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:

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<reasoning>

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).
</reasoning>

<answer>
<mcq-correct>Krypton (Kr)</mcq-correct>
<mcq-option>Radon (Rn)</mcq-option>
<mcq-option>Argon (Ar)</mcq-option>
<mcq-option>Xenon (Xe)</mcq-option>
<mcq-option>Neon (Ne)</mcq-option>
</answer>

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Answer:

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<reasoning>

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).
</reasoning>

<answer>
<mcq-correct>Krypton (Kr)</mcq-correct>
<mcq-option>Radon (Rn)</mcq-option>
<mcq-option>Argon (Ar)</mcq-option>
<mcq-option>Xenon (Xe)</mcq-option>
<mcq-option>Neon (Ne)</mcq-option>
</answer>

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