Answer:
[tex]\frac{v_{2}}{v_{1}}=2[/tex].
Explanation:
The average kinetic energy per molecule of a ideal gas is given by:
[tex]\bar{K}=\frac{3k_{B}T}{2}[/tex]
Now, we know that [tex]\bar{K} = (1/2)m\bar{v}^{2}[/tex]
Before the absorption we have:
[tex](1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}[/tex] (1)
After the absorption,
[tex](1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}[/tex] (2)
If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)
[tex]\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}[/tex]
[tex]\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}[/tex]
[tex]\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}[/tex]
[tex]\frac{v_{2}}{v_{1}}=\sqrt{4}[/tex]
Therefore the ratio will be [tex]\frac{v_{2}}{v_{1}}=2[/tex]
I hope it helps you!
Answer:
The ratio of v₂/v₁ is 2
Explanation:
Here we have Charles law which can be presented as
P₁/T₁ = P₂/T₂
Therefore,
P₂ = T₂ × P₁/T₁
Also from the kinetic theory of gases we have;
[tex]v_{rms} = \sqrt{\frac{3\times R\times T}{MW} }[/tex]
Where:
[tex]v_{rms}[/tex] = Rms speed
R = Universal gas constant
T = Temperature in Kelvin
MW = Molecular weight
We therefore have for the before and after speeds as
[tex]v_1 = \sqrt{\frac{3\times R\times T_1}{MW} }[/tex] and [tex]v_2 = \sqrt{\frac{3\times R\times T_2}{MW} }[/tex]
Therefore,
[tex]\frac{v_2}{v_1 } = \frac{\sqrt{\frac{3\times R\times T_2}{MW} }}{\sqrt{\frac{3\times R\times T_1}{MW} }} = \sqrt{\frac{\frac{3\times R\times T_2}{MW}}{\frac{3\times R\times T_1}{MW}} } = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{v_2}{v_1 } = \sqrt{\frac{T_2}{T_1} } = \sqrt{\frac{1340}{335} } =\sqrt{\frac{4}{1} } = 2[/tex]
∴ v₂/v₁ = 2.
