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3 years ago

5

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 23

5 K to 470K. If v1 is the average speed of the gas molecules before the absorption of heat and v2 their average speed after the absorption of heat, what is the ratio v2 v1

1 answer:

vichka [17]3 years ago

7 0

Answer:

V2 / V1 = √(2)

Explanation:

From kinetic theory of gas,

V = √(3RT/M)

v = speed of the gas

R = ideal gas constant

T = Temperature of the gas

M = molar mass of the gas

V = √(3RT/M)

V₂ / V₁ = √(3RT₂/M) / √(3RT₁/M)

3R / M = 3R / M = 1

V₂ / V₁ = √(T₂ / T₁)

V₂ / V₁ = √(470 / 235)

V₂ / V₁ = √(2)

The ratio of the two velocities V₂ / V₁ = √(2)

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