Question

At what temperature the molecules of nitrogen will have the same rms. velocity as the molecules of Oxygen at \(127^{\circ} \mathrm{C}\). (A) \(273^{\circ} \mathrm{C}\) (B) \(350^{\circ} \mathrm{C}\) (C) \(77^{\circ} \mathrm{C}\) (D) \(457^{\circ} \mathrm{C}\)

Short answer

The temperature at which the molecules of Nitrogen will have the same rms velocity as the molecules of Oxygen at 127°C is \(T_{N_2} = 77^{\circ} \mathrm{C}\) (Option C).

Step-by-step solution

Understanding rms velocity formula

We need to first recall the formula for rms velocity, which is: \(v_{rms} = \sqrt{\frac{3RT}{M}} \) where: - \(v_{rms}\) is the root-mean-square velocity, - R is the universal gas constant (8.31 J/(mol K)), - T is the temperature in Kelvin, and - M is the molar mass of the given gas in kg/mol.

Converting given temperature to Kelvin

We are given that the temperature of the Oxygen gas is 127°C. To use the rms velocity formula, we need to convert this temperature to Kelvin: (127 + 273) K = 400 K

Calculate rms velocity of Oxygen

Using the rms velocity formula, we calculate the rms velocity of Oxygen at 127°C (400 K). The molar mass of Oxygen (O₂) is 32 g/mol or 0.032 kg/mol. \(v_{O_2} = \sqrt{\frac{3 × 8.31 × 400}{0.032}}\)

Setting up an equation for Nitrogen's temperature

To find the temperature at which Nitrogen has the same rms velocity as Oxygen, we set up an equation: \(v_{O_2} = v_{N_2}\) We know that the molar mass of Nitrogen (N₂) is 28 g/mol or 0.028 kg/mol. Using the rms velocity formula for both gases and setting their velocities equal, we get: \(\sqrt{\frac{3 × 8.31 × 400}{0.032}} = \sqrt{\frac{3 × 8.31 × T_{N_2}}{0.028}}\)

Solve the equation for Nitrogen's temperature

Now we solve the equation for \(T_{N_2}\): \(\frac{3 × 8.31 × 400}{0.032} = \frac{3 × 8.31 × T_{N_2}}{0.028}\) \(T_{N_2} = \frac{400 × 0.028}{0.032}\) \(T_{N_2} = 350\, \text{K}\)

Convert Nitrogen's temperature to Celsius

Now we convert the temperature 350 K back to Celsius: (350 - 273)°C = 77°C Thus, we find that the temperature at which Nitrogen has the same rms velocity as Oxygen at 127°C is: \(T_{N_2} = 77^{\circ} \mathrm{C}\) The correct option is (C) 77°C.

Key concepts

Root-mean-square velocity

Root-mean-square velocity (rms velocity) is a concept that helps us understand the movement of gas molecules. It's a type of average velocity that considers the energy evenly distributed among all molecules in a gas. Calculating this velocity involves using the square root of the average of the squares of the velocities. The formula used is: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

• \(v_{rms}\): Root-mean-square velocity.
• \(R\): Universal gas constant, 8.31 J/(mol K).
• \(T\): Temperature in Kelvin.
• \(M\): Molar mass in kg/mol.

Using rms velocity, scientists can compare the energy and speed of different gases under various conditions. In our problem, we needed to find the temperature where nitrogen and oxygen molecules move at the same rms velocity.

Ideal gas law

The ideal gas law is a fundamental principle that describes the behavior of gases. It helps to relate temperature, pressure, and volume of an ideal gas, and is expressed as: \[ PV = nRT \] Where \(P\) stands for pressure, \(V\) stands for volume, \(n\) stands for the amount of substance in moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. In the context of rms velocity, the ideal gas law helps us understand how temperature relates to molecular speed. While the ideal gas law is not directly used in the exercise, it hovers in the background, ensuring that all calculations about gases, like the determination of rms velocity, are grounded on this essential principle. Understanding this connection helps solidify why temperature adjustments impact rms velocity as well as other gas properties.

Molar mass

Molar mass is key when working with gases, especially in the root-mean-square velocity formula. It refers to the mass of one mole of a given substance and is expressed in g/mol or kg/mol. Different gases have different molar masses, which affect their rms velocities. For example:

• Oxygen (O₂) has a molar mass of 32 g/mol or 0.032 kg/mol.
• Nitrogen (N₂) has a molar mass of 28 g/mol or 0.028 kg/mol.

In our problem, knowing the molar masses of both oxygen and nitrogen allowed us to set the equation for their rms velocities. The differences in molar mass mean that at the same temperature, they would usually have different rms velocities. However, adjusting the temperature balances this out so both gases can reach the same speed.

Temperature conversion

Temperature conversion is crucial when working with scientific formulas involving gases. Many formulas, like the rms velocity formula, use temperature in Kelvin. Therefore, we often need a conversion from Celsius to Kelvin, which follows the simple rule:

• Kelvin = Celsius + 273

In our exercise, we converted the initial temperature of the oxygen gas from 127°C to 400 K because the rms velocity formula requires Kelvin. Similarly, after calculating nitrogen's temperature in Kelvin (350 K), we converted back to Celsius to find it at 77°C. These conversions ensure accurate calculations and align with the scientific standards for temperature measurements while working with gases.