HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.

[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]

The temperature of the gas molecules is calculated as follows;

[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]

where;

k is Boltzmann constant

[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]

The number of gas molecules per unit volume is calculated as follows;

[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]

The pressure of the gas molecule is calculated as follows;

[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]

Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.

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