Data: T2 = 273 K, M01 (oxygen) = 32 × 10-3 kg/mol,
M02 (hydrogen) = 4 × 10-3 kg/mol,
`"v"_"rms" = sqrt("3RT"/"M"_0)`
The rms speed of oxygen molecules, `"v"_1 = sqrt("3RT"_1/"M"_01)` and that of helium molecules,
`"v"_2 = sqrt("3RT"_2/"M"_02)`
when v1 = v2 ,
`sqrt("3RT"_1/"M"_01) = sqrt("3RT"_2/"M"_02)`
`∴ "T"_1/"M"_01 = "T"_2/"M"_02`
∴ Temperature, T1 = `"M"_01/"M"_02. "T"_2 = ((32 xx 10^-3)(273))/(4 xx 10^-3)`
= 2184 K
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Concept:
Root Mean Square Speed
- It is defined as the square root of the mean of squares of the speed of different molecules.
- The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
- The rms speed of any homogeneous gas sample is given by:
Where R = universal gas constant, T = temperature and M = Molecular mass
Calculation:
Let velocity of oxygen = Vo
Let velocity of hydrogen = Vh
Molecular weight of oxygen Mo = 32
Molecular weight of hydrogen Mh = 2
Let the required temperature be T0 =
Temperature of hydrogen TH = 150 °C = 150 + 273 = 423 K
Now
Equating
Squaring and cancling constant terms
we get
⇒ To = 16 Th
⇒ To = 16 × 423 K = 6768 K
This temperature in °C will be 6768 - 273 = 6495°C
So, the correct option is 6495°C.
Mistake Points
Aspirants make take the temperature in °C. It should be taken in Kelvin, the SI unit for any calculation purpose.
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In this article, we shall study to find r.ms. speed of gas molecules, density of gas, pressure exerted by the gas using kinetic theory of gases.
Example 01:
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.
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 =?
Solution:
Ans: r.m.s. velocity of hydrogen molecule is 1842 m/s 0r 1.842 km/s
Example 02:
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.
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 =?
Solution:
Ans: r.m.s. velocity of nitrogen molecule is 493.1 m/s
Example 03:
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.
Given: Density = ρ = 1.44 kg/m3, P = 1.013 x 105 N/m2.
To Find: r.m.s. speed = C =?
Solution:
Ans: r.m.s. velocity of oxygen molecule is 459.4 m/s
Example 04:
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.
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 =?
Solution:
Ans: density of He is 0.1798 kg/m3
Example 05:
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.
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 =?
Solution:
Ans: Pressure of oxygen = 105 N/m2
Example 06:
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?
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 =?
Solution:
Ans: The ratio of r.m.s. speed is 1.08:1
Example 07:
At what temperature will the R.M.S. speed of the molecules of gas be three times its value at N.T.P.?
Given: Condition = N.T.P., T1 = 273 K, C2 = 3C1
To Find: Temperature = T2 =?
Solution:
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 08:
At what temperature will the R.M.S. speed of the molecules of gas be four times its value at N.T.P.?
Given: Condition = N.T.P., T1 = 273 K, C2 = 4C1
To Find: Temperature = T2 =?
Solution:
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 09:
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?
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 =?
Solution:
Part – II:
Given: Condition = N.T.P., T1 = 273 K, C1 = 497 m/s, C2= 994 m/s
To Find: Temperature = T2 =?
Solution:
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.
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, Temperature = T2 = 27 oC= 27 = 273 = 300 K
To Find: r.m.s. speed = C2 =?
Solution:
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.
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, Temperature = T2 = 127 oC= 127 = 273 = 400 K
To Find: r.m.s. speed = C2 =?
Solution:
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.
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, Temperature = T2 = 225 oC= 225 + 273 = 498 K
To Find: r.m.s. speed = C2 =?
Solution:
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?
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, Temperature = T2 = 200 oC= 200 + 273 = 473 K
To Find: r.m.s. speed = C2/ C1=?
Solution:
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.
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
Solution:
(C2)2 = (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.
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 =?
Solution:
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.
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 =?
Solution:
Ans: r.m.s speed of nitrogen at 340 K is 548 m/s
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