Question

The molecules of a given mass of a gas have r.m.s. velocity of 200m s⁻1 at 27^∘C and 1.0×10⁵N m⁻2 pressure. When the temperature and pressure of the gas are respectively 127^∘C and 0.05×10⁵N m⁻2, the r.m.s. velocity of its molecules in m s⁻1 is:

• A. \(100\sqrt{2}\)
• B. \(\dfrac{400}{\sqrt{3}}\)
• C. \(\dfrac{100\sqrt{2}}{3}\)
• D. (100)/(3)

Hint:

For a given gas, r.m.s. speed depends only on absolute temperature: vᵣms ∝ √(T) Pressure changes do not affect molecular speed.

Answer 1

The Correct Option is B

Step 1: The r.m.s. speed of gas molecules is given by vᵣms=√((3RT)/(M)) Hence, vᵣms ∝ √(T) and is independent of pressure.
Step 2: Initial temperature: T₁ = 27^∘C = 300K Final temperature: T₂ = 127^∘C = 400K
Step 3: Using proportionality: (v₂)/(v₁) = √((T₂)/(T₁)) = √((400)/(300)) = √((4)/(3))
Step 4: Hence, v₂ = 200 × √((4)/(3)) = \frac400√(3)m s⁻1