Respuesta :
55.44 K is the temperature that need to change in order for the neon gas to be 39.0% faster than the RMS speed of the argon.
Given
RMS speed of argon = 1.20 km/sec = 1.20 ×[tex]10^{3}[/tex] m/sec
RMS speed of neon = 39% faster than the speed of argon
= RMS speed of argon + [tex]\frac{39}{100}[/tex] RMS speed of argon
= RMS speed of argon ( 1 +0.39)
RMS speed of neon = 1.39 times of RMS speed of argon
Hence [tex]\frac{v_{ne} }{v_{ar} } = \frac{1.39}{1}[/tex] … (1)
The atomic mass of argon is 39.95 g/mol = 39.95×[tex]10^{-3}[/tex] Kg/mol
and the atomic mass of neon is 20.18 g/mol = 20.18 ×[tex]10^{-3}[/tex] Kg/mol
According to the formula of root-mean-square velocities of gas molecules
[tex]v_{rms} = \sqrt{3RT/M}[/tex] where,
[tex]v_{rms}[/tex] = root-mean-square velocity
M = molar mass of gas (in kg per mole)
R = Molar gas constant = 8.314 J/mol/K
T = temperature (in kelvin)
[tex]v_{arg }[/tex] = [tex]\sqrt{3RT_{arg}/M }[/tex]
[tex]v_{arg} ^{2}[/tex] = [tex]3RT_{arg} /M[/tex]
1.20 ×1.20 ×[tex]10^{6\\[/tex] = 3 ×8.314×[tex]T_{arg}[/tex] /39.95 ×[tex]10^{-3}[/tex]
[tex]T_{arg}[/tex] = 1.44 × 39.95 ×[tex]10^{3}[/tex] / 24.942
[tex]T_{arg}[/tex] = 2306.47 K
Now,
RMS velocity is directly proportional to [tex]\sqrt{T}[/tex] and inversely proportional to [tex]\sqrt{M}[/tex]
[tex]v_{rms}[/tex] [tex]\alpha[/tex] [tex]\sqrt{T}[/tex] and [tex]v_{rms} \frac{1}{\alpha } \sqrt{M}[/tex]
[tex]\frac{v_{ne} }{v_{ar} }[/tex] = [tex]\sqrt{\frac{T_{ne} }{T_{ar} } }[/tex]× [tex]\sqrt{\frac{M_{ar} }{M_{ne} } }[/tex]
According to equation 1
[tex]\sqrt{\frac{T_{ne} }{T_{ar} } }[/tex] × [tex]\sqrt{\frac{M_{ar} }{M_{ne} } }[/tex] = 1.39
[tex]T_{ne}[/tex] = 1.39 ×1.39 × 2306.47 × 20.18 / 39.95
[tex]T_{ne}[/tex] = 2251.03K
Change in temp ΔT = 2306.47-2251.03= 55.44 K
Hence, 55.44 K is the temperature that need to change in order for the neon gas to be 39.0% faster than the RMS speed of the argon.
Learn more about Root-mean-square velocity here https://brainly.com/question/25806742
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