Subject: Physics

Consider one mole of an ideal gas at pressure P, volume V and temperature T. From the equation of state for ideal gas, we have

$$PV = RT \dots (i)$$

Again from kinetic theory of gases, the pressure of gas

$$ P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

where M is the molar mass of the gas

$$ \therefore PV= \frac {1}{3} M \bar c^2 \dots (ii)$$

From equation (i) and (ii), we get

$$\frac {1}{3}M \bar c^2 = RT $$

$$\text{or,}\frac {1}{2}M \bar c^2 = \frac {3}{2} RT \dots (iii)$$

Here, \(\frac {1}{2} M c^2\) is the average kinetic energy of one mole. So, average kinetic energy per mole of the gas is \(\frac {3}{2} RT\). If the mass of a molecule is m, and N is the number of molecules in one mole of gas, equation (iii) can be written as

$$\frac {1}{2} mN \bar c^2 = \frac {3}{2} RT$$

$$\text{or,}\frac {1}{2}m \bar c^2 = \frac {3}{2} \frac {R}{N}T =\frac {3}{2} KT$$

where, \(\frac {R}{N} = K \) is the Boltzman constant. It's value is \( 1.38 \times 10^{-23} J mole^{-1} K^{-1} \) .

Here \(\frac{1}{2} m\bar c^2\) is the average kinetic energy per molecule of the gas. So,

$$ \text{average kinetic energy} =\frac {3}{2} KT $$

From the Kinetic theory of gases, the average kinetic energy of gas is \(\frac {1}{2}M \bar c^2 = \frac {3}{2} KT\). This shows that the average kinetic energy of the gas is directly proportional to the absolute temperature. Thus, the temperature of the gas gives a measure of an average kinetic energy of the gas.

$$\text {If} T = 0, \frac {1}{2} m\bar c^2 = 0$$

Thus at absolute zero temperature, the molecules of the gas are at completely rest.

Root mean square speed of gas molecule is a square root of mean square of velocities of all gas molecules. i.e

$$c_{rms} = \sqrt {\frac{c_1^2 + c_2^2 + c_3^2 \dots + c_n^2 }{n}}$$

$$(c_{rms})^2 =\frac{c_1^2 + c_2^2 + c_3^2 \dots + c_n^2 }{n}$$

$$\;\;\;\;(c_{rms})^2 = \bar c^2\;\;\;\;\;\;\;\;\;\;\; [\because \bar c^2 = \frac{c_1^2 + c_2^2 + c_3^2 \dots + c_n^2 }{n}$$

The pressure exerted by the gas is

$$p = \frac{1}{3} \rho \bar c^2$$

$$\text {or,} \sqrt {\bar c^2} = \sqrt {\frac {3p}{\rho }} $$

$$\text {or,} c_{rms} = \sqrt {\frac {3p}{\rho }} \dots (i) $$

Again, from average kinetic energy per mole of a gas, we have

$$\frac {1}{2} M \bar c^2 = \frac {3}{2} RT$$

$$\text {or,} \sqrt {\bar c^2} = \sqrt {\frac {3RT}{M}} $$

$$\text {or,} c_{rms} = \sqrt {\frac {3RT}{M}} \dots (ii) $$

Again from \(\frac {1}{2} mc^2 = \frac {3}{2} KT \), we have

$$\text {or,} \sqrt {\bar c^2} = \sqrt {\frac {3KT}{m}} $$

$$\text {or,} c_{rms} = \sqrt {\frac {3kT}{m}} \dots (iii) $$

From equation (ii) and (iii), we see that

$$c_{rms} \propto \sqrt T $$

Thus, the root mean square speed of the gas is directly proportional to the square root of the absolute temperature.

The average kinetic energy per mole of a gas is

$$\frac {1}{2} M \bar c^2 = \frac {3}{2} RT$$

where M is molar mass of the gas,

$$\text {or,} \sqrt {\bar c^2} = \sqrt {\frac {3RT}{M}} $$

$$\text {or,} c_{rms} = \sqrt {\frac {3RT}{M}} $$

- The root mean square speed or velocity of the molecules of a given gas is directly proportional to the square root of the absolute temperature,

$$c_{rms} \propto \sqrt T $$ - The root mean square speed or velocity of the different gases at the same temperature is inversely proportional to the square root of molar mass of the gases,

$$c_{rms} \propto \sqrt {\frac {1}{M}}$$

This shows that c_{rms} vary for different gases, all the gases have different molar masses.

Root mean square speed of gas molecule is a square root of mean square of velocities of all gas molecules.

The root mean square speed or velocity of the molecules of a given gas is directly proportional to the square root of the absolute temperature.

The root mean square speed or velocity of the different gases at the same temperature is inversely proportional to the square root of molar mass of the gases.

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