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Standard XII

Physics

Root Mean Square Speed

At room tempe...

Question

At room temperature ${N H}_{3}$ gas at one atm and $H C l$ gas at $P$ atm are allowed to effuse through identical pin holes to the opposite ends of a glass tube $1 m$ long and uniform cross-section. A white deposit is observed at a distance of $60 c m$ from the $H C l$ end. What is the value of $P$ ?

3.197 atm

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2.198 atm

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4.197 atm

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1.197 atm

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Solution

The correct option is B 2.198 atm
Using Graham's law of diffusion
$\frac{r_{H C l}}{4_{N H_{3}}} = \sqrt{\frac{M_{N H_{3}}}{M_{H C l}}} \times \frac{P_{H C l}}{P_{N H_{3}}}$
$\frac{60}{40} = 0.628 \times P_{H C l}$
$P_{H C l} = 2.198 a t m$

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At room temperature NH3 gas at one atm and HCl gas at P atm are allowed to effuse through identical pin holes to the opposite ends of a glass tube 1m long and uniform cross-section. A white deposit is observed at a distance of 60cm from the HCl end. What is the value of P ?

The correct option is B 2.198 atmUsing Graham's law of diffusionrHCl4NH3=√MNH3MHCl×PHClPNH36040=0.628×PHClPHCl=2.198 atm

The correct option is B 2.198 atm
Using Graham's law of diffusion
$\frac{r_{H C l}}{4_{N H_{3}}} = \sqrt{\frac{M_{N H_{3}}}{M_{H C l}}} \times \frac{P_{H C l}}{P_{N H_{3}}}$
$\frac{60}{40} = 0.628 \times P_{H C l}$
$P_{H C l} = 2.198 a t m$