex Academy · Accepted Answer

Correct answer: D. (7)/(4)T

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

U_{mix} = U_1 + U_2

(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 (

C_{V} = \dfrac{3}{2}R

).

(n_1 + n_2) T_{mix} = n_1 T_1 + n_2 T_2

Substituting the given values

n_1 = 2

,

T_1 = T

,

n_2 = 6

, and

T_2 = 2T

:

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

8 T_{mix} = 2T + 12T

8 T_{mix} = 14T

T_{mix} = \dfrac{14}{8}T = \dfrac{7}{4}T

Answer:

\dfrac{7}{4}T

If 2 mole of an ideal monoatomic gas at temperature T , is mixed...

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