1.

If the temperature and pressure of a gas is doubled the mean free path of the gas molecules

Answer»

<P>remains same
doubled
tripled
quadruples

Solution :MEAN FREE path of the MOLECULE
`gamma=(kT)/(sqrt2pid^(2)P`
If T and P are double, `gamma=(2kT)/(sqrt2pid^(2)(2P))=(kT)/(sqrt2pid^(2)P)`


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