Question

The density of a gas at normal pressure and 0°C
temperature is 1.2 kg m Calculate : (i) the root-
mean-square velocity of the molecules of the gas at
0°C, (ii) the temperature at which the velocity will
become three times the initial velocity. (Normal
pressure = 1.0 x 109 Nm2).​

Answer 1

Given info : The density of a gas at normal pressure and 0°C temperature is 1.2 kg/m³

To find :

1. the root mean square velocity of the molecule of the gas at 0°C.
2. the temperature at which velocity becomes three times the initial velocity.

solution : root mean square velocity of the molecules is given by, v = √{3P/ρ}

P is atmospheric pressure i.e., 1.01 × 10⁵ N/m²

ρ is density of gas i.e., ρ = 1.2 kg/m³

so v = √{3 × 1.01 × 10⁵/1.2}

= 502.5 m/s

we know, root mean square velocity is also given by, v = √{3RT/M}

i.e., v ∝ √T

means, v₁/v₂ = √(T₁/T₂)

here 3v₁ = v₂ , T₁ = 0°C = 273 K

⇒v₁/3v₁ = √(273/T₂)

⇒9 × 273 = T₂

⇒T₂ = 2457 K

Therefore temperature of the gas would be 2457K.