Question

(d) A gas has P-105 Pa. V-0.3 m3 and T--300 K. The area un- der the Maxwell-Boltzmann distribution (as described in the notes) between v 0 → 1730 m s-1 is 0.81. Calculate how many molecules have speed v1730 ms-1

Answer 1

Given, gas has Pressure (P) = 10⁵ Pa , Volume (V) = 0.3 m³ and Temperature (T) = 300 K.

The ideal gas equation is

PV = nRT

Where, n is no of moles of gas and

R = 8.314 J/mol.K is universal gas constant

So, for the given gas

  

Given, gas has Pressure (P) = 105 Pa, Volume (V) = 0.3 m3 and Temperature (T) = 300 K. The ideal gas equation is PV = nRT Where, n is no of moles of gas and R = 8.314 J/mol. K is universal gas constant So, for the given gas n = PV/RT = 10^5 times 0.3/8.314 times 300 = 12.027 almostequalto 12 mol So, 12 moles of gas is present in the system. We know that, 1 mol = 6.022 x 1023 particle So, total no of molecules present in the system = 12.027 x 6.022 x 1023 = 72.43 x 10^23 Now, the area under the Maxwell-Boltzmann (MB) distribution of a certain system from zero up to a certain molecular speed (v), represents the probability of finding a molecule with a molecular speed smaller or equal than v in that system. According to the question, The area under the MB distribution between v = 0 to 1730 m/s is 0.81.

So, 12 moles of gas is present in the system.

We know that , 1 mol = 6.022 x 10²³ particle

So, total no of molecules present in the system = 12.027 x 6.022 x 10²³ = 72.43 x 10²³

Now, the area under the Maxwell-Boltzmann (MB) distribution of a certain system from zero up to a certain molecular speed (v), represents the probability of finding a molecule with a molecular speed smaller or equal than v in that system.

According to the question, The area under the MB distribution between v = 0 to 1730 m/s is 0.81.

i.e. the probability of finding a molecule with a molecular speed 0 to 1730 m/s is 0.81.

So, probability of finding a molecule with a molecular speed greater than 1730 m/s is = 1 -0.81 = 0.19

i.e. 19 % molecules have molecular speed greater than 1730 m/s.

or, the no of molecules with a molecular speed greater than 1730 m/s

   $=72.43\times 10^{23}\times \frac{19}{100} = 13.76 \times 10^{23}$

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