Respuesta :
Answer:
The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Explanation:
Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg
The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)
where u = initial rms speed of the gas = 850 m/s
Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J
Then, 3600 J is added to the gas,
New kinetic energy of the gas = 4335 + 3600 = 7935 J
New kinetic energy of Helium atoms = (1/2)(m)(v²)
where v = final rms speed of the gas = ?
7935 = (1/2)(0.012)(v²)
v² = (7935×2)/0.012
v² = 1,322,500
v = 1150 m/s
Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Hope this Helps!!!
The rms speed of the gas molecule after the addition of heat energy is 1,150 m/s.
RMS speed has been the root mean square velocity of the molecule. It provides the value to the kinetic energy of the molecule.
The moles of gas = 3 moles
Mass of the gas = moles [tex]\times[/tex] molecular weight
Mass of the gas = 3 [tex]\times[/tex] 4
Mass of the He gas = 12 grams.
1000 grams = 1 kg
12 grams = 0.012 kg.
Intital kinetic velocity = [tex]\rm \dfrac{1}{2}\;mu^2[/tex]
The given value of initial velocity (u) = 850m/s
The initial kinetic (rms) energy = [tex]\rm \dfrac{1}{2}\;\times\;0.012\;\times\;(850)^2[/tex]
The initial kinetic (rms) energy = 4335 J.
Since the heat is gained by the molecule, the new energy will be the sum of initial and gained energy.
New rms energy = 4335 J + 3600 J
New rms energy = 7935 J
The rms speed can be given by:
Kinetic energy = [tex]\rm \dfrac{1}{2}\;mv^2[/tex]
7935 J = [tex]\rm \dfrac{1}{2}\;\times\;0.012\;\times\;(v)^2[/tex]
[tex]\rm v^2[/tex] = 1,322,500
v = [tex]\rm \sqrt{1,322,500}[/tex]
v = 1,150 m/s
The rms speed of the gas molecule after the addition of heat energy is 1,150 m/s.
For more information about rms speed, refer to the link:
https://brainly.com/question/15514391?
