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Question: The RMS speed of \({{N}_{2}}\) molecule at STP (P=1 atm; T=0\(^\circ\)C) is……. (the density of \({...

The RMS speed of N2{{N}_{2}} molecule at STP (P=1 atm; T=0∘^\circC) is…….
(the density of N2{{N}_{2}} in these conditions is 1.25 kg/m3m^3)
A. 493 m/s
B. 390 m/s
C. 290 m/s
D. 590 m/s

Explanation

Solution

Using the formula for the energy of gas particles, we will assume that all the particles have same kinetic energy and then solve for the velocity that all particles must have so that their combined kinetic energy is equal to the energy as given by the formula. The velocity that we get is termed as root mean square velocity or Vrms{{V}_{rms}}.

Formula used:
PV=23EPV=\dfrac{2}{3}E

Complete answer:
First, we will find the root mean square speed of the N2{{N}_{2}} molecule in the gas. It can be found by using the formula
PV=23E=23×12mvrms2PV=\dfrac{2}{3}E=\dfrac{2}{3}\times\dfrac{1}{2}m{{v}_{rms}}^{2}
Here, E is the total kinetic energy of all the gas molecules.
P is the pressure of gas which in this case is 1 atm = 101325 Pascals
m is the mass of the gas (in kg)

& {{v}_{rms}}^{2}=\dfrac{3PV}{m} \\\ & \dfrac{m}{V}=\rho \\\ & \therefore {{v}_{rms}}^{2}=\dfrac{3P}{\rho } \\\ \end{aligned}$$ Here ρ is the density of the gas, which is given as 1.25 kg/m3 . So, $${{v}_{rms}}=\sqrt{\dfrac{3P}{\rho }}=\sqrt{\dfrac{3\times 101325}{1.25}}=493$$m/s **Hence, the correct option is (a) 493 m/s** **Additional Information:** Root mean square speed of the gas is a measure of average kinetic energy possessed by the molecules of the gas. It only depends on the temperature of the gas and is completely independent of pressure, volume and the nature of gas for an ideal gas. This interpretation of the relation of temperature with the average energy of molecules shows complete consistency between the kinetic theory of gases and ideal gas equation along with the various laws based on it. **Note:** Take care to take R in SI units i.e. J/mol.K; T in Kelvin and M in kilograms. Calculation mistakes must be avoided. Alternatively, we can use the formula $${{v}_{rms}}^{2}=\sqrt{\dfrac{3RT}{M}}$$, here R is the universal gas constant, T is the temperature in Kelvin and M is the molar mass in kg. $${{v}_{rms}}^{2}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3\times 8.314\times 273.15}{0.028}}=493$$m/s. As we can see both the formulae give the same result.