RMS Speeds: Essential Concepts for Competitive Exam Prep

RMS Speeds

Q. A gas at 300 K has an RMS speed of 400 m/s. What will be its RMS speed at 600 K?

A. 400 m/s
B. 400 sqrt(2) m/s
C. 800 m/s
D. 200 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.

Correct Answer: B — 400 sqrt(2) m/s

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Q. A gas at 300 K has an RMS speed of 500 m/s. What will be its RMS speed at 600 K?

A. 500 m/s
B. 707 m/s
C. 1000 m/s
D. 250 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, v_rms at 600 K = 500 * sqrt(600/300) = 500 * sqrt(2) ≈ 707 m/s.

Correct Answer: B — 707 m/s

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Q. A gas has an RMS speed of 500 m/s. If the molar mass of the gas is 0.02 kg/mol, what is the temperature of the gas?

A. 250 K
B. 500 K
C. 1000 K
D. 2000 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Substituting v_rms = 500 m/s and M = 0.02 kg/mol gives T = 500 K.

Correct Answer: B — 500 K

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Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol? (R = 8.314 J/(mol K))

A. 500 K
B. 600 K
C. 700 K
D. 800 K

Solution

Using v_rms = sqrt(3RT/M), we solve for T: T = (v_rms^2 * M) / (3R) = (1000^2 * 0.044) / (3 * 8.314) = 700 K.

Correct Answer: C — 700 K

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Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol?

A. 300 K
B. 400 K
C. 500 K
D. 600 K

Solution

Using v_rms = sqrt(3RT/M), we rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T approximately 500 K.

Correct Answer: C — 500 K

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Q. At what temperature will the RMS speed of a gas be 300 m/s if its molar mass is 28 g/mol?

A. 300 K
B. 600 K
C. 900 K
D. 1200 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T. Setting v_rms = 300 m/s and M = 28 g/mol, we find T = (M * v_rms^2)/(3R) = 600 K.

Correct Answer: B — 600 K

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Q. At what temperature will the RMS speed of a gas be 600 m/s if its molar mass is 0.02 kg/mol?

A. 300 K
B. 600 K
C. 900 K
D. 1200 K

Solution

Using v_rms = sqrt(3RT/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T = (600^2 * 0.02) / (3 * 8.314) = 900 K.

Correct Answer: C — 900 K

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Q. Calculate the RMS speed of a gas with molar mass 0.028 kg/mol at 300 K. (R = 8.314 J/(mol K))

A. 500 m/s
B. 600 m/s
C. 700 m/s
D. 800 m/s

Solution

Using v_rms = sqrt(3RT/M), we find v_rms = sqrt(3 * 8.314 * 300 / 0.028) = 600 m/s.

Correct Answer: B — 600 m/s

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Q. For a gas at 300 K, if the RMS speed is 500 m/s, what will be the RMS speed at 600 K?

A. 500 m/s
B. 707 m/s
C. 1000 m/s
D. 250 m/s

Solution

RMS speed is proportional to the square root of temperature, so v_rms at 600 K = 500 * sqrt(600/300) = 707 m/s.

Correct Answer: B — 707 m/s

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Q. For a gas at 300 K, what is the RMS speed if the molar mass is 0.028 kg/mol?

A. 500 m/s
B. 600 m/s
C. 700 m/s
D. 800 m/s

Solution

Using v_rms = sqrt(3RT/M), we calculate v_rms = sqrt(3 * 8.314 * 300 / 0.028) which gives approximately 600 m/s.

Correct Answer: B — 600 m/s

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Q. For a gas at a certain temperature, if the molar mass is halved, what happens to the RMS speed?

A. Increases by a factor of 2
B. Increases by a factor of sqrt(2)
C. Decreases by a factor of 2
D. Remains the same

Solution

RMS speed is inversely proportional to the square root of molar mass. Halving the molar mass increases the RMS speed by a factor of sqrt(2).

Correct Answer: B — Increases by a factor of sqrt(2)

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Q. For a gas at a constant temperature, if the molar mass is halved, what happens to the RMS speed?

A. Increases by a factor of sqrt(2)
B. Increases by a factor of 2
C. Decreases by a factor of 2
D. Remains the same

Solution

The RMS speed is inversely proportional to the square root of the molar mass. If the molar mass is halved, the RMS speed increases by a factor of sqrt(2), which is approximately 1.414, but in terms of doubling the speed, it is considered to increase by a factor of 2.

Correct Answer: B — Increases by a factor of 2

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Q. For a gas mixture, how is the RMS speed calculated?

A. Using the average molar mass of the mixture
B. Using the molar mass of the heaviest gas
C. Using the molar mass of the lightest gas
D. Using the molar mass of the most abundant gas

Solution

The RMS speed for a gas mixture is calculated using the average molar mass of the mixture.

Correct Answer: A — Using the average molar mass of the mixture

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Q. For a gas with a molar mass of 32 g/mol at 273 K, what is the RMS speed?

A. 300 m/s
B. 400 m/s
C. 500 m/s
D. 600 m/s

Solution

Using v_rms = sqrt(3RT/M), we find v_rms = sqrt(3 * 8.314 * 273 / 0.032) = 300 m/s.

Correct Answer: A — 300 m/s

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Q. For a gas with a molar mass of 32 g/mol at a temperature of 300 K, what is the RMS speed?

A. 273 m/s
B. 400 m/s
C. 500 m/s
D. 600 m/s

Solution

Using the formula v_rms = sqrt((3RT)/M), where R = 8.314 J/(mol·K), M = 0.032 kg/mol, and T = 300 K, we find v_rms ≈ 400 m/s.

