Physics, 23.10.2019 16:50 hamaama

The temperature near the surface of the earth is 295 k. an argon atom (atomic mass = 39.948 u) has a kinetic energy equal to the average translational kinetic energy and is moving straight up. if the atom does not collide with any other atoms or molecules, how high up would it go before coming to rest? assume that the acceleration due to gravity is constant throughout the ascent.

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Answer from: scasemere

Atom Argon would high up 9403.761 m before coming to rest

Further explanation

Energy is the ability to do work

Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:

$\large {\boxed {\bold {KE = \frac {1} {2} mv ^ 2}}}$

So for two objects that have the same speed, the greater the mass of the object, the greater the kinetic energy

Whereas for a stationary object it has kinetic energy = 0

While the energy produced from its position is called potential energy (PE)

which can be formulated as:

PE = m. g. h ⇒ gravitational potential energy

The sum of kinetic and potential energy is called mechanical energy

ME = KE + PE

The position of the object when near the surface PE = 0

Because objects (argon atoms) have kinetic energy equal to the average translational kinetic energy, then :

$\rm KE=\dfrac{3}{2}kT$

K = average kinetic energy per molecule of gas (J)

k = Boltzmann's constant (1.38.10⁻²³ J / K)

T = temperature (K)

So, near the surface :

ME = KE

ME = 3/2 kT

ME = 3/2 .1.38.10⁻²³. 295

ME = 6.11.10⁻²¹

atomic mass of Argon = 39,948 u

because 1 u = 1.66.10⁻²⁷ kg, the Argon mass:

= 39,948.1.66.10⁻²⁷

= 6.63.10⁻²⁶ kg

before coming to rest: KE = 0, because ME = constant then:

ME = PE

6.11.10⁻²¹ = m. g. h (g = 9.8 m / s²)

6.11.10⁻²¹ = 6.63.10⁻²⁶. 9.8. h

$\rm h=\dfrac{6.11.10^{-21}}{6.63.10^{-26}.9.8}\\\\h=9403.761~m$

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