A mixture of 8gm of helium and 14gm of nitrogen is enclosed in a vessel of constant volume at 300K. The quantity of heat absorbed by the mixture to double the root mean velocity of its molecules is – (R = universal gas constant)
Text Solution
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Q = Δ U + W
W = 0 since volume is constant
Q = Δ U
V rms = 
U mix =U 1 + U 2
U mix = n 1 
T + n 2 
T = (n 1 + n 2 ) (C v ) mix T
(U f ) – (U i ) = nC v (T 2 – T 1 ) = (n 1 + n 2 ) (C v ) mix (T 2 –T 1 )
V ′ rms = 2V rms ⇒ T ′ = 4T
n 1 = 
= 2; n 2 = 
= 1/2
(C v )mix = 
U f – U i = (n 1 
+ n 2 
) (T 2 – T 1 )
= 
R [1200 – 300]
× R × 100 = 3825 R
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