One mole of helium at $300 K$ and two moles of nitrogen (rigid diatomic) at $450 K$ are mixed in a rigid, adiabatic container. Treating the gases as ideal, the final temperature is nearest to

Answer: B

With no heat or work exchanged with the surroundings,

Δ U_{t o t a l} = 0

, so

1 (1.5 R) (T - 300) + 2 (2.5 R) (T - 450) = 0

. Then

1.5 (T - 300) + 5 (T - 450) = 0

, giving

T = (450 + 2250) /6.5 = 415.4 K

. Weighting by moles alone ignores the different

C_{V}

values.