Respuesta :
Answer: (a) The total translational kinetic energy of the gas is [tex]35 \times 10^{7} J[/tex].
(b) The energy as heat providing to the gas so that the average speed of its molecules increases to double is [tex]8.7 \times 10^{7} J[/tex].
Explanation:
Given: Average speed of molecules = [tex]5 \times 10^{3}[/tex] 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 = [tex]\frac{1}{2}mv^{2}[/tex]
where,
m = mass
v = velocity
Substitute the values into above formula as follows.
[tex]Translational K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 28 \times (5 \times 10^{3})^{2}\\= 35 \times 10^{7} J[/tex]
(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.
[tex]E \propto (2v)^{2}\\or, E \propto 4v^{2}\\v^{2} \propto \frac{E}{4}[/tex]
This means that velocity gets one-fourth times the energy of its molecules.
Therefore, energy will be calculated as follows.
[tex]Energy = \frac{35 \times 10^{7}}{4}\\= 8.7 \times 10^{7} J[/tex]
Therefore, we can conclude that
(a) The total translational kinetic energy of the gas is [tex]35 \times 10^{7} J[/tex].
(b) The energy as heat providing to the gas so that the average speed of its molecules increases to double is [tex]8.7 \times 10^{7} J[/tex].
