Question

If the velocity of gas molecules is doubled, then it’s kinetic energy will?( hint formula for kinetic energy is KE=.5 x mv2)

Answer 1

Answer:

D quadruple

Explanation:

$E = \frac{1}{2}mv^2\\E'=\frac{1}{2}m(2v)^2=\frac{1}{2}m4 v^2 = 4(\frac{1}{2}mv^2)=4E$

Answer 2

Answer:

(D) quadruple

Explanation:

$KE_{1} = \frac{1}{2}m v^{2}$ -----equation 1

where;

KE₁ is the initial kinetic energy

m is the gas molecule mass

v is the velocity of gas molecules

⇒ When the velocity of gas molecules is doubled (2v)

$KE_{2} = \frac{1}{2}m (2v)^{2}$

$KE_{2} = \frac{1}{2}m*4v^{2}$

$KE_{2} = 4(\frac{1}{2}mv^2)$ -------equation 2

Recall; From equation 1;

$KE_{1} = \frac{1}{2}m v^{2}$

Substitute in KE₁ in equation 2

KE₂ = 4(KE₁)

Therefore, the kinetic energy will quadruple