The sample of an ideal gas is in a tank of constant volume the sample absorbs heat energy so that its temperature changes from 300 to 600 K if v is the average speed of the gas molecules before the absorption of heat and v is their average speed after the absorption of heat what is the ratio of v2/v1

Explanation

Given that

Temperature changes from

Initial temperature T1 = 300K

Final temperature T2 = 600K

v1 = average speed of gas molecule before absorption of heat

v2 = average speed of gas molecule after absorption of heat

Ratio v2/v1 =?

Related to average speed,

Vrms = √(3RT/M)

We will take 3, T and M as constant,

Then,

Vrms ∝ √T

Then,

Vrms / √T = k

So,

V1 / √T1 = V2 / √T2

Cross multiply

V1 • √T2 = V2 • √T1

Make V2 / V1 subject of formula

V2 / V1 = √T2 / √ T1

V2 / V1 = √600 / √300

We know from indices that

√a / √b = √(a/b)

Then,

V2 / V1 = √(600 / 300)

V2 / V1 = √2

Then, The ratio v2 / v1 is √2

So, if T is double, then, V increases by a factor of √2

(75°f − 32) × 5/9 = 23.889°c