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Q.Deduce

(a) Graham's law and
(b) Dalton's law from Kinetic gas equation.
Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2025Subjective· 4mImportance★★★★★
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The kinetic gas equation gives urmsu_{rms} in terms of molar mass, from which Graham's law of diffusion follows directly; and because the pressure contributions of different, non-interacting gas molecules in a mixture simply add up, Dalton's law of partial pressures follows too.

The kinetic gas equation for nn moles of an ideal gas (molar mass MM) is:

PV=13Murms2(per mole)PV = \frac{1}{3}Mu_{rms}^2 \quad \text{(per mole)}

  1. Deriving Graham's Law of Diffusion: Since PV=RTPV = RT for one mole of an ideal gas, equating with the kinetic expression: 13Murms2=RT  ⟹  urms=3RTM\frac{1}{3}Mu_{rms}^2 = RT \implies u_{rms} = \sqrt{\frac{3RT}{M}} The rate of diffusion (or effusion) of a gas is proportional to the average speed of its molecules, i.e. r∝urmsr \propto u_{rms}. At the same temperature and pressure, for two gases 1 and 2: r1r2=urms,1urms,2=M2M1\frac{r_1}{r_2} = \frac{u_{rms,1}}{u_{rms,2}} = \sqrt{\frac{M_2}{M_1}} Since density d∝Md \propto M at constant T,PT,P, this is equivalently r1r2=d2d1\dfrac{r_1}{r_2}=\sqrt{\dfrac{d_2}{d_1}} — this is exactly Graham's law of diffusion: the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density).
  2. Deriving Dalton's Law of Partial Pressures: Consider a mixture of several non-reacting ideal gases (1, 2, 3, ...) occupying the same container of volume VV at temperature TT. Kinetic theory assumes gas molecules do not interact with each other (except during elastic collisions), so each gas's molecules independently contribute to the pressure exerted on the container walls exactly as if the other gases were absent. For gas ii alone in the same volume VV, it would exert a pressure (its partial pressure): PiV=13niMiui,rms2P_iV = \frac{1}{3}n_iM_iu_{i,rms}^2 …

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