Question

Calculate the densities (in grams per liter) of the following gases at \(75^{\circ} \mathrm{F}\) and \(1.33\) bar. (a) argon (b) ammonia (c) acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\)

Short answer

Question: Calculate the densities of Argon, Ammonia, and Acetylene gases at a temperature of 75°F and a pressure of 1.33 bar. Answer: The densities of the gases at 75°F and 1.33 bar are: (a) Argon: 2.24 g/L (b) Ammonia: 0.89 g/L (c) Acetylene: 1.38 g/L

Step-by-step solution

Convert Pressure and Temperature to Appropriate Units

First, we need to convert the given pressure from bar to Pascals (Pa) and the temperature from Fahrenheit to Kelvin. Pressure conversion: $$ 1.33 \,\text{bar} \times \frac{100000\,\text{Pa}}{1\,\text{bar}} = 133000\,\text{Pa} $$ Temperature conversion: $$ T(K) = \frac{75 - 32}{1.8} + 273.15 = 297.59\,K $$

Find the Molar Mass of Each Gas

We need to find the molar mass (M) for each gas in grams per mole (g/mol): (a) Argon (Ar): $$ M_{Ar} = 39.95\,\text{g/mol} $$ (b) Ammonia (NH\(_3\)): $$ M_{NH_3} = 14.01 + 3 \times 1.01 = 17.04\,\text{g/mol} $$ (c) Acetylene (C\(_2\)H\(_2\)): $$ M_{C_2H_2} = 2 \times 12.01 + 2 \times 1.01 = 26.04\,\text{g/mol} $$

Calculate Density Using the Ideal Gas Law Formula

Now, we can find the densities of the gases using the ideal gas law formula for density: $$ \rho = \frac{PM}{RT} $$ with \(R = 8.314 \,\text{J/(mol K)}\). (a) Argon (Ar): \begin{align*} \rho_{Ar} &= \frac{133000\,\text{Pa} \times 39.95\,\text{g/mol}}{8.314\,\text{J/(mol K)} \times 297.59\,K} \\ &= 2241.22\,\text{g/m}^3 \\ &= 2.24\,\text{g/L} \end{align*} (b) Ammonia (NH\(_3\)): \begin{align*} \rho_{NH_3} &= \frac{133000\,\text{Pa} \times 17.04\,\text{g/mol}}{8.314\,\text{J/(mol K)} \times 297.59\,K} \\ &= 888.83\,\text{g/m}^3 \\ &= 0.89\,\text{g/L} \end{align*} (c) Acetylene (C\(_2\)H\(_2\)): \begin{align*} \rho_{C_2H_2} &= \frac{133000\,\text{Pa} \times 26.04\,\text{g/mol}}{8.314\,\text{J/(mol K)} \times 297.59\,K} \\ &= 1376.93\,\text{g/m}^3 \\ &= 1.38\,\text{g/L} \end{align*} Thus, the densities of the gases at \(75^{\circ}\mathrm{F}\) and \(1.33\) bar are: (a) Argon: 2.24 g/L (b) Ammonia: 0.89 g/L (c) Acetylene: 1.38 g/L

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and amount of an ideal gas through the formula: \[ PV = nRT \]
Where P represents pressure, V is volume, n is the number of moles of gas, R is the ideal or universal gas constant, and T is the temperature in Kelvin. When dealing with gas density calculations, the ideal gas law can be modified to express the density (\(\rho\)) as a function of the pressure (P), molar mass (M), and temperature (T). This adaptation looks like the following equation:\[ \rho = \frac{PM}{RT} \]
The beauty of this formula is that it allows for the computation of gas density under various conditions of pressure and temperature, assuming the gas behaves ideally. In educational practice, it's important to help students remember to always use the proper units when using the ideal gas law, which often requires conversion of units.

Conversion of Units

When solving physics or chemistry problems, converting units into a consistent set is crucial for accuracy. As seen in the original exercise, pressure was converted from bar to Pascals, and the temperature from Fahrenheit to Kelvin.

Conversion of pressure units is straightforward using the conversion factor that 1 bar equals 100,000 Pascals (Pa), as done for the exercise. Temperature conversions require a bit more attention as you have to use the formula for converting Fahrenheit (\(F\)) to Kelvin (\(K\)), which is \[ T(K) = \frac{F - 32}{1.8} + 273.15 \]
Students should familiarize themselves with common conversion factors and practice converting units until it becomes second nature. A helpful tip for remembering these conversions is to keep a conversion chart handy or memorize a few key factors.

Molar Mass Computation

The molar mass of a substance corresponds to the mass of one mole of that substance, typically expressed in grams per mole (g/mol). This intrinsic property is a pivotal component when calculating the density of gases. To find the molar mass of a compound, like ammonia (NH₃) or acetylene (C₂H₂), you should sum the atomic masses of each element in the molecule. For instance:
- Ammonia has one nitrogen atom and three hydrogen atoms. Hence, its molar mass is the sum of one nitrogen atom's mass (~14.01 g/mol) and three hydrogen atoms' masses (3 times ~1.01 g/mol).
- Acetylene consists of two carbon atoms and two hydrogen atoms. So, its molar mass is two times the mass of carbon (~12.01 g/mol) plus two times the mass of hydrogen.
Remember, understanding how atomic masses contribute to the overall molar mass of a compound allows for better grasping of stoichiometry and related calculations like gas density.

Density of Gases

Gas density is a measure of mass per unit volume and is typically expressed in grams per liter (g/L) when discussing gases. Density can be calculated using the modified ideal gas law formula, integrating molar mass and limiting the equation to the density of a given substance. Through \[ \rho = \frac{PM}{RT} \],where \(\rho\) stands for density, the equation shows us how the density of a gas is directly proportional to its pressure (P) and molar mass (M), and inversely proportional to the temperature (T) when the gas constant (R) is a known value.

Recalling the exercise, once you have the pressure in Pascals and the temperature in Kelvin, along with the computed molar mass, all that's left is to input these into the formula to find the gas density. Always check that your units match the units needed for the constants in the formula. This directly leads to understanding that gases can have variable densities depending on their molecular makeup and the conditions they are found in.