Question

estimate pressure of air molecules at 273 K , if mean square speed is 500m2/s2 and density of air is 6 kg/m3

Answer 1

estimate pressure of air molecules at 273 K , if mean square speed is 500m2/s2 and density of air is 6 kg/m3

solution : we know from kinetic theory of gases, square mean root velocity of gas , $V_{rms}=\sqrt{\frac{3P}{\rho}}$

given, mean square , $V_{rms}^2$= 500m²/s²

taking square root of it to get rms velocity.

i.e., $V_{rms}=\sqrt{500}=10\sqrt{5}$

density , ρ = 6 kg/m³

so, 10√5 = √{3P/6 }

squaring both sides,

500 = 3P/6

or, 500 × 6/3 = P

or, P = 1000 N/m²

hence, pressure of air molecule is 1000 N/m² or Pascal.

Answer 2

Answer:

hence, pressure of air molecule is 1000 N/m² or Pascal.

Explanation: