The root mean square velocity ‘c’ of a gas molecule is given by the following relations:

(1)

(2)

(3)

The first relation gives the r.m.s. speed in terms of Boltzman’s constant ‘k’ and the molecular mass ‘m’. The second one gives the r.m.s. speed in terms of universal gas constant ‘R’ and the molar mass ‘M’. The third one gives the r.m.s. speed in terms of the pressure and density of the gas.

Note that the r.m.s. speed is directly proportional to the square root of the absolute temperature of the gas and also that the r.m.s. speed of the molecules of a given gas is constant at a constant temperature. At constant temperature, even if you change the pressure, the r.m.s. speed is unchanged (as given by the third relation) since the density ‘ρ’ is directly proportional to the pressure ‘P’.

The gas molecules obey Maxwell’s distribution law and the most probable velocity is given by

Now, consider the following M.C.Q.:

If you want to write this in a mathematical step, you have v/v' = √(250/750) from which you obtain the final r.m.s. speed v'.

Now, let us discuss the folowing question:

(You can dispense with these steps and simply argue that the temperature must be halved when the pressure is halved at constant volume). Since the RMS speed is directly proportional to the square root of the absolute temperature, the final speed is 11.2×1/√2 = 7.9 km/s.

(1)

**c = √(3kT/m)**(2)

**c = √(3RT/M)**(3)

**c = √(3P/ρ)**The first relation gives the r.m.s. speed in terms of Boltzman’s constant ‘k’ and the molecular mass ‘m’. The second one gives the r.m.s. speed in terms of universal gas constant ‘R’ and the molar mass ‘M’. The third one gives the r.m.s. speed in terms of the pressure and density of the gas.

Note that the r.m.s. speed is directly proportional to the square root of the absolute temperature of the gas and also that the r.m.s. speed of the molecules of a given gas is constant at a constant temperature. At constant temperature, even if you change the pressure, the r.m.s. speed is unchanged (as given by the third relation) since the density ‘ρ’ is directly proportional to the pressure ‘P’.

The gas molecules obey Maxwell’s distribution law and the most probable velocity is given by

**c**_{mp}= √(2kT/m) = √(2RT/M) = √(2P/ρ).Now, consider the following M.C.Q.:

**Nitrogen gas molecules have r.m.s.speed ‘v’ at -23˚C. Their r.m.s speed at 477˚C will be**

(a) 4.55v (b) 20.7v (c) 9v (d) 1.732v (e) 1.414v

Note that the temperatures are given in degree Celsius. Convert them to the Kelvin scale to obtain 250K and 750K. The rise in the temperature is 3 times. Since the r.m.s. speed is directly proportional to the the square root of the absolute temperature, the final speed is √3 times the initial speed. So, the answer is 1.732v.(a) 4.55v (b) 20.7v (c) 9v (d) 1.732v (e) 1.414v

If you want to write this in a mathematical step, you have v/v' = √(250/750) from which you obtain the final r.m.s. speed v'.

Now, let us discuss the folowing question:

**Root mean square speed of oxygen gas molecules at a certain temperature is 11.2 km/s. The gas is cooled so that the pressure is halved without any change in its density. The final value of the root mean square speed is**

(a) 11.2 km/s (b) 7.9km/s (c) 6.4km/s (d) 5.6 km/s (e) 2.8km/s

As the density is unchanged, the volume of the gas is constant. Applying Charles law (P/T = P'/T'), we have P/T = (P/2)/T' from which T' = T/2.(a) 11.2 km/s (b) 7.9km/s (c) 6.4km/s (d) 5.6 km/s (e) 2.8km/s

(You can dispense with these steps and simply argue that the temperature must be halved when the pressure is halved at constant volume). Since the RMS speed is directly proportional to the square root of the absolute temperature, the final speed is 11.2×1/√2 = 7.9 km/s.

You can arrive at the answer in no time if you remember the expression for pressure ‘P’ in the form, P = ⅓ρc2. Since the density ‘ρ’ is unchanged, the RMS speed ‘c’ should become 1/√2 times the initial value when the pressure becomes half the initial value.

Let us consider another question:

Therefore, T

Consider now the following simple question which is often found in test papers:

**At what temperature will the molecules of nitrogen have the same r.m.s. speed as the molecules of oxygen at 27˚C?**

(a) -10.5˚C (b) -34.5˚C (c) -262.5˚C (d) 262.5˚C (e) 23.62˚C

Since the r.m.s. speed is given by c = √(3RT/M) we have √(3RT(a) -10.5˚C (b) -34.5˚C (c) -262.5˚C (d) 262.5˚C (e) 23.62˚C

_{n}/M_{n}) =√(3RT_{o}/M_{n}) where the suffix ‘n’ and suffix ‘o’ refer to nitrogen and oxygen respectively.Therefore, T

_{n}/28 = 300/32. Note that the molar mass of nitrogen is 28 and that of oxygen is32 and we have substituted the temperature in Kelvin. This yields the value 262.5K as the temperature of nitrogen molecules. The value in Celsius scale is -10.5˚C.Consider now the following simple question which is often found in test papers:

**If ‘v’ and ‘c’ are respectively the velocity of sound in a gas and the r.m.s. speed of the gas molecules at a particular temperature and γ is the ratio of specific heats, then**

(a) v = c (b) v = √γc (c) v = c√(γ/3) (d) v = √(γc/3) (e) v = √(3γ/c)(a) v = c (b) v = √γc (c) v = c√(γ/3) (d) v = √(γc/3) (e) v = √(3γ/c)

The correct option is (c) since we have v = √(γP/ρ) [Newton-Laplace equation] and c = √(3P/ρ). Therefore, v/c = √(γ/3) from which v = c√(γ/3).

To conclude this post, let me ask you a simple question: Do you know

**why the moon does not possess an atmosphere?**Many of you might be knowing the**reason:**The escape velocity on the moon’s surface is 2.4 km/s only (compared to 11.2 km/s for the earth) because of the smaller values of acceleration due to gravity and the radius in the case of the moon, compared to the values in the case of the earth. [Remember v=√(2gR)]. Molecular speeds on the moon can attain values significant compared to 2.4 km/s and hence gas molecules can escape into the outer space.
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