Question

A hydrogen molecule (diameter $= 1.0 \times 10^{- 8} cm$ ) travelling at the $R M S$ speed, escapes from a $2000 K$ furnace into a chamber containing cold argon atoms (diameter $= 3.0 \times 10^{- 8} cm$ ) at a density $2.0 \times 10^{19}$ atoms per ${cm}^{3}$ .

$(A)$ What is the speed of the hydrogen molecule?

$(B)$ If it collides with an argon atom, what is the closest distance, their centers can be, considering each as spherical?

$(C)$ What is the initial number of collisions per second experienced by the hydrogen molecule? (Assume the argon atoms to be stationary)

Accepted Answer

$(A)$

The root mean square speed is given by

$v_{rms} = \sqrt{\frac{3 R T}{M}}$

The molar mass of hydrogen is $2000 K$ .

Hence,

$v_{rms} = \sqrt{\frac{3 (8.31 J mol K^{- 1}) (2000 K)}{2.02 \times 10^{- 3} kg {mol}^{- 1}}} = 5.0 \times 10^{3} m s^{- 1}$

$(B)$

When the surfaces of the spheres that represent an $H_{2}$ molecule and an $Ar$ atom will touch, the distance between their centers is the sum of their radii.

$d = r_{1} + r_{2} = 0.5 \times 10^{- 8} cm + 1.5 \times 10^{- 8} cm = 2.0 \times 10^{- 8} cm$

$(C)$

The argon atoms are initially at rest.

Hence, in time $t$ , the hydrogen atom collides with all the argon atoms in a cylinder of radius $d$ and length $v t$ (where $v$ is its speed).

The number of collisions is $\frac{π d^{2} v t N}{V}$ .

where,

$\frac{N}{V}$ is the concentration of argon atoms.

The number of collisions per unit time is,

$\frac{π d^{2} v N}{V} = π {(2.0 \times 10^{- 10} m)}^{2} (5.0 \times 10^{3} m s^{- 1}) (2.0 \times 10^{25} m^{- 3}) = 1.25 \times 10^{10} collisions s^{- 1}$

Important Questions on The Kinetic Theory of Gases

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