At what temp do hydrogen molecules, H2, have the same rms speed as nitrogen molecules, N2, at 455 C? At what temperature do hydrogen molecules have the same average kinetic energy?

So Tony’s wrong, the answer is most definitely not 455 C. You need to use the rms molecular speed equation, u=(3(RT)/Mm)^1/2. R=molar has constant, which is usually 8.31, T=temperature, and Mm= molar mass, but use kg/mol, instead of the normal g/mol. You can solve part A of the problem now by pluggin values in the find the rms speed of N2 at 455C which is about 805 m/s. Use that and plug it into the equation again, but leave T as the unknown value, and remember to switch the bottom mole mass value to H2, and you’ll get around 52 K for the answer. I don’t know how to solve part B either, so someone please help?

To find the temperature at which hydrogen (H2) and nitrogen (N2) molecules have the same root mean square (rms) speed at 455°C, we can use the formula:

rms speed = √(3RT/M)

where:
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (K)
- M is the molar mass of the gas in kilograms (kg/mol)

The molar mass of hydrogen (H2) is approximately 2.016 g/mol, and the molar mass of nitrogen (N2) is approximately 28.013 g/mol. Converting these values to kilograms:

M(H2) = 2.016 g/mol = 0.002016 kg/mol
M(N2) = 28.013 g/mol = 0.028013 kg/mol

Let's determine the temperature at which the rms speed of hydrogen molecules equals the rms speed of nitrogen molecules at 455°C:

1. Convert 455°C to Kelvin:
T = 455 + 273.15
= 728.15 K

2. Calculate the rms speed for hydrogen (H2) at 728.15 K:
v(H2) = √(3RT/M(H2))
= √((3 × 8.314 J/(mol·K) × 728.15 K) / 0.002016 kg/mol)

3. Calculate the rms speed for nitrogen (N2) at 728.15 K:
v(N2) = √(3RT/M(N2))
= √((3 × 8.314 J/(mol·K) × 728.15 K) / 0.028013 kg/mol)

The temperature at which hydrogen molecules (H2) have the same rms speed as nitrogen molecules (N2) at 455°C is when the rms speeds are equal:

v(H2) = v(N2)

To find the temperature at which hydrogen molecules have the same average kinetic energy, we can use the formula:

average kinetic energy = (3/2)kT

where:
- k is the Boltzmann constant (1.38 × 10^(-23) J/K)

Set the average kinetic energies for hydrogen (H2) and nitrogen (N2) equal to each other and solve for T:

(3/2)kT(H2) = (3/2)kT(N2)

Canceling out the common terms:

T(H2) = T(N2)

Now, let's calculate the temperature at which hydrogen molecules have the same average kinetic energy.

To determine the temperature at which hydrogen molecules (H2) have the same root mean square (rms) speed as nitrogen molecules (N2) at a given temperature, we can use the equation of kinetic energy. The average kinetic energy of a gas molecule can be calculated using the following equation:

KE = (3/2) * k * T

where KE is the average kinetic energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), and T is the temperature in Kelvin.

First, we need to convert 455°C to Kelvin by adding 273.15:

T_N2 = 455 + 273.15 = 728.15 K

To find the temperature at which hydrogen molecules have the same rms speed as nitrogen molecules, we can equate the rms speed equations of the two gases:

vrms_H2 = vrms_N2

The rms speed can be calculated using the following equation:

vrms = sqrt((3 * k * T) / m)

where vrms is the root mean square speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the gas.

The molar mass of hydrogen (H2) is 2 g/mol, and the molar mass of nitrogen (N2) is 28 g/mol.

Now we can set up and solve the equation:

sqrt((3 * k * T_H2) / m_H2) = sqrt((3 * k * T_N2) / m_N2)

Using m_H2 = 2 g/mol and m_N2 = 28 g/mol in grams:

sqrt((3 * (1.38 x 10^-23 J/K) * T_H2) / 2) = sqrt((3 * (1.38 x 10^-23 J/K) * 728.15 K) / 28)

Simplifying the equation:

sqrt((3 * T_H2) / 2) = sqrt((3 * 728.15) / 28)

(3 * T_H2) / 2 = (3 * 728.15) / 28

Now, solve for T_H2:

T_H2 = ((3 * 728.15) / 28) * (2 / 3)

T_H2 ≈ 247.40 K

Therefore, at approximately 247.40 Kelvin (or -25.75°C), hydrogen molecules (H2) will have the same root mean square speed as nitrogen molecules (N2) at 455°C.

To find the temperature at which hydrogen molecules have the same average kinetic energy as nitrogen molecules, we can equate their kinetic energy equations:

(3/2) * k * T_H2 = (3/2) * k * T_N2

Now we can solve for T_H2:

T_H2 = (T_N2 * m_N2) / m_H2

Using T_N2 = 728.15 K, m_N2 = 28 g/mol, and m_H2 = 2 g/mol:

T_H2 = (728.15 * 28) / 2

T_H2 ≈ 10207.20 K

Therefore, at approximately 10207.20 Kelvin (or 994 °C), hydrogen molecules (H2) will have the same average kinetic energy as nitrogen molecules (N2) at 455°C.

Dont worry bro i feel you the answer is 455C.