A molecule of mass m of an ideal gas collides with the wall of the vessel with the velocity v and returns back with the same velocity- The change in the linear momentum of the molecule will be-4mv8mv2mv-2mv

Question

Toppr Admin · Accepted Answer

-Initial momentum -xA0-mv-Final momentum -xA0-x2212-mv-Change in momentum -xA0-x2212-mv-x2212-mv-x2212-2mv

A molecule of mass $m$ of an ideal gas collides with the wall of the vessel with the velocity $v$ and returns back with the same velocity. The change in the linear momentum of the molecule will be:
$4 m v$
$8 m v$
$2 m v$
$- 2 m v$

Initial momentum $= m v$

Final momentum $= - m v$

Change in momentum $= - m v - (m v) = - 2 m v$

A molecule of mass m of an ideal gas collides with the wall of the vessel with the velocity v and returns back with the same velocity. The change in the linear momentum of the molecule will be:4mv8mv2mv−2mv

Initial momentum =mv Final momentum =−mv Change in momentum =−mv−(mv)=−2mv

A molecule of mass $m$ of an ideal gas collides with the wall of the vessel with the velocity $v$ and returns back with the same velocity. The change in the linear momentum of the molecule will be:
$4 m v$
$8 m v$
$2 m v$
$- 2 m v$

Initial momentum $= m v$

Final momentum $= - m v$

Change in momentum $= - m v - (m v) = - 2 m v$