Q) Considering the value of ideal gas constant in S. I. unit, find the volume of 35g O2 at 27°C and 72

cm Hg pressure. Later, if we keep this pressure constant, the r. m. s velocity of this oxygen molecules

become double at a certain temperature. Calculate the value of this temperature.

V = 0.0283 m³ = 28300 cm³

T₂ = 1200 K

Explanation:

The volume of the gas can be determined by using General Gas Equation:

PV = nRT

where,

P = Pressure of Gas = (72 cm of Hg) (1333.2239 Pa/cm of Hg) = 95992.12 Pa

V = Volume of Gas = ?

n = no. of moles = mass/molar mass = (35 g) / (32 g/mol) = 1.09 mol

R = General Gas Constant = 8.314 J / mol. k

T = Temperature of Gas = 27°C + 273 = 300 k

Therefore,

(95992.12 Pa) (V) = (1.09 mol) (8.314 J/mol. k) (300 k)

V = 2718.678 J/95992.12 Pa

V = 0.0283 m³ = 28300 cm³

The Kinetic Energy of gas molecule is given as:

K. E = (3/2) (KT)

Also,

K. E = (1/2) (mv²)

Comparing both equations, we get:

(3/2) (KT) = (1/2) (mv²)

v² = 3KT/m

v = √ (3KT/m)

where,

v = r. m. s velocity

K = Boltzamn Constant

T = Absolute Temperature

m = mass of gas molecule

At T₁ = 300 K, v = v₁

v₁ = √ (3K*300/m)

v₁ = √ (900 K/m)

Now, for v₂ = 2v₁ (double r. m. s velocity), T₂ = ?

v₂ = 2v₁ = √ (3KT₂/m)

using value of v₁:

2√ (900 K/m) = √ (3KT₂/m)

4 (900) = 3 T₂

T₂ = 1200 K