Problems Based on Kinetic Theory of Gases

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/m3 . Density of mercury  = 13600 kg/m3, g = 9.8 m/s2 .
  • Solution:
  • Given: Density of hydrogen =  ρ = 8.957 x 10-2 kg/m3, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9. 8 N/m2, Density of Mercury = 13600 kg/m3, g = 9.8 m/s2 .
  • 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/m3, condition N.T.P., P = 1.013 x 105 N/m2.
  • 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 105 N/m2 (N.T.P.) given the density of Oxygen is 1.44 kg/m3.
  • Solution:
  • Given: Density =  ρ = 1.44  kg/m3, P = 1.013 x 105 N/m2.
  • 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 105 N/m2, Density r.m.s. speed = C = 1300 m/s .
  • To Find: Density =  ρ =  ?

Ans: density of He is 0.1798 kg/m3

Example – 5:

  • Determine the pressure of oxygen at 0 °C if the density of oxygen at NTP is 1.44 kg/m3 and r.m.s speed of the molecule at NTP is 456.4 m/s.
  • Solution:
  • Given: Density =  ρ = 1.44  kg/m3, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 105 N/m2, r.m.s. speed = C = 456.4 m/s
  • To Find: Pressure of gas = P = ?

Ans: Pressure of oxygen = 105 N/m2

Example – 6:

  • Two gases are at temperatures of 77 oC and 27 oC. What is the ratio of the R.M.S. velocities of the molecules of the two gases?
  • Solution:
  • Given: temperature of first gas = T1 = 77 oC = 77 + 273 = 350 K, temperature of second gas = T2 = 27 oC = 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., T1 = 273 K, C2 = 3C1
  • To Find: Temperature = T2 =?

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., T1 = 273 K, C2 = 4C1
  • To Find: Temperature = T2 =?

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., TN = 273 K, CN = 497 m/s, PN = 1.013 x 105 N/m2, For oxygen: Condition = N.T.P., TN = 273 K, PH = 1.013 x 105 N/m2.
  • To Find: r.m.s. speed of hydrogen = CH =?

  • Part – II:
  • Given: Condition = N.T.P., T1 = 273 K,  C1 = 497 m/s, C2= 994 m/s
  • To Find: Temperature = T2 =?

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/m3.
  • Solution:
  • Given: For nitrogen: Condition = N.T.P. T1 = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 105 N/m2, density of oxygen = ρ = 1.44 kg/m3,  Tempearture = T2 = 27 oC= 27 = 273 = 300 K
  • To Find: r.m.s. speed = C2 =?

Now C1 = 459.4 m/s, T1 = 273 K, = T2 = 27 oC= 27 = 273 = 300 K, C2 = ?

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/m3.
  • Solution:
  • Given: For nitrogen: Condition = N.T.P. T1 = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 105 N/m2, density of oxygen = r = 1.44 kg/m3,  Tempearture = T2 = 127 oC= 127 = 273 = 400 K
  • To Find: r.m.s. speed = C2 =?

Now C1 = 459.4 m/s, T1 = 273 K, = T2 = 127 oC= 127 = 273 = 400 K, C2 = ?

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/m5. and 1 atmosphere = 1.013 x 105 N/m2 .
  • Solution:
  • Given: For nitrogen: Condition = N.T.P. T1 = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 105 N/m2, density of oxygen = ρ = 1.44 kg/m3,  Tempearture = T2 = 225 oC= 225 + 273 = 498 K
  • To Find: r.m.s. speed = C2 =?

Now C1 = 459.4 m/s, T1 = 273 K, = T2 = 127 oC= 127 = 273 = 400 K, C2 = ?

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/m3 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. T1 = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 105 N/m2, density of oxygen = ρ = 0.178 kg/m3,  Tempearture = T2 = 200 oC= 200 + 273 = 473 K
  • To Find: r.m.s. speed = C2/ C1=?

Now T1 = 273 K, = T2 = 200 oC= 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 T1 = 27 oC = 27 + 273 = 300 K, Required speed at temperature = T2 = 127 oC = 127 + 273 = 400 K,
  • To Find: C2 = ?  and (C2)2

(C2)=(577.4)2 = 3.33 x 105 m2/s2

Ans : r.m.s. velocity is 577.4 m/s, and mean square velocity is 3.33 x 105 m2/s2

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 = CH = 1.84 km/s, molecular mass MO = 32, MH = 2, temperature TH = TO = 273 K, pressure PH = PO = 1.013 x 105 N/m2.
  • To Find: r.m.s. velocity of oxygen molecule = CO =?

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 = CO = 493 m/s, molecular mass MO = 32, MHe = 4,
    temperature THe
     = TO, Pressure PHe = PO.
  • To Find: r.m.s. velocity of helium molecule =  CHe=?

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 = CO1 = 459.3 m/s, TO1 = 273 K, TO2 = 340 K  = TN , MO = 32, MN = 28
  • To Find: CN =?

r.m.s. velocity

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

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