A mixture of 4 g helium and 28 g of nitrogen is enclosed in a vessel of constant volume at 300 K. The quantity of heat absorbed by the mixture to double the root-mean-square velocity of its molecules

Question

A mixture of 4 g helium and 28 g of nitrogen is enclosed in a vessel of constant volume at 300 K. The quantity of heat absorbed by the mixture to double the root-mean-square velocity of its molecules is ( R = universal gas constant )

TestHub · Accepted Answer

4 g He = 1 mol , 28 g N 2 ​ = 1 mol . Doubling v r m s ​ means T becomes 4 T = 1200 K , so Δ T = 900 K . Q = ( 1 × 2 3 R ​ + 1 × 2 5 R ​ ) × 900 = 3600 R .

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