Question

question 11

the particles of a particular monoatomic noble gas have a root-mean-square (vrms) velocity of 413.0 meters/sec at 300.0 °c. what is this gas?

krypton (kr)
radon (rn)
argon (ar)
xenon (xe)
neon (ne)

Explanation:

Identify the given values and formula

We are given the root-mean-square velocity \(v_{\text{rms}} = 413.0\text{ m/s}\) and the temperature \(T = 300.0\text{ }^\circ\text{C}\).
The formula for the root-mean-square velocity of a gas is:

$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$

where:

• \(R = 8.314\text{ J/(mol}\cdot\text{K)}\) is the ideal gas constant.
• \(T\) is the absolute temperature in Kelvin.
• \(M\) is the molar mass of the gas in kilograms per mole (\(\text{kg/mol}\)).

Convert temperature to Kelvin

To use the gas constant \(R\), we must convert the temperature from Celsius to Kelvin:

$$T = 300.0 + 273.15 = 573.15\text{ K}$$

Rearrange the formula to solve for molar mass

Squaring both sides of the \(v_{\text{rms}}\) equation gives:

$$v_{\text{rms}}^2 = \frac{3RT}{M}$$

Solving for \(M\):

$$M = \frac{3RT}{v_{\text{rms}}^2}$$

Calculate the molar mass

Substitute the known values into the rearranged equation:

$$M = \frac{3 \times 8.314\text{ J/(mol}\cdot\text{K)} \times 573.15\text{ K}}{(413.0\text{ m/s})^2}$$

$$M = \frac{14295.42}{170569}\text{ kg/mol}$$

$$M \approx 0.08381\text{ kg/mol}$$

Convert the molar mass to grams per mole (\(\text{g/mol}\)):

$$M \approx 0.08381\text{ kg/mol} \times 1000\text{ g/kg} = 83.81\text{ g/mol}$$

Identify the noble gas

We compare our calculated molar mass of \(83.81\text{ g/mol}\) with the molar masses of the noble gases listed in the options:

• Neon (Ne): \(20.18\text{ g/mol}\)
• Argon (Ar): \(39.95\text{ g/mol}\)
• Krypton (Kr): \(83.80\text{ g/mol}\)
• Xenon (Xe): \(131.29\text{ g/mol}\)
• Radon (Rn): \(222\text{ g/mol}\)

The calculated molar mass matches Krypton (Kr).

Answer:

• Krypton (Kr) (Correct answer)
• Radon (Rn)
• Argon (Ar)
• Xenon (Xe)
• Neon (Ne)