#### Example – 1:

- Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. given : Density of Hydrogen at N.T.P. = ρ = 8.957 x 10
^{-2}kg/m^{3}. Density of mercury = 13600 kg/m^{3}, g = 9.8 m/s^{2}. **Solution:****Given:**Density of hydrogen = ρ = 8.957 x 10^{-2}kg/m^{3}, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9. 8 N/m^{2}, Density of Mercury = 13600 kg/m^{3}, g = 9.8 m/s^{2}.**To Find:**r.m.s. speed = C =?

**Ans : **r.m.s. velocity of hydrogen molecule is 1842 m/s 0r 1.842 km/s

#### Example – 2:

- Find the R.M.S. velocity of Nitrogen molecules at N.T.P. given that the density of Nitrogen at this temperature is 1.25 g/litre.
**Solution:****Given:**Density = ρ = 1.25 g/litre = 1.25 x 1 = 1.25 kg/m^{3}, condition N.T.P., P = 1.013 x 10^{5}N/m^{2}.**To Find:**r.m.s. speed = C =?

**Ans : **r.m.s. velocity of nitrogen molecule is 493.1 m/s

**Example – 3:**

- Calculate the R.M.S. velocity of Oxygen molecules at a pressure of 1.013 x 10
^{5}N/m^{2}(N.T.P.) given the density of Oxygen is 1.44 kg/m^{3}. **Solution:****Given:**Density = ρ = 1.44 kg/m^{3}, P = 1.013 x 10^{5}N/m^{2}.**To Find:**r.m.s. speed = C =?

**Ans : **r.m.s. velocity of oxygen molecule is 459.4 m/s

**Example – 4:**

- Calculate the density of He at N.T.P. given that R.M.S. velocity of He molecules at N.T.P. is 1300 m/s.
**Solution:****Given:**Condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, Density r.m.s. speed = C = 1300 m/s .**To Find:**Density = ρ = ?

**Ans: **density of He is 0.1798 kg/m^{3}

**Example – 5:**

- Determine the pressure of oxygen at 0 °C if the density of oxygen at NTP is 1.44 kg/m
^{3}and r.m.s speed of the molecule at NTP is 456.4 m/s. **Solution:****Given:**Density = ρ = 1.44 kg/m^{3}, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, r.m.s. speed = C = 456.4 m/s**To Find:**Pressure of gas = P = ?

**Ans:** Pressure of oxygen = 10^{5} N/m^{2}

#### Example – 6:

- Two gases are at temperatures of 77
^{o}C and 27^{o}C. What is the ratio of the R.M.S. velocities of the molecules of the two gases? **Solution:****Given:**temperature of first gas = T_{1}= 77^{o}C = 77 + 273 = 350 K, temperature of second gas = T_{2}= 27^{o}C = 27 + 273 = 300 K**To Find:**Ratio of r.m.s speeds =?

**Ans: **The ratio of r.m.s. speed is 1.08:1

#### Example – 7:

- At what temperature will the R.M.S. speed of the molecules of a gas be three times its value at N.T.P.?
**Solution:****Given:**Condition = N.T.P., T_{1}= 273 K, C_{2}= 3C_{1}**To Find:**Temperature = T_{2}=?

**Ans:** At a temperature of 2184 °C the R.M.S. speed of the molecules of

a gas is three times its value at N.T.P.

#### Example – 8:

- At what temperature will the R.M.S. speed of the molecules of a gas be four times its value at N.T.P.?
**Solution:****Given:**Condition = N.T.P., T_{1}= 273 K, C_{2}= 4C_{1}**To Find:**Temperature = T_{2}=?

**Ans:** At a temperature of 4095 °C the R.M.S. speed of the molecules of

a gas is four times its value at N.T.P.

#### Example – 9:

- The R.M.S. velocity of Nitrogen molecules at N.T.P. is 497 m/s. Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. At what temperature will the R.M.S. velocity of Nitrogen molecules be 994 m/s?
**Solution:****Part – I:****Given:**For nitrogen: Condition = N.T.P., T_{N}= 273 K, C_{N}= 497 m/s, P_{N}= 1.013 x 10^{5}N/m^{2}, For oxygen: Condition = N.T.P., T_{N}= 273 K, P_{H}= 1.013 x 10^{5}N/m^{2}.**To Find:**r.m.s. speed of hydrogen = C_{H}=?

**Part – II:****Given:**Condition = N.T.P., T_{1}= 273 K, C_{1}= 497 m/s, C_{2}= 994 m/s**To Find:**Temperature = T_{2}=?

**Ans : **r.m.s. speed of hydrogen molecules at NTP1860 m/s,

At a temperature of 819oC the speed of nitrogen molecules at NTP 994 m/s,

#### Example – 10:

- Calculate the R.M.S. velocity of Oxygen molecules at 27 °C. The density of Oxygen at N.T.P. 1.44 kg/m
^{3}. **Solution:****Given:**For nitrogen: Condition = N.T.P. T_{1}= 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, density of oxygen = ρ = 1.44 kg/m^{3}, Tempearture = T_{2}= 27^{o}C= 27 = 273 = 300 K**To Find:**r.m.s. speed = C_{2}=?

Now C_{1} = 459.4 m/s, T_{1} = 273 K, = T_{2} = 27 ^{o}C= 27 = 273 = 300 K, C_{2} = ?

