Question

At \(44.0^{\circ} \mathrm{C}\) and \(1.23 \times 10^{-2}\) atm the density of a gas is \(1.32 \times 10^{-5} \mathrm{~g} / \mathrm{cm}^{3} .(a)\) Find \(v_{\mathrm{rms}}\) for the gas molecules. (b) Using the ideal gas law, find the number of moles per unit volume (molar density) of the gas. ( \(c\) ) By combining the results of (a) and \((b)\), find the molar mass of the gas and identify it.

Short answer

The step-by-step solution provides a detailed method to calculate \(v_{rms}\), molar density and the molar mass of the gas. After calculating these values it allows one to identify the gas.

Step-by-step solution

Conversion of Variables to SI Units

The first step is to convert temperature from Celsius to Kelvin and pressure from atm to Pa. The conversion factors are \(T (K) = T(° C) + 273.15\) and \(P (Pa) = P (atm) \times 1.01 \times 10^{5} \). So, the variables convert to \(T = 44 + 273.15 = 317.15 K\) and \(P = 1.23 \times 10^{-2} \times 1.01 \times 10^{5} = 1242.3 Pa\).

Apply the formula to Calculate \(v_{rms}\)

Utilize the formula \(v_{rms}=\sqrt{\frac{3kT}{m}}\). Rewriting this in terms of density ρ (density equals to mass/volume), the formula becomes \(v_{rms}=\sqrt{\frac{3kT}{ρ}}\). Substituting the values \(k = 1.38 \times 10^{-23} J/K\), \(T = 317.15 K\) and \(ρ = 1.32 \times 10^{-5} g/cm^3 = 1.32 \times 10^{-2} kg/m^3\), we calculate the root mean square speed.

Use Ideal Gas Law to Find Number of Moles

We apply the ideal gas law \(PV = nRT\), solve it for n to get the molar density \(n=\frac{PV}{RT}\). Here, \(R = 8.314 J/(mol.K)\). Substituting the known values \(P = 1242.3 Pa\), \(V = 1 m^3\), \(R = 8.314 J/(mol.K)\) and \(T = 317.15 K\), it allows us to calculate the number of moles per unit volume.

Calculation of Molar Mass

Molar mass (\(M\)) is the mass per mole of the substance. In this step, using the density (\(ρ\)) and molar density (\(n\)), molar mass can be calculated using the formula \(M = \frac{\rho}{n}\). After calculating the value, the gas can be identified by comparing the calculated molar mass with the known molar masses of gases

Key concepts

Root Mean Square Velocity

The root mean square velocity, often abbreviated as 'RMS velocity', is a way to compute the average speed of gas particles in a sample. It plays an essential role in understanding the behavior of gases within the Ideal Gas Law context.

It is derived using the formula \(v_{rms}=\sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature in Kelvin, and \(m\) represents the mass of a single molecule of the gas. Instead of focusing on the speed of one particle, the RMS velocity considers the collective motion of particles, giving a statistical view of their kinetic energy.

Nevertheless, in step 2 of our solution, the formula is reorganized in relation to density, which makes it \(v_{rms}=\sqrt{\frac{3kT}{\rho}}\). This substitution allows us to address the exercise's first question more directly, by using the known temperature and density, presenting students with a clear path towards calculating the gas molecule's RMS velocity.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of its particles, whether they are atoms, molecules, or other chemical units. It's generally expressed in grams per mole (g/mol) and is fundamental in stoichiometry and other chemical calculations.

To calculate the molar mass in the context of a gas, as per step 4 in the solution provided, we use the formula \(M = \frac{\rho}{n}\), where \(\rho\) is the gas's density and \(n\) is the molar density — the number of moles per unit volume. The density gives us the mass per unit volume, and dividing this by the number of moles per unit volume isolates the mass per mole — which is, by definition, the molar mass.

This operation's result helps us identify the gas by comparing the computed molar mass with known values. Students might find it helpful to have a periodic table or another reference of molar masses for common substances at hand to facilitate the identification.

Molar Density

Molar density is determined by the number of moles of a substance within a given volume and is typically expressed in moles per cubic meter (mol/m³) in the SI system. In the gas law context, it offers a bridge between the macroscopic scale of gas properties and the microscopic scale of molecular quantities.

Following step 3 of the exercise's solution path, the ideal gas law \(PV = nRT\) is rearranged to \(n=\frac{PV}{RT}\), isolating the molar density (\(n\)). This practical yet potent rearrangement takes the pressure (\(P\)), volume (\(V\)), gas constant (\(R\)), and absolute temperature (\(T\)) into account. For student exercises, understanding molar density can scaffold the learning of how to relate conditions of pressure and temperature to the number of particles and types of gases involved.

Unit Conversion

Unit conversion is a fundamental skill in chemistry and physics, allowing for the standardization of measurements to facilitate calculation and comparison. During the problem-solving process, as demonstrated in step 1 of the solution, temperature and pressure values are converted into SI units, which are the international standard for scientific measurements.

The given temperature is converted from degrees Celsius to Kelvin using the relation \(T(K) = T(°C) + 273.15\). Pressure is converted from atmospheres (atm) to Pascals (Pa) with the conversion factor \(P(Pa) = P(atm) \times 1.01 \times 10^{5}\). These conversions are critical for the integrity of the calculations that follow since the formulas used are derived based on these SI units.

Students often struggle with unit conversion. It is crucial to double-check units in final answers and make sure they correspond to those requested in the question or consistent throughout the calculation.