Question

Some ideal monoatomic gas A in an enclosure has a pressure P and the temparature T. Another ideal monoatomic gas B enclosed in a container of the same volume has a pressure 2P and temparature T/2. the ratio of average kinetic energy of molecule of gas A to gas B is

Answer 1

Answer:

2

Explanation:

Number of moles of gas (n) remains constant because the volume of gas is constant.

As the container has fixed volume.

Also, Average kinetic energy of molecules

$K.E.=\frac{3}{2} k_B T$

⇒K.E.∝T

therefore,

$\frac{K.E._A}{K.E._B} =\frac{T_A}{T_B}$

$\frac{K.E._A}{K.E._B} =\frac{T}{\frac{T}{2} }\\\frac{K.E._A}{K.E._B}=2$