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Q.

A container contains helium gas at $0 ° C$ . What is the average linear momentum of a helium molecule of mass $6.64 \times 1 0^{- 27}$ kg in the container?

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a

$18 \times 1 0^{- 24} {kg m s}^{- 1}$

b

$8 \times 1 0^{- 24} {kg m s}^{- 1}$

c

$8 \times 1 0^{- 23} {kg m s}^{- 1}$

d

$18 \times 1 0^{- 23} {kg m s}^{- 1}$

answer is A.

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Detailed Solution

Average linear momentum is given by $p_{a v g} = m v_{a v g}$
But $v_{a v g} = \sqrt{\frac{8}{π} \frac{R T}{M}}$
$p_{a v g} = 6.64 \times 1 0^{- 27} \sqrt{\frac{8}{π} \times \frac{8.314 \times 273}{4 \times 10^{- 3}}} = 8 \times 1 0^{- 24} {kg m s}^{- 1}$ .

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A container contains helium gas at 0 °C. What is the average linear momentum of a helium molecule of mass 6.64×10–27 kg in the container?