Solved Examples of Kinetic Theory of Gases

Problem 1 (JEE Main):-

Calculate the number of molecules in 2 $\times$ 10^-6 m³ of a perfect gas at 27°C and at a pressure of 0.01 m of mercury. Mean K.E of a molecule at 27°C = 4 $\times$ 10^-11 J and g = 9.8 ms^-2.

Solution:-

P = (1/3) (M/V) C² or PV = (1/3) MC²

But M = (m) (n)

Where ‘m’ is mass of one molecule and n is the number of molecules.

PV = (1/3) (m) C²

n = 3PV/mC² = [3/2 PV] / [(1/2) mC²]

Here, P = 0.01 m of mercury = 0.01 $\times$ 13600 $\times$ 9.8Nm^-2

V = 2 $\times$ 10^-6 m³ = 1332.8 Nm^-2

½ mc² = 4 $\times$ 10^-11J

$n = \frac{3}{2}\times \frac{1332.8\times 2\times 10^{-6}}{4\times 10^{-11}}$ $n = \frac{3}{2}\times \frac{1332.8\times 2\times 10^{-6}}{4\times 10^{-11}} = 9.996\times 10^{7}$

Thus from the above observation we conclude that, the number of molecules would be 9.996 $\times$ 10⁷.

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Problem 2:-

You are throwing a birthday party and decide to fill the room with helium balloons. You also want to have a few larger balloons to put at the door. The smaller balloons are filled occupy 0.240 m³ when the pressure inside them is 0.038 atm and the temperature of the room is 70° F. What pressure should you fill the larger balloons to so that they occupy 0.400 m³?

Solution:-

The temperature of the room is assumed to be held constant, so it is extraneous information. Since you are dealing with volume and pressure, you would use Boyle's Law.

In accordance to data,

P₁= 0.038 atm, V₁= 0.240 m³

V₂=0.400 m³

We have to find out V₂.

Substitute the vale of P₁, V₁ and V₂ in equation P₁V₁=P₂V₂

P₁V₁=P₂V₂

(0.038 atm)(0.240 m³) = P₂(0.400 m³)

0.00912 = 0.400 P₂ (units cancel out so that pressure will be in atm)

P₂ = 0.0228 atm

From the above observation we conclude that, the pressure would be 0.0228 atm.

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Problem 3:-

Find the RMS speed of a sample of neon gas at 80° F.

Solution:-

First convert Fahrenheit to Celsius.

°C = (°F- 32)/1.8

= (80 -32)/1.8

= 48/1.8

= 26.6

Convert Celsius to Kelvin.

K = °C + 273.15

= 26.6 + 273.15

= 299.75 K

Substitute the known information into the equation for RMS speed and solv, we get,.

v_rms = √3RT/M_x

= √3(8.3144) (299.75)/ (20.179)

= 19.2 m/s

From the above observation we conclude that, the RMS speed of a sample of neon gas at 80° F would be 19.2 m/s.

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