Question

Calculate the most probable speed, average speed & root mean square speed for gas molecule at 300K .the mass of gas molecule is 5×1/10000000000000000000000000000 kg &Kb = 1.38 ×1/1000000000000000000000000 J/know.

Answer 1

Answer on Question #66756-Physics-Molecular Physics-Thermodynamics

Calculate the most probable speed, average speed & root mean square speed for gas molecule at 300K .the mass of gas molecule is $5 \times 1 / 100000000000000000000000000000000000000000000000000000000000000000000000000$ kg

Solution

The root mean square speed is

$$v _ {r m s} = \sqrt {\frac {3 R T}{M _ {m}}} = \sqrt {\frac {3 R T}{m \cdot N _ {A}}} = \sqrt {\frac {3 \cdot 8 . 3 1 \frac {J}{K \cdot m o l} \cdot 3 0 0 K}{5 \cdot 1 0 ^ {- 2 8} k g \cdot 6 . 0 2 \cdot 1 0 ^ {2 3} \frac {1}{m o l}}} = 4 9 8 5 \frac {m}{s}.$$

The average speed is

$$v _ {a v e} = \sqrt {\frac {8 R T}{\pi M _ {m}}} = \sqrt {\frac {8 R T}{\pi m \cdot N _ {A}}} = \sqrt {\frac {8 \cdot 8 . 3 1 \frac {J}{K \cdot m o l} \cdot 3 0 0 K}{5 \pi \cdot 1 0 ^ {- 2 8} k g \cdot 6 . 0 2 \cdot 1 0 ^ {2 3} \frac {1}{m o l}}} = 4 5 9 2 \frac {m}{s}.$$

The most probable speed is

$$v _ {p} = \sqrt {\frac {2 R T}{M _ {m}}} = \sqrt {\frac {2 R T}{m \cdot N _ {A}}} = \sqrt {\frac {2 \cdot 8 . 3 1 \frac {J}{K \cdot m o l} \cdot 3 0 0 K}{5 \cdot 1 0 ^ {- 2 8} k g \cdot 6 . 0 2 \cdot 1 0 ^ {2 3} \frac {1}{m o l}}} = 4 0 7 0 \frac {m}{s}.$$

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