Problem 134 - Chemistry The Science in Context

Question

The density of Ne at \(760 \mathrm{mmHg}\) is exactly half its density at 2.00 atm at constant temperature. Is the root-meansquare speed of Ne at 2.00 atm half, twice, or the same as \(u_{\mathrm{rms}, \mathrm{Ne}}\) at \(760 \mathrm{mmH} \mathrm{g} ?\)

Step-by-Step Solution

Answer

Answer: The root-mean-square speed of Ne at 2.00 atm is greater by a factor of √2 compared to the root-mean-square speed of Ne at 760 mmHg (1 atm).

Step 1: Convert units

Convert the pressure from mmHg to atm for easier comparison. We know that 1 atm = 760 mmHg. Therefore, 760 mmHg = 1 atm.

Step 2: Apply the Ideal Gas Law

We will use the Ideal Gas Law in the form \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rewrite the Ideal Gas Law as \(P = \rho RT / M\), where ρ is the density and M is the molar mass of the gas. This equation establishes a relationship between pressure, density, and temperature.

Step 3: Compare densities at different pressures

According to the problem, the density of Ne at 1 atm (760 mmHg) is exactly half its density at 2.00 atm. Let the density at 1 atm be denoted as ρ1 and the density at 2 atm be ρ2. We have ρ2 = 2ρ1.

Step 4: Calculate the ratio of \(u_{\mathrm{rms}}\)

The root-mean-square speed, \(u_{\mathrm{rms}}\), is given by \(\sqrt{3RT / M}\). To find the ratio of the root-mean-square speeds at 2 atm and 1 atm, we will use the equation \(u_{\mathrm{rms}, \mathrm{2 atm}}\) / \(u_{\mathrm{rms}, \mathrm{1 atm}}\) = \(\sqrt{\frac{3R\cdot T_{2}}{M}}\) / \(\sqrt{\frac{3R\cdot T_{1}}{M}}\). Considering that T remains constant, the ratio simplifies to: \(u_{\mathrm{rms}, \mathrm{2 atm}}\) / \(u_{\mathrm{rms}, \mathrm{1 atm}}\) = \(\sqrt{T_{2}}\) / \(\sqrt{T_{1}}\).

Step 5: Use the relationship between pressure, density, and temperature

Since we have already established the relationship between pressure, density, and temperature, we can plug in the densities at 2 atm and 1 atm: \(T_{2}\)/ \(T_{1}\) = \((\frac{\rho_{2}}{\rho_{1}})^{2}P_{1}/P_{2}\) = \((\frac{2\rho_{1}}{\rho_{1}})^{2}\frac{1}{2}\) = 2.

Step 6: Calculate the ratio of root-mean-square speeds

Plug in the values of \(T_{2}\)/ \(T_{1}\) into the equation for the ratio of root-mean-square speeds: \(u_{\mathrm{rms}, \mathrm{2 atm}}\) / \(u_{\mathrm{rms}, \mathrm{1 atm}}\) = \(\sqrt{2}\). Thus, the root-mean-square speed of Ne at 2.00 atm is greater by a factor of \(\sqrt{2}\) compared to the root-mean-square speed of Ne at 760 mmHg (1 atm).

Key Concepts

Root-Mean-Square Speed

When thinking about the motion of gas particles, one important measure is their average speed. However, instead of a simple average, we often use the root-mean-square (RMS) speed, which is more representative of the typical speed of particles in a gas. RMS speed is given by the formula \( u_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} \), where:

• \( R \) is the gas constant,
• \( T \) is the temperature in Kelvin, and
• \( M \) is the molar mass of the gas.

This formula shows that RMS speed depends on both the temperature and the molar mass. As the temperature increases, molecules move faster, so \( u_{\mathrm{rms}} \) increases. Similarly, lighter molecules have a higher \( u_{\mathrm{rms}} \) compared to heavier ones. In the exercise, when comparing the RMS speed of Neon (Ne) at different pressures, the key insight is that the temperature remains constant. Thus, changes in RMS speed only depend on pressure differences and density ratios due to the Ideal Gas Law.

Gas Density

Density of a gas is a crucial concept when dealing with the Ideal Gas Law, as it helps link together pressure, temperature, and volume. The Ideal Gas Law in the form \( PV = nRT \) can be re-arranged to express pressure in terms of gas density: \( P = \frac{\rho RT}{M} \), where \( \rho \) is the density given by \( \frac{mass}{volume} \), and \( M \) is the molar mass of the gas.

In this context, if pressure or temperature changes, the density can also change, which in turn affects other properties like the root-mean-square speed. In the exercise, the problem states that the density at 2 atm is twice as much as the density at 1 atm. This directly uses the gas law relationship, as an increase in pressure at a constant temperature generally results in an increase in density, assuming the gas behaves ideally. Understanding this relationship helps solve problems where density varies due to changes in other states of the gas.

Pressure Units Conversion

Pressure is a measure of force exerted per unit area and is crucial in calculations involving gases. In scientific studies, it's common to encounter different units of pressure like Pascal (Pa), atmospheres (atm), and millimeters of mercury (mmHg). Correctly converting between these units is essential for accurate results.

• 1 atm = 760 mmHg
• 1 atm is approximately 101325 Pa

The problem exercise starts by requiring a conversion from mmHg to atm since the given densities correlate through pressure. Utilizing consistent units makes it easier to work with the Ideal Gas Law and further simplifies calculations. By converting 760 mmHg to 1 atm in the solution, the problem aligns the densities at consistent pressure units, demonstrating how conversions are indispensable in chemistry and physics problem-solving.