IIT-JEE 2009 · Paper 2 · Chemistry Q13

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ChemistryInteger+4 / -1Question 13 of 57

At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is

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Answer and worked solution

Official answer: 4

Equate the speeds.
$\sqrt{\frac{3R(400)}{40}}=\sqrt{\frac{2R(60)}{M_Y}}\implies30=\frac{120}{M_Y}\implies M_Y=4.$

Answer: 4

Topics:Kinetic theory of gases

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IIT-JEE 2009 · Paper 2, Chemistry question 13