At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at 126 oC? (Hint: The molar mass of hydrogen atoms is 1.008 g/mol and of nitrogen atoms is 14.007 g/mol. The molar mass of H2 is twice the molar mass of hydrogen atoms, and similarly for N2.) The answer is in degree C.
7 years ago
6 years ago
nbknk
11 months ago
To find the temperature at which the root-mean-square (rms) speed of nitrogen molecules is equal to the rms speed of hydrogen molecules, we can use the formula:
rms speed = sqrt((3RT) / (M))
where:
- rms speed is the root-mean-square speed
- R is the ideal gas constant (8.314 J/(mol•K))
- T is the temperature in Kelvin
- M is the molar mass of the gas in kg/mol
First, let's convert the given temperature of 126 oC to Kelvin:
T = 126 oC + 273.15 = 399.15 K
Next, let's calculate the rms speed for hydrogen and nitrogen using their respective molar masses.
For hydrogen (H2):
M(H2) = 2 * 1.008 g/mol = 2.016 g/mol = 0.002016 kg/mol
For nitrogen (N2):
M(N2) = 2 * 14.007 g/mol = 28.014 g/mol = 0.028014 kg/mol
Now we can plug the values into the formula and solve for T:
rms(H2) = sqrt((3 * R * T) / M(H2))
rms(N2) = sqrt((3 * R * T) / M(N2))
To find the temperature where these values are equal, we can set them equal to each other.
sqrt((3 * R * T) / M(H2)) = sqrt((3 * R * T) / M(N2))
Squaring both sides of the equation:
(3 * R * T) / M(H2) = (3 * R * T) / M(N2)
The R, 3, and T cancel out, leaving us with:
1 / M(H2) = 1 / M(N2)
Solving for T:
T = (M(N2) / M(H2)) * 399.15 K
Substituting the values:
T = (0.028014 kg/mol) / (0.002016 kg/mol) * 399.15 K ≈ 5545.19 K
Converting back to Celsius:
T ≈ 5545.19 K - 273.15 ≈ 5272.04 oC
Therefore, the temperature at which the root-mean-square speed of nitrogen molecules is equal to the root-mean-square speed of hydrogen molecules at 126 oC is approximately 5272.04 oC.