At what temperature will the rms velocity of oxygen be 3/2times of its value at NTP.

To determine the temperature at which the root mean square (rms) velocity of oxygen is 1.5 times its value at Normal Temperature and Pressure (NTP), we first need to understand the relationship between temperature and rms velocity. The rms velocity of a gas is given by the formula:

Understanding rms Velocity

The rms velocity (\(v_{rms}\)) can be expressed as:

\(v_{rms} = \sqrt{\frac{3kT}{m}}\)

Where:

• k is the Boltzmann constant.
• T is the absolute temperature in Kelvin.
• m is the mass of a gas molecule.

Values at NTP

At NTP, the standard temperature is 273.15 K. For oxygen (O₂), the molar mass is approximately 32 g/mol, which translates to a molecular mass of:

\(m = \frac{32 \text{ g/mol}}{N_A} = \frac{32 \times 10^{-3} \text{ kg}}{6.022 \times 10^{23}} \approx 5.31 \times 10^{-26} \text{ kg}\)

Now, we can calculate the rms velocity at NTP:

\(v_{rms, NTP} = \sqrt{\frac{3k(273.15)}{m}}\)

Finding the New Temperature

We want the rms velocity at the new temperature (\(T_f\)) to be 1.5 times that at NTP:

\(v_{rms, T_f} = 1.5 \times v_{rms, NTP}\)

Substituting the rms velocity formula into this equation gives:

\(\sqrt{\frac{3kT_f}{m}} = 1.5 \times \sqrt{\frac{3k(273.15)}{m}}\)

Squaring both sides to eliminate the square root leads to:

\(\frac{3kT_f}{m} = (1.5)^2 \times \frac{3k(273.15)}{m}\)

Notice that \(3k/m\) cancels out from both sides:

\(T_f = (1.5)^2 \times 273.15\)

Calculating the Final Temperature

Now, we can compute \(T_f\):

\(T_f = 2.25 \times 273.15 \approx 615.5 \text{ K}\)

Final Thoughts

Thus, the temperature at which the rms velocity of oxygen is 1.5 times its value at NTP is approximately 615.5 K. This illustrates how temperature directly influences the speed of gas molecules, which is a fundamental concept in thermodynamics and kinetic theory.