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:
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:
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For a given gas, r.m.s. speed depends only on absolute temperature:
vᵣms ∝ √(T)
Pressure changes do not affect molecular speed.
Updated On: Mar 19, 2026
- \(100\sqrt{2}\)
- \(\dfrac{400}{\sqrt{3}}\)
- \(\dfrac{100\sqrt{2}}{3}\)
- (100)/(3)
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The Correct Option is B
Solution and Explanation
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
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