Question

The density of neon will be highest at (a) STP (b) \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) (c) \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) (d) \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\)

Short answer

The density of neon is highest at \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\).

Step-by-step solution

Understanding Density and Ideal Gas Law

Density is the mass per unit volume of a substance. For gases, it can be determined using the ideal gas law \( PV = nRT \) and the definition of density \( \rho = \frac{m}{V} \). Rearranging the ideal gas law for density, we have \( \rho = \frac{MP}{RT} \), where \( M \) is the molar mass of the gas, \( P \) is pressure, \( R \) is the gas constant, and \( T \) is temperature in Kelvin. The density is directly proportional to pressure and inversely proportional to temperature.

Convert Temperatures to Kelvin

We need to ensure that all temperatures are in Kelvin for calculations. - For STP (Standard Temperature and Pressure), \( T = 273 \text{K} \).- For \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\), \( T = 273 \text{K} \).- For \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), \( T = 273 + 273 = 546 \text{K} \).- For \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\), \( T = 273 + 273 = 546 \text{K} \).

Calculate Density Using Ideal Gas Law

Substitute values for pressure \( (P) \), temperature \( (T) \), and molar mass \( (M) \) of neon \(20.18 \text{ g/mol}\) into the rearranged ideal gas equation \( \rho = \frac{MP}{RT} \).- At STP \( (P = 1 \, \text{atm}, T = 273 \, \text{K}) \) : \( \rho = \frac{20.18 \times 1}{0.0821 \times 273} \).- At \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) \( (P = 2 \, \text{atm}, T = 273 \, \text{K}) \) : \( \rho = \frac{20.18 \times 2}{0.0821 \times 273} \).- At \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) \( (P = 1 \, \text{atm}, T = 546 \, \text{K}) \) : \( \rho = \frac{20.18 \times 1}{0.0821 \times 546} \).- At \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) \( (P = 2 \, \text{atm}, T = 546 \, \text{K}) \) : \( \rho = \frac{20.18 \times 2}{0.0821 \times 546} \).

Compare Densities

Compute the densities:- At STP: \( \rho \approx 0.90 \, \text{g/L} \).- At \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\): \( \rho \approx 1.81 \, \text{g/L} \).- At \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\): \( \rho \approx 0.45 \, \text{g/L} \).- At \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\): \( \rho \approx 0.90 \, \text{g/L} \).Comparing these, the highest density is at \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\).

Key concepts

Density of gases

Density measures how much mass is contained within a given volume of a gas. Unlike solids and liquids, a gas's density can change significantly based on external conditions, particularly pressure and temperature. In the context of gases, density is usually represented as \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.

The Ideal Gas Law, represented by the equation \( PV = nRT \), plays a crucial role in understanding gas density. Here, \( P \) denotes the pressure of the gas, \( V \) is its volume, \( n \) refers to the number of moles of gas, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. By integrating these concepts, we can rearrange the formula to relate density directly to pressure and temperature as \( \rho = \frac{MP}{RT} \),where \( M \) is the molar mass of the gas. This equation helps illustrate that pressure and temperature significantly influence gas density.

Effects of temperature and pressure

Temperature and pressure have noticeable effects on the volume and, therefore, the density of a gas. Understanding these effects is key to fully grasping how gases behave under different conditions.

• Temperature: As temperature increases, the kinetic energy of gas molecules rises, causing them to spread out and occupy more space. This increase in volume, if pressure is constant, leads to a decrease in density. Lowering the temperature has the opposite effect; molecules lose energy and come closer together, increasing density.
• Pressure: An increase in pressure compresses gas molecules into a smaller volume. This compression results in a higher density, given that temperature remains constant. Conversely, reducing pressure allows them to expand, resulting in lower density.

The interplay between temperature and pressure is crucial when calculating the real-world behavior of gases. Balancing these two factors can alter a gas's density significantly, illustrating the need for careful consideration when making related calculations.

Calculation of gas density

Calculating the density of a gas involves using the Ideal Gas Law to connect its physical properties. Here's a simple breakdown of how this calculation is performed:

• Convert Temperature to Kelvin: All temperature values must be in Kelvin to ensure accurate calculations. The conversion is straightforward: simply add 273 to the Celsius temperature.
• Rearrange the Ideal Gas Law: Use the rearranged formula \( \rho = \frac{MP}{RT} \),where \( \rho \) is density, \( M \) is molar mass, \( P \) is pressure, \( R \) is the gas constant (approximately 0.0821 L·atm/mol·K), and \( T \) is temperature in Kelvin.
• Substitute Known Values: Input the known values for each factor into the formula and solve for \( \rho \). These include the pressure, temperature, and molar mass specific to the gas being studied.
• Comparison: Evaluate the results to determine how varying the temperature and pressure affects gas density in specific conditions.

By following this process, it's possible to accurately predict and adjust the density of gases such as neon in various scenarios, helping frame concept comprehension in scientific studies.