Respuesta :
Answer:
[tex]C_{mp}=332.8046\ m/s[/tex]
[tex]C_{avg}=375.4542\ m/s[/tex]
[tex]C_{rms}=407.6007\ m/s[/tex]
Explanation:
The expression for the most probable speed is:
[tex]C_{mp}=\sqrt {\dfrac {2RT}{M}}[/tex]
R is Gas constant having value = 8.314 J / K mol
M is the molar mass of gas
Given that : The gas is Carbon dioxide (Corrected and assumed)
Molar mass of [tex]CO_2[/tex] = 44.01 g/mol
Also, 1 g = 0.001 kg
So, Molar mass of [tex]CO_2[/tex] = 0.04401 kg/mol
Temperature = 20°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Thus,
[tex]C_{mp}=\sqrt {\frac{2\times 8.314\times 293.15}{0.04401}}[/tex]
[tex]C_{mp}=\sqrt{\frac{4874.4982}{0.04401}}[/tex]
[tex]C_{mp}=\sqrt{110758.87752}[/tex]
[tex]C_{mp}=332.8046\ m/s[/tex]
The expression for the mean speed is:
[tex]C_{avg}=\sqrt {\dfrac{8RT}{\pi M}}[/tex]
R is Gas constant having value = 8.314 J / K mol
M is the molar mass of gas
So, Molar mass of [tex]CO_2[/tex] = 0.04401 kg/mol
T = 293.15 K
Thus,
[tex]C_{avg}=\sqrt{\frac{8\times 8.314\times 293.15}{\frac{22}{7}\times 0.04401}}[/tex]
[tex]C_{avg}=\sqrt{\frac{19497.9928}{0.13831}}[/tex]
[tex]C_{avg}=375.4542\ m/s[/tex]
The expression for the root mean square speed is:
[tex]C_{rms}=\sqrt {\dfrac {3RT}{M}}[/tex]
R is Gas constant having value = 8.314 J / K mol
M is the molar mass of gas
Molar mass of [tex]CO_2[/tex] = 0.04401 kg/mol
Temperature 293.15 K
Thus,
[tex]C_{rms}=\sqrt{\frac{3\times 8.314\times 293.15}{0.04401}}[/tex]
[tex]C_{rms}=\sqrt{\frac{7311.7473}{0.04401}}[/tex]
[tex]C_{rms}=407.6007\ m/s[/tex]
