First slide

Heat given

Question

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 $30^{0} C$ . Then ‘Q’ is equal to

Easy

A

55R
B

200R
C

325R
D

285R

Solution

$For the mixture, C_{v} = \frac{3 \times \frac{3 R}{2} + 2 \times \frac{5 R}{2}}{(3 + 2)} = \frac{19}{10} R$
$∴ Q = (3 + 2) \times \frac{19 R}{10} \times 30 Joule$
= 285 R

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