The molecular velocities of two gases at the same temperature are u1 and u2
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Answer:
The kinetic energy can be shown as
K = (1 /2) m V^2
Where, m is mass of gas particle
V is volume of gas particle
K is kinetic energy.
In a random motion of gas, all the particles of gas are having equivalent energy in them which make them to collide with each other.
The kinetic energy of 1st gas particle is equivalent to 2nd particle can be shown as:
K_1 =K_2
(1 /2) m_1 U_1^2 = (1 /2) m_2 U _2^2
m_1 U _1^2 = m_2 U _2^2
Thus, option b is correct
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