If, at a particular temperature, the average speed of CH4 molecules is 1000 mine/hr, what would be the average speed of CO2 molecules at the same temperature?
To find the average speed of CO2 molecules at the same temperature, we can use the concept of the root mean square (rms) speed of gas molecules. The rms speed relates to the temperature and molar mass of the gas.
The formula to calculate the rms speed is:
v = √((3RT) / M)
Where:
v is the rms speed of the gas molecules
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (K)
M is the molar mass of the gas in kilograms per mole (kg/mol)
First, let's find the molar mass of CH4 (methane) and CO2 (carbon dioxide):
Molar mass of CH4 (methane) = 12.01 g/mol (carbon atomic mass) + 4(1.01 g/mol) = 16.04 g/mol
Molar mass of CO2 (carbon dioxide) = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Next, we need to convert the molar masses from grams to kilograms:
Molar mass of CH4 (methane) = 16.04 g/mol ÷ 1000 = 0.01604 kg/mol
Molar mass of CO2 (carbon dioxide) = 44.01 g/mol ÷ 1000 = 0.04401 kg/mol
Now, let's plug in the values into the rms speed formula:
For CH4 molecules:
v_CH4 = √((3RT) / M_CH4)
For CO2 molecules:
v_CO2 = √((3RT) / M_CO2)
Since both CH4 and CO2 are at the same temperature, the value of T (temperature) is the same in both equations.
To find the ratio of the rms speeds (v_CO2 / v_CH4) at the same temperature, we divide the two equations:
(v_CO2 / v_CH4) = (√((3RT) / M_CO2)) ÷ (√((3RT) / M_CH4))
Now, substituting the values:
R = 8.314 J/(mol·K)
T is given in Kelvins
M_CH4 = 0.01604 kg/mol
M_CO2 = 0.04401 kg/mol
Calculating the ratio will give us the average speed of CO2 molecules relative to CH4 molecules at the same temperature.