A mixture of 4 gm helium and 28 gm of nitrogen is enclosed in a vessel of constant volume at 300 K. The quantity of heat absorbed by the mixture to increase root mean velocity of its molecules by 50%

Question

A mixture of 4 gm helium and 28 gm of nitrogen is enclosed in a vessel of constant volume at 300 K. The quantity of heat absorbed by the mixture to increase root mean velocity of its molecules by 50% is (R is universal gas constant)

TestHub · Accepted Answer

v r m s ​ ∝ T ​ : a factor 1.5 × means T → 2.25 T , so Δ T = 1.25 × 300 = 375 K . C V mi x ​ = 2 1 × 2 3 ​ R + 1 × 2 5 ​ R ​ = 2 R per mole, n = 2 mol total. Δ Q = n C V mix ​ Δ T = 2 × 2 R × 375 = 1500 R .

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