Name: Graham's Diffusion Law
Uploaded: 2019-07-19
Duration: 9 min 43 s
Description: Gases diffuse at a different rate. Watch this video to understand Graham's diffusion law, which deals with the diffusion of gases. Have fun learning!

Question

Deduce any three gas laws from kinetic gas equation.

Accepted Answer

Deduction of Boyle's law from kinetic gas equation:

According to kinetic gas equation, $PV = \frac{1}{3} {mnc}^{2}$ $= \frac{2}{3} \times \frac{1}{2} {mnc}^{2}$ $= \frac{2}{3} \times \frac{1}{2} {Mc}^{2}$ ----(i) where, ( $M = m \times n$ , is the total mass of the gas).

But, kinetic energy $E_{k}$ $= \frac{1}{2} {Mc}^{2}$ therefore, equation (i) becomes, $PV = \frac{2}{3} E_{k}$ .

According to one of the postulates of the kinetic theory of gases, $E_{k} \propto$ Absolute Temperature ( $T$ ) or $E_{k} = k T$ -----(ii), where, $k$ is the proportionality constant. On substituting the value of $E_{k}$ in equation (i), we get $PV = \frac{2}{3} \times kT$ -------(iii). In this equation, $\frac{2}{3}$ and $k$ are constants therefore, if $T$ is kept constant, $(\frac{2}{3} kT)$ also becomes constant hence, the equation (iii) can be written as $PV = Constant$ , which is Boyle's law.

Deduction of Charles' law from kinetic gas equation:

According to kinetic gas equation, $PV = \frac{1}{3} {mnc}^{2}$ $= \frac{2}{3} \times \frac{1}{2} {mnc}^{2}$ $= \frac{2}{3} \times \frac{1}{2} {Mc}^{2}$ ----(i) where, ( $M = m \times n$ , is the total mass of the gas).

But, kinetic energy $E_{k}$ $= \frac{1}{2} {Mc}^{2}$ therefore, equation (i) becomes, $PV = \frac{2}{3} E_{k}$ .

According to one of the postulates of the kinetic theory of gases, $E_{k} \propto$ Absolute Temperature ( $T$ ) or $E_{k} = k T$ -----(ii), where, $k$ is the proportionality constant. On substituting the value of $E_{k}$ in equation (i), we get $PV = \frac{2}{3} \times kT$ -------(iii).

Equation (iii) can be written as, $\frac{V}{T} = \frac{2}{3} \times \frac{k}{P}$ ------(iv), $\frac{2}{3}$ and $k$ are constants hence, if $P$ is kept constant, according to equation (iv) $\frac{V}{T} = Constant$ , which is Charles' law.

Deduction of Dalton's law of partial pressures from kinetic gas equation:

Let us consider only two gases. According to kinetic gas equation, $PV = \frac{1}{3} {mnc}^{2}$ or $P = \frac{1}{3 V} {mnc}^{2}$

Now, if only the first gas is enclosed in the vessel of volume $V$ , the pressure exerted would be $P_{1} = \frac{1}{3 V} m_{1} n_{1} {c_{1}}^{2}$ . Again, if only the second gas is enclosed in the same vessel (so that $V$ is constant), then the pressure exerted would be, $P_{2} = \frac{1}{3 V} m_{2} n_{2} {c_{2}}^{2}$ .

Lastly, if both the gases are enclosed together in the same vessel then since the gases do not react with each other, their molecules behave independent of each other. Hence, the total pressure exerted would be,

$P =$ $\frac{1}{3 V} m_{1} n_{1} {c_{1}}^{2} +$ $\frac{1}{3 V} m_{2} n_{2} {c_{2}}^{2}$ $= P_{1} + P_{2}$ . Similarly, if more than two gases are present, then it can be proved that $P = P_{1} + P_{2} + P_{3} + \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$ this is Dalton's law of partial pressures.

Deduce any three gas laws from kinetic gas equation.

Important Questions on States of Matter : Gases and Liquids

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