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.

2571.4375

nbknk

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.