First slide

Kinetic theory of ideal gases

Question

A gas molecule at the surface of earth happens to have rms speed for the gas at 0^oC. Suppose it went straight up without colliding with other molecules, how high would it rise? Mass of the molecule is $4.65 \times 10^{- 26} k g .$ Boltzmann’s constant $= 1.38 \times 10^{- 23} J K^{- 1}$ .

Difficult

A

12.2 km
B

4.1 km
C

6.3 km
D

3.2 km

Solution

$\begin{array}{l} v_{RMS} = \sqrt{\frac{3 k_{B} T}{M}} = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 273}{4.65 \times 10^{- 26}}} \\ v_{RMS} = 493 {ms}^{- 1} \end{array}$
Particle moves upwards under the influence of gravity and stops (temporarily) after some time.
$\begin{array}{l} h = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 g} = \frac{(493)^{2}}{2 \times 10} \\ \Rightarrow h = 1.22 \times 10^{4} m \end{array}$

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