Respuesta :
The rms speed of a gas is given by
[tex]v_{rms} = \sqrt{ \frac{3RT}{M_m} } [/tex] (1)
where
R is the gas constant
T is the absolute temperature of the gas
[tex]M_m[/tex] is the molar mass of the gas
For the oxygen molecules in the problem, T=300 K, and the molar mass of molecular oxygen is [tex]M_m = 32 g/mol = 0.032 kg/mol[/tex], therefore the rms speed of the molecules is
[tex]v_{rms}= \sqrt{ \frac{3 (8.31 J/molK)(300 K)}{0.032 kg/mol} }=483.4 m/s [/tex]
We want the atoms of helium to have the same rms speed. So we can re-arrange equation (1), and using [tex]v_{rms}=483.4 m/s[/tex] and [tex]M_m=4 g/mol=0.004 kg/mol[/tex] (molar mass of helium) in order to find the helium atom temperature T:
[tex]T= \frac{v_{rms}^2 M_m}{3 R}= \frac{(483.4 m/s)^2(0.004 kg/mol)}{3 (8.31 J/mol K)} =37.5 K [/tex]
[tex]v_{rms} = \sqrt{ \frac{3RT}{M_m} } [/tex] (1)
where
R is the gas constant
T is the absolute temperature of the gas
[tex]M_m[/tex] is the molar mass of the gas
For the oxygen molecules in the problem, T=300 K, and the molar mass of molecular oxygen is [tex]M_m = 32 g/mol = 0.032 kg/mol[/tex], therefore the rms speed of the molecules is
[tex]v_{rms}= \sqrt{ \frac{3 (8.31 J/molK)(300 K)}{0.032 kg/mol} }=483.4 m/s [/tex]
We want the atoms of helium to have the same rms speed. So we can re-arrange equation (1), and using [tex]v_{rms}=483.4 m/s[/tex] and [tex]M_m=4 g/mol=0.004 kg/mol[/tex] (molar mass of helium) in order to find the helium atom temperature T:
[tex]T= \frac{v_{rms}^2 M_m}{3 R}= \frac{(483.4 m/s)^2(0.004 kg/mol)}{3 (8.31 J/mol K)} =37.5 K [/tex]
Temperature of Helium atom is 37.5 K
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Further explanation
Let's recall Work Done by Ideal Gas and Root-Mean-Square Speed of Gas formula as follows:
[tex]\boxed{ W = \int {P} \, dV }[/tex]
where:
P = Gas Pressure ( Pa )
V = Gas Volume ( m³ )
[tex]\texttt{ }[/tex]
[tex]\boxed{ v_{rms} = \sqrt{ \frac{3RT}{M} }}[/tex]
where:
v_rms = root mean square speed ( m/s )
R = gas constant ( J/mol K )
T = temperature ( K )
M = molar mass of gas ( kg / mol )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
molar mass of Helium = M_He = 4 gram/mol
molar mass of Oxygen = M_O₂ = 32 gram/mol
temperature of Oxygen = T_O₂ = 300 K
Asked:
temperature of Helium = T_He = ?
Solution:
[tex]v_{He} = v_{O_2}[/tex]
[tex]\sqrt{ \frac{3RT_{He}}{M_{He}} } = \sqrt{ \frac{3RT_{O_2}}{M_{O_2}} }[/tex]
[tex]\sqrt{ \frac{T_{He}}{M_{He}} } = \sqrt{ \frac{T_{O_2}}{M_{O_2}}}[/tex]
[tex]\sqrt{ \frac{T_{He}}{4 }} = \sqrt{ \frac{300}{32}}[/tex]
[tex]\frac{T_{He}}{4 } = \frac{300}{32}}[/tex]
[tex]T_{He} = \frac{4}{32} \times 300[/tex]
[tex]\boxed{T_{He} = 37.5 \texttt{ K}}[/tex]
[tex]\texttt{ }[/tex]
Learn more
- Buoyant Force : https://brainly.com/question/13922022
- Kinetic Energy : https://brainly.com/question/692781
- Volume of Gas : https://brainly.com/question/12893622
- Impulse : https://brainly.com/question/12855855
- Gravity : https://brainly.com/question/1724648
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Answer details
Grade: High School
Subject: Physics
Chapter: Thermodynamics

