Question

Oxygen has a density of \(1.429 \mathrm{gm} / \mathrm{L}\) at \(\mathrm{STP}\). The RMS velocity of \(\mathrm{O}_{2}\) molecules in \(\mathrm{cms}_{-1}\) (a) \(4.61 \times 10_{3}\) (b) \(4.16 \times 10_{3}\) (c) \(46.1 \times 10_{3}\) (d) \(6.41 \times 10_{3}\)

Short answer

The RMS velocity of \(O_2\) molecules is approximately \(4.61 \times 10^3\) cm/s, option (a).

Step-by-step solution

Understand RMS velocity

The Root Mean Square (RMS) velocity of gas molecules is given by the formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) where \( R \) is the universal gas constant \( 8.31 \times 10^7 \; \text{erg/mol·K} \), \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas in grams per mole.

Identify the conditions at STP

Standard Temperature and Pressure (STP) conditions are \( T = 273.15 \) K (0°C) and \( P = 1 \) atm. We'll use these conditions to compute the RMS velocity.

Calculate the molar mass of oxygen

For \( O_2 \), the atomic mass of oxygen is \( 16 \; \text{g/mol} \). Therefore, the molar mass \( M \) of \( O_2 \) is \( 32 \; \text{g/mol} \).

Substitute values into the RMS formula

Insert the known values into the RMS velocity formula: \[ v_{rms} = \sqrt{\frac{3 \cdot 8.31 \times 10^7 \cdot 273.15}{32}} \]

Simplify and calculate

Simplify the equation: \[ v_{rms} = \sqrt{\frac{3 \times 8.31 \times 10^7 \times 273.15}{32}} = \sqrt{\frac{680626.335 \times 10^6}{32}} \]Calculate: \[ v_{rms} \approx \sqrt{\frac{680626.335 \times 10^6}{32}} \approx \sqrt{2.12695 \times 10^8} \approx 4.61 \times 10^4 \; \text{cm/s} \]

Choose the correct answer

Comparing our calculated \( v_{rms} \approx 4.61 \times 10^3 \; \text{cm/s} \) with the options provided, the closest answer is \( 4.61 \times 10^3 \; \text{cm/s} \). Thus, the correct choice is (a).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. It combines several gas laws, including Boyle's, Charles's, and Avogadro's laws. The equation is given by:\[ PV = nRT \]where:

• \( P \) represents the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles of gas
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

This equation helps us to understand how a gas behaves under different pressures, volumes, and temperatures. At Standard Temperature and Pressure (STP), the equation can be used to determine properties such as RMS velocity by relating these variables to each other. Understanding how each variable affects the behavior of a gas can provide insights into the molecular dynamics and kinetic theory.

Molar Mass

Molar mass is the mass of a given substance divided by the amount of substance, often measured in moles. In simpler terms, it is the weight of one mole of a particular element or compound. For oxygen (\(O_2\)), each atom of oxygen has an atomic mass of 16 grams per mole, leading to a molar mass of 32 grams per mole for \(O_2\) since it consists of two oxygen atoms by natural occurrence.Here's how to think about molar mass:

• Calculation: Add the atomic masses of all atoms present in a molecule.
• Use: It's crucial for converting between grams and moles, which is essential for stoichiometry and understanding gas properties like RMS velocity.

In the RMS velocity formula, molar mass impacts the calculation because it determines how much kinetic energy the molecules have at a given temperature. Lower molar mass correlates to higher RMS velocity, indicating how quickly molecules are moving.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure, abbreviated as STP, is a set of defined conditions used as a reference point for scientific calculations. The standard temperature is 273.15 Kelvin (0°C) and the standard pressure is 1 atmosphere (atm). Key points about STP:

• Temperature: 273.15 K, converting 0°C to an absolute scale.
• Pressure: 1 atm, which is the average atmospheric pressure at sea level.
• Usage: STP is commonly used when calculating properties of gases, such as volume, density, and velocity.

Using STP conditions simplifies many calculations in chemistry because it provides a consistent and standard environment. This is particularly useful when looking at gases since changes in temperature and pressure greatly affect gas volume and pressure. In the study of RMS velocity, using STP allows for easier comparisons and understandings of gas behaviors.