Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at $0^{0}$ C. Mass of a helium molecule = 6.64 $\times 10^{- 27}$ kg and Boltzmann constant 1.38 $\times 10^{- 23}$ J $K^{- 1}$ .

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Solution

M = $4 \times 10^{- 3}$ kg

$V_{a v g} = \frac{\sqrt{8 R T}}{\sqrt{π M}}$

= $\frac{\sqrt{8 \times 8.3 \times 273}}{\sqrt{3.14 \times 4 \times 10^{- 3}}}$

= 1201.35

Momentum

= M $\times V_{a v g}$

= 6.64 $\times 10^{- 27} \times$ 1201.35

= 7.97 $\times 10^{- 24}$

= 8 $\times 10^{- 24}$ kg - m/s

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Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 00C. Mass of a helium molecule = 6.64 × 10−27 kg and Boltzmann constant 1.38 ×10−23 J K−1.

M = 4×10−3 kg Vavg=√8RT√πM = √8×8.3×273√3.14×4×10−3 = 1201.35 Momentum = M×Vavg = 6.64 ×10−27× 1201.35 = 7.97 ×10−24 = 8 ×10−24 kg - m/s

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at $0^{0}$ C. Mass of a helium molecule = 6.64 $\times 10^{- 27}$ kg and Boltzmann constant 1.38 $\times 10^{- 23}$ J $K^{- 1}$ .