Answer:
2790 Pa
Explanation:
Given wavelength λ= 5μm
temperature T= 400 K
cross section of collision σ= 0.28 nm^2
molar mass = 39.9 g/mole
pressure = [tex]P= \frac{RT}{\sqrt{2}N_A\sigma\lambda }[/tex]
putting values we get
=[tex] \frac{8.314\times400}{\sqrt{2}\times6.022\times10^{23}\times0.28\times10^{-18}\times5\times10^{-6} }[/tex]
⇒P = 2790 J/m^3
the partial pressure are argon atoms expected= 2790 Pa