Correct Answer: B — 400 m/s

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Q. For a gas with molar mass M at temperature T, what is the relationship between RMS speed and molar mass?

A. v_rms is directly proportional to M
B. v_rms is inversely proportional to M
C. v_rms is independent of M
D. v_rms is proportional to M^2

Solution

The RMS speed is given by v_rms = sqrt((3RT)/M). This shows that v_rms is inversely proportional to the square root of the molar mass M.

Correct Answer: B — v_rms is inversely proportional to M

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Q. For a gas with molar mass M, what is the relationship between RMS speed and molar mass?

A. v_rms is directly proportional to M
B. v_rms is inversely proportional to M
C. v_rms is independent of M
D. v_rms is proportional to M^2

Solution

The RMS speed is inversely proportional to the square root of the molar mass (v_rms = sqrt((3RT)/M)). Thus, as molar mass increases, RMS speed decreases.

Correct Answer: B — v_rms is inversely proportional to M

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Q. For a gas with molar mass M, what is the relationship between RMS speed and molecular mass?

A. v_rms is directly proportional to M
B. v_rms is inversely proportional to M
C. v_rms is independent of M
D. v_rms is proportional to M^2

Solution

The RMS speed is inversely proportional to the square root of the molar mass (v_rms = sqrt((3RT)/M)). Thus, as molar mass increases, RMS speed decreases.

Correct Answer: B — v_rms is inversely proportional to M

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Q. For a monoatomic ideal gas, the RMS speed is given by which of the following expressions?

A. sqrt((3kT)/m)
B. sqrt((3RT)/M)
C. Both of the above
D. None of the above

Solution

Both expressions are valid for calculating the RMS speed of a monoatomic ideal gas.

Correct Answer: C — Both of the above

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Q. For an ideal gas, if the temperature is increased, what happens to the RMS speed?

A. Increases
B. Decreases
C. Remains constant
D. Depends on the gas

Solution

The RMS speed increases with temperature as v_rms = sqrt(3RT/M) shows that it is directly proportional to the square root of temperature T.

Correct Answer: A — Increases

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Q. If the molar mass of a gas is doubled, how does its RMS speed change?

A. Increases by sqrt(2)
B. Decreases by sqrt(2)
C. Remains the same
D. Increases by 2

Solution

RMS speed is inversely proportional to the square root of molar mass, so if M is doubled, v_rms decreases by sqrt(2).

Correct Answer: B — Decreases by sqrt(2)

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Q. If the molar mass of a gas is doubled, how does the RMS speed change?

A. Increases by sqrt(2)
B. Decreases by sqrt(2)
C. Remains the same
D. Increases by 2

Solution

The RMS speed is inversely proportional to the square root of the molar mass. If M is doubled, v_rms decreases by sqrt(2).

Correct Answer: B — Decreases by sqrt(2)

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Q. If the molar mass of a gas is halved, what happens to its RMS speed?

A. Increases by a factor of sqrt(2)
B. Increases by a factor of 2
C. Decreases by a factor of sqrt(2)
D. Remains the same

Solution

If the molar mass is halved, the RMS speed increases by a factor of sqrt(2) because RMS speed is inversely proportional to the square root of molar mass.

Correct Answer: A — Increases by a factor of sqrt(2)

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Q. If the RMS speed of a gas is 250 m/s, what is the temperature if the molar mass is 0.028 kg/mol?

A. 100 K
B. 200 K
C. 300 K
D. 400 K

Solution

Using v_rms = sqrt(3RT/M), we can rearrange to find T = (v_rms^2 * M) / (3R) = 300 K.

Correct Answer: C — 300 K

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Q. If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the temperature of the gas?

A. 300 K
B. 600 K
C. 900 K
D. 1200 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M)/(3R). Plugging in the values gives T = 600 K.

Correct Answer: B — 600 K

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Q. If the RMS speed of a gas is 300 m/s at 300 K, what will be its RMS speed at 600 K?

A. 300 m/s
B. 600 m/s
C. 300√2 m/s
D. 600√2 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 300 * sqrt(2) m/s.

Correct Answer: C — 300√2 m/s

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Q. If the RMS speed of a gas is 300 m/s at 400 K, what will be the RMS speed at 200 K?

A. 150 m/s
B. 300 m/s
C. 600 m/s
D. 100 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 200 K, the RMS speed will be 300 * sqrt(200/400) = 150 m/s.

Correct Answer: A — 150 m/s

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Q. If the RMS speed of a gas is 300 m/s at 400 K, what will be the RMS speed at 800 K?

A. 300 m/s
B. 600 m/s
C. 424 m/s
D. 848 m/s

Solution

RMS speed is proportional to the square root of temperature. v_rms at 800 K = 300 * sqrt(800/400) = 300 * sqrt(2) = 600 m/s.

Correct Answer: B — 600 m/s

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Q. If the RMS speed of a gas is 300 m/s, what is the RMS speed of the same gas at double the temperature?

A. 300 m/s
B. 600 m/s
C. 300√2 m/s
D. 600√2 m/s

Solution

The RMS speed is proportional to the square root of the temperature. If the temperature is doubled, the RMS speed increases by a factor of sqrt(2). Therefore, the new RMS speed will be 300 * sqrt(2), which is approximately 600 m/s.

Correct Answer: B — 600 m/s

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Q. If the RMS speed of a gas is 400 m/s and its molar mass is 16 g/mol, what is the temperature of the gas?

A. 200 K
B. 400 K
C. 800 K
D. 1600 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T: T = (M * v_rms^2) / (3R). Substituting M = 0.016 kg/mol and v_rms = 400 m/s gives T = 400 K.

Correct Answer: B — 400 K

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RMS Speeds

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