Question

If the density of nitrogen is 1.25 kg/m3

at a pressure of 105

Pa, find the root mean square

velocity of oxygen molecules​

Answer 1

Answer:

Vrms=√252

Explanation:

Vrms=√[3RT/M]

now 3RT/M = 3(PV)/M (as PV=nRT where n=1)

= 3P/d (as d=M/V ,where d=density)

= 3(105)/1.25

= 252

therefore Vrms=√252=15.875 m/s

Answer 2

Answer:

The rms velocity of oxygen molecules is approximately 461.2 m/s under these conditions.

Explanation:

The root mean square (rms) velocity of gas molecules can be calculated using the following formula:

v = √(3kT/m)

where v is the rms velocity, k is the Boltzmann constant (1.38 × 10^-23 J/K), T is the absolute temperature in kelvin, and m is the molar mass of the gas in kilograms.

To find the rms velocity of oxygen molecules, we need to first determine the temperature and molar mass of oxygen. We are not given any information about the temperature or pressure of the oxygen gas, so we will assume that it is at the same temperature and pressure as the nitrogen gas. This is known as the "ideal gas assumption," which assumes that gases behave similarly under the same conditions.

The molar mass of oxygen is 32.00 g/mol or 0.032 kg/mol. Therefore, the mass of one oxygen molecule is:

m = 0.032 kg/mol / (6.022 × 10^23 molecules/mol) = 5.31 × 10^-26 kg/molecule

Now we need to determine the temperature of the gas. We can use the ideal gas law to relate pressure, volume, and temperature:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant (8.31 J/mol·K), and T is the absolute temperature in kelvin.

We are not given the volume or number of moles of gas, but we can assume that the nitrogen and oxygen gases are in thermal equilibrium and have the same temperature. Therefore, we can write:

P(N2) = P(O2)

where P(N2) is the pressure of nitrogen and P(O2) is the pressure of oxygen. Solving for P(O2), we get:

P(O2) = P(N2) = 105 Pa

Now we can use the ideal gas law to solve for the temperature of the gas:

T = PnR/V

We can rewrite this equation as:

T = PV/nR

Since the pressure and number of moles of gas are the same for nitrogen and oxygen, we can write:

T(N2) = T(O2)

Substituting in the values for pressure and density of nitrogen, we get:

T = P(N2) / ρ(N2) R

T = 105 Pa / (1.25 kg/m^3) × 8.31 J/mol·K

T = 8402 K

Now we can calculate the rms velocity of oxygen molecules using the formula:

v = √(3kT/m)

Substituting in the values we have calculated, we get:

v = √(3 × 1.38 × 10^-23 J/K × 8402 K / 5.31 × 10^-26 kg/molecule)

v ≈ 461.2 m/s

Therefore, the rms velocity of oxygen molecules is approximately 461.2 m/s under these conditions.

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