How can we define tho motion of the atoms of a weightless gas? (for example, In ISS, air has no weight, so is the motion of atoms in that air different because it is weightless?)
ALL gases have mass.........
You will to qualify your question. All matter, including gases, possess mass. That helium and hydrogen balloons float (IN AIR) is a manifestation of very old physical principles known since antiquity, and relates to relative densities. See old Archimedes principle.
The main difference is that gas motion is more uniform, and the assumption that motion in each direction is equivalent (as it would be for a homogeneous gas) is even better than on earth.
So basically, the assumptions made to derive the kinetic theory of gases don't change, and relevant definitions don't change; they just become more valid.
THE CARTESIAN DIRECTIONS ARE LINEARLY INDEPENDENT
In considering the effects of (presumably) no gravity, only the vertical direction of motion is affected. Consider Newton's equations of motion:
#vecF_x = m(d^2x)/(dt^2) = mveca_x# #" "bb((1))#
#vecF_y = m(d^2y)/(dt^2) = mveca_y# #" "bb((2))#
#vecF_z = m(d^2z)/(dt^2) = mvecg# #" "bb((3))#
#m#is mass in #"kg"#, #vecF_q#is the force in #"N"#along the #q#direction, #veca_q#is acceleration in #"m/s"^2#in the #q#direction, and #vecg#is the gravitational constant in #"m/s"^2#.
Of these, if
In addition, the ISS was made to have
So, you can think of it as a more kinematically-balanced earthen atmosphere (ordinary pressures, roughly room temperature, but ~ no accelerated downwards influence on momentum).
INFLUENCES ON GAS MOTION & IDEALITY
This increases the ideality of the gases involved, since the vertical component of intermolecular forces is lessened, while the horizontal component is never affected by gravity.
In theory then, what we're saying is that the gas motion is more isotropic than usual, i.e. the assumption that the root-mean square speed is...
#v_(RMS) = sqrt(1/N sum_(j=1)^(N) (v_(jx)^2 + v_(jy)^2 + v_(jz)^2)) ~~ sqrt(1/(N) sum_(j=1)^(N) v_(j)^2)#,
#1/N sum_(j=1)^(N) v_(jx)^2 ~~ 1/N sum_(j=1)^(N) v_(jy)^2 ~~ 1/N sum_(j=1)^(N) v_(jz)^2 -= 1/(3N) sum_(j=1)^(N) v_(j)^2#, and #N#is the number of gas particles,
...is now a better assumption. In other words, the motion in all directions is now more uniform.
Overall, the main difference is that gas motion is more uniform, and the assumption that motion in each direction is equivalent is even better than on earth.