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 :

Step-by-step solution

Using the principle of conservation of internal energy for the mixture of ideal gases: U_ mix = U₁ + U₂ (n₁ + n₂) C_ V T_ mix = n₁ C_ V T₁ + n₂ C_ V T₂ Since both gases are monoatomic, their molar heat capacities at constant volume are equal ( C_ V = 3 2 R ). (n₁ + n₂) T_ mix = n₁ T₁ + n₂ T₂ Substituting the given values n₁ = 2 , T₁ = T , n₂ = 6 , and T₂ = 2T : (2 + 6) T_ mix = 2(T) + 6(2T) 8 T_ mix = 2T + 12T 8 T_ mix = 14T T_ mix = 14 8 T = 7 4 T Answer: 7 4 T