Question

calculate the root mean square velocity of molecule of hydrogen gas at room temperture 27c and calculate the mean kinetick energy of one molecule and one mole

Answer 1

Answer on Question #43116 – Physics – Molecular Physics | Thermodynamics

Question.

Calculate the root mean square velocity of molecule of hydrogen gas at room temperature 27c and calculate the mean kinetic energy of one molecule and one mole.

Given:

$$T = 27{}^{\circ} \mathrm{C} = 300 \, \mathrm{K}$$ $$m(H_2) = 3.34 \cdot 10^{-27} \, \mathrm{kg}$$ $$M(H_2) = 2 \cdot 10^{-3} \, \frac{\mathrm{kg}}{\mathrm{mole}}$$ $$\nu = 1 \, \mathrm{mole}$$

Find:

$$\bar{\nu} = ?$$ $$E_k(1 \, \mathrm{molecule}) = ?$$ $$E_k(1 \, \mathrm{mole}) = ?$$

Solution.

From thermodynamics it's known that the average kinetic energy of the gas is proportional to temperature:

$$E_k = \frac{3}{2} k T$$

$k = 1.38 \cdot 10^{-23} \, \frac{\mathrm{J}}{\mathrm{K}}$ is the Boltzmann's constant;

$T$ is the temperature.

$$R = k \cdot N_a$$

$R = 8.31 \, \frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}$ the gas constant;

$N_a = 6.02 \cdot 10^{23} \, \mathrm{mole}^{-1}$ is the Avogadro constant.

So, the kinetic energy of 1 mole is:

$$E_k = \frac{3}{2} k T \cdot N_a = \frac{3}{2} R T$$

On the other hand, from classical mechanics we know that the kinetic energy of a particle is:

$$E _ {k} = \frac {m}{2} \bar {v} ^ {2}$$

Therefore, the kinetic energy of $N$ particles is:

$$E _ {k} = N \frac {m}{2} \bar {v} ^ {2}$$

There $N_{a}$ molecules in 1 mole of the substance $\rightarrow$ $N = N_{a}$

The kinetic energy of 1 mole is:

$$E _ {k} = N _ {a} \frac {m}{2} \bar {v} ^ {2}$$

So,

$$E _ {k} = N _ {a} \frac {m}{2} \bar {v} ^ {2} = \frac {3}{2} R T \rightarrow \bar {v} ^ {2} = \frac {3 R T}{m N _ {a}}$$ $$\nu = \frac {m}{M} = \frac {N}{N _ {a}}$$

$\nu$ is the amount of substance;

$M$ is the molar mass.

If $\nu = 1$ mole, then $M = mN_{a}$

Thus,

$$\bar {v} ^ {2} = \frac {3 R T}{M} \rightarrow \bar {v} = \sqrt {\frac {3 R T}{M}}$$

Let calculate all desired values.

The root mean square velocity is:

$$\bar {v} = \sqrt {\frac {3 R T}{M}} = \sqrt {\frac {3 \cdot 8.31 \cdot 300}{2 \cdot 10 ^ {- 3}}} = 1934 \frac {m}{s}$$

The kinetic energy of 1 molecule:

$$E _ {k} (1 \text{ molecule}) = \frac {m}{2} \bar {v} ^ {2} = \frac {3}{2} k T = \frac {3}{2} 1.38 \cdot 10 ^ {- 23} \cdot 300 = 6.21 \cdot 10 ^ {- 21} J$$

The kinetic energy of 1 mole:

$$E _ {k} (1 \text{ mole}) = N _ {a} E _ {k} (1 \text{ molecule}) = N _ {a} \frac {m}{2} \bar {v} ^ {2} = \frac {3}{2} R T = \frac {3}{2} 8.31 \cdot 300 = 3739.5 J$$

Answer.

The root mean square velocity is:

$$\bar {v} = \sqrt {\frac {3 R T}{M}} = 1934 \frac {m}{s}$$

The kinetic energy of 1 molecule:

$$E _ {k} (1 \text { molecule}) = \frac {m}{2} \bar {v} ^ {2} = \frac {3}{2} k T = 6.21 \cdot 10 ^ {- 21} J$$

The kinetic energy of 1 mole:

$$E _ {k} (1 \text { mole}) = N _ {a} E _ {k} (1 \text { molecule}) = N _ {a} \frac {m}{2} \bar {v} ^ {2} = \frac {3}{2} R T = 3739.5 J$$

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