an ideal gas molecule is present at 27 degree celsius .By how many degree centigrade its temperature should be raised so that its Vrms, Vmp and Vav all may double
a-900°C
b-81°C
Answers
Answer:
The initial temperature of an ideal gas molecule, T = 27℃ = 27 + 273 K = 300 K
We know,
The relation between the velocities of an ideal gas molecule is,
Vrms : Vmp : Vav = √[3RT/M] : √[2RT/M] : √[8RT/πM] ….. (i)
We are given that, Vrms, Vmp & Vav are doubled.
Except for the temperature “T” all other terms in the formulas are constant and the overall change in the velocity will have the same effect to the temperature for all the 3 velocity formulas, so we will consider any one of the velocity formulas for the ideal gas molecule from eq. (i),
Final Vrms = 2*√[3RT/M] = √[4*3RT/M] = √[3R(4*300)/M] = √[3R(1200)/M]
∴ Final temperature, after Vrms is doubled = 1200 K = 1200 – 273 = 927℃
Thus,
In order to double the Vrms, Vav & Vmp, the temperature should be raised by
= Final temperature – Initial temperature
= (927 – 27) ℃
= 900 ℃