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Q.

Three moles of a monoatomic gas are mixed with two moles a diatomic gas. When ‘Q’ amount of heat is supplied to the mixture at constant volume, temperature of the mixture is increased by 300C. Then ‘Q’ is equal to

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a

55R

b

200R

c

325R

d

285R

answer is D.

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Detailed Solution

For the mixture, Cv = 3×3R2+2×5R23+2 = 1910R∴ Q = 3+2×19R10×30   Joule           = 285 R
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