Physics, 21.06.2021 14:00 mjam85877

The average speed of molecules of a 0.1 mole nitrogen gas in a container is 5103 m/s.
a/ Determine the total translational kinetic energy of the gas.
b/ Compute the energy as heat providing to the gas so that the average speed of its molecules increases to
double.

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Answer from: irenecupcake4348

(a) The total translational kinetic energy of the gas is $35 \times 10^{7} J$ .

(b) The energy as heat providing to the gas so that the average speed of its molecules increases to double is $8.7 \times 10^{7} J$ .

Explanation:

Given: Average speed of molecules = $5 \times 10^{3}$ m/s

Moles = 0.1 mol

(a) As the give gas is nitrogen so its mass is 28 g/mol.

Formula to calculate translational kinetic energy is as follows.

Total translational K.E = $\frac{1}{2}mv^{2}$

where,

m = mass

v = velocity

Substitute the values into above formula as follows.

$Translational K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 28 \times (5 \times 10^{3})^{2}\\= 35 \times 10^{7} J$

(b) As the energy is directly proportional to the square of velocity. So, when average speed of molecules increases to double then relation between energy and velocity will be as follows.

$E \propto (2v)^{2}\\or, E \propto 4v^{2}\\v^{2} \propto \frac{E}{4}$

This means that velocity gets one-fourth times the energy of its molecules.

Therefore, energy will be calculated as follows.

$Energy = \frac{35 \times 10^{7}}{4}\\= 8.7 \times 10^{7} J$

Therefore, we can conclude that

(a) The total translational kinetic energy of the gas is $35 \times 10^{7} J$ .

(b) The energy as heat providing to the gas so that the average speed of its molecules increases to double is $8.7 \times 10^{7} J$ .

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