**Ans: **The R.M.S. velocity of Oxygen molecules at 27 °C is 481.6 m/s

#### Example – 11:

- Compute the R.M.S. velocity of Oxygen molecules at 127 °C. Density of Oxygen at N.T.P. = 1.44 kg/m
^{3}. **Solution:****Given:**For nitrogen: Condition = N.T.P. T_{1}= 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, density of oxygen = r = 1.44 kg/m^{3}, Tempearture = T_{2}= 127^{o}C= 127 = 273 = 400 K**To Find:**r.m.s. speed = C_{2}=?

Now C_{1} = 459.4 m/s, T_{1} = 273 K, = T_{2} = 127 ^{o}C= 127 = 273 = 400 K, C_{2} = ?

**Ans: **The R.M.S. velocity of Oxygen molecules at 127 °C is 556.1 m/s.

#### Example – 12:

- Calculate the R.M.S. velocity of Oxygen molecules at 225 °C. The density of oxygen at NTP is 1.42 kg/m
^{5}. and 1 atmosphere = 1.013 x 10^{5}N/m^{2}. **Solution:****Given:**For nitrogen: Condition = N.T.P. T_{1}= 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, density of oxygen = ρ = 1.44 kg/m^{3}, Tempearture = T_{2}= 225^{o}C= 225 + 273 = 498 K**To Find:**r.m.s. speed = C_{2}=?

Now C_{1} = 459.4 m/s, T_{1} = 273 K, = T_{2} = 127 ^{o}C= 127 = 273 = 400 K, C_{2} = ?

**Ans: **The R.M.S. velocity of Oxygen molecules at 127 °C is 624.8 m/s

#### Example – 13:

- The density of a gas is 0.178 kg/m
^{3}at N.T.P. Find the R.M.S. velocity of gas molecules. By what factor will the velocity of molecules increase at 200 °C? **Solution:****Given:**For nitrogen: Condition = N.T.P. T_{1}= 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10^{5}N/m^{2}, density of oxygen = ρ = 0.178 kg/m^{3}, Tempearture = T_{2}= 200^{o}C= 200 + 273 = 473 K**To Find:**r.m.s. speed = C_{2}/ C_{1}=?

Now T_{1} = 273 K, = T_{2} = 200 ^{o}C= 200 + 273 = 473 K

**Ans: **The r.m.s. velocity of the gas molecule at NTP is 1.306 km/s,

The r.m.s. velocity will increase by a factor of 1.316 at 200 °C.

#### Example – 14:

- R M.S. velocity of oxygen molecules at 27 °C is 500 m/s. Calculate the R.M.S. and mean square velocities of oxygen molecules at 127 °C.
- Solution:
- Given: C1 = 500 m/s at temperature T
_{1}= 27^{o}C = 27 + 273 = 300 K, Required speed at temperature = T_{2}= 127^{o}C = 127 + 273 = 400 K, - To Find: C
_{2}= ? and (C_{2})^{2}

(C_{2})^{2 }=(577.4)^{2} = 3.33 x 10^{5} m^{2}/s^{2}

**Ans : **r.m.s. velocity is 577.4 m/s, and mean square velocity is 3.33 x 10^{5} m^{2}/s^{2}

Example – 15:

- Taking the R.M.S. velocity of Hydrogen molecules at N.T.P. as 1.84 km/s, calculate the R.M.S. velocity of Oxygen molecules at N.T.P. Molecular weights of Oxygen and Hydrogen are 32 and 2 respectively.
**Solution:****Given:**r.m.s. velocity of hydrogen = C_{H}= 1.84 km/s, molecular mass M_{O}= 32, M_{H}= 2, temperature T_{H}= T_{O}= 273 K, pressure P_{H}= P_{O}= 1.013 x 10^{5}N/m^{2}.**To Find:**r.m.s. velocity of oxygen molecule = C_{O}=?

**Ans : **The R.M.S. velocity of Oxygen molecules at N.T.P. is 0.46 km/s

#### Example – 16:

- R.M.S. speed of Oxygen molecules is 493 m/s at a certain temperature. Calculate the R.M.S. speed of helium molecules at the same temperature. Molecular weights of Oxygen and Helium are 32 and 4 respectively.
**Solution:****Given:**r.m.s. velocity of oxygen = C_{O}= 493 m/s, molecular mass M_{O}= 32, M_{He }= 4,

temperature T_{He}= T_{O}, Pressure P_{He}= P_{O}.**To Find:**r.m.s. velocity of helium molecule = C_{He}=?

**Ans:** The r.m.s. speed of helium molecule is 1394.2 m/s

#### Example – 17:

- R.M.S. speed of Oxygen molecules at N.T.P. is 459.3 m/s. Find the R.M.S. speed of Nitrogen molecules at 340 K. Molecular weights of Oxygen and Nitrogen are respectively 32 and 28.
**Solution:****Given:**r.m.s. speed of oxygen = C_{O1}= 459.3 m/s, T_{O1}= 273 K, T_{O2}= 340 K = T_{N}, M_{O}= 32, M_{N}= 28**To Find:**C_{N}=?

**Ans:** r.m.s speed of nitrogen at 340 K is 548 m/s