Question

20\. Use the ideal gas law to complete the following table for propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) gas. \begin{tabular}{llcll} \hline & Pressure & Volume & Temperature & Moles & Grams \\ \hline (a) \(18.9 \mathrm{psi}\) & \(0.886 \mathrm{~L}\) & \(22^{\circ} \mathrm{C}\) & & \\ (b) \(633 \mathrm{~mm} \mathrm{Hg}\) & \(1.993 \mathrm{~L}\) & & 0.0844 & \\ \hline (c) \(1.876 \mathrm{~atm}\) & & \(75^{\circ} \mathrm{F}\) & 2.842 & \\ \hline (d) & \(2244 \mathrm{~mL}\) & \(13^{\circ} \mathrm{C}\) & & 47.25 \\ \hline \end{tabular}

Short answer

In this exercise, we used the ideal gas law formula to find the missing information for each scenario. Remember to convert all the units to the appropriate form before applying the formula. a) Scenario A: We found the number of moles to be 0.0484 mol and the mass to be 2.13 g. b) Scenario B: We found the temperature to be 13.73 °C (286.88 K) and the mass to be 3.72 g. c) Scenario C: We found the volume to be 36.42 L and the mass to be 125.29 g. d) Scenario D: We found the pressure to be 1.438 atm and the number of moles to be 1.071 mol.

Step-by-step solution

a) Find moles and grams for Scenario A

First, we need to convert the given temperature from Celsius to Kelvin and the pressure from psi to atm: \(T_a = 22°C + 273.15 = 295.15 K\) \(P_a = 18.9 psi * (1 atm / 14.7 psi)= 1.286 atm\) Now, replacing the values in the ideal gas law formula, and solving for the number of moles (n): \(PV = nRT\) \((1.286 atm)(0.886 L) = n(0.0821 atm L/mol K)(295.15 K)\) \(n_a = \frac{(1.286)(0.886)}{(0.0821)(295.15)}\) \(n_a = 0.0484 \thinspace mol\) Next, convert moles into grams: Molar mass of C3H8 = (3 * 12.01) + (8 * 1.01) = 44.09 g/mol \(mass_{a} = moles_{a} \times molar\thinspace mass\) \(mass_{a} = 0.0484 \thinspace mol \times 44.09 \thinspace g/mol \) \(mass_{a} = 2.13 \thinspace g\)

b) Find temperature and grams for Scenario B

First, we need to convert the given pressure from mm Hg to atm: \(P_b = 633 mm Hg * (1 atm / 760 mm Hg) = 0.833 atm\) Now, replacing the values in the ideal gas law formula, and solving for the temperature (T): \(PV = nRT\) \((0.833 atm)(1.993 L) = (0.0844 mol)(0.0821 atm L/mol K)T\) \(T_b = \frac{(0.833)(1.993)}{(0.0844)(0.0821)}\) \(T_b = 286.88 K\) To convert back into Celsius: \(T_b = 286.88 K - 273.15 = 13.73 °C\) Next, find grams of propane: \(mass_{b} = moles_{b} \times molar\thinspace mass\) \(mass_{b} = 0.0844 \thinspace mol \times 44.09 \thinspace g/mol \) \(mass_{b} = 3.72 \thinspace g\)

c) Find volume and grams for Scenario C

First, we need to convert the given temperature from Fahrenheit to Kelvin: \(T_c = (75°F - 32) * \frac{5}{9} + 273.15 = 297.04 K\) Now, replacing the values in the ideal gas law formula, and solving for the volume (V): \(PV = nRT\) \((1.876 atm)V = (2.842 mol)(0.0821 atm L/mol K)(297.04 K)\) \(V_c = \frac{(2.842)(0.0821)(297.04)}{(1.876)}\) \(V_c = 36.42 \thinspace L\) Next, find grams of propane: \(mass_{c} = moles_{c} \times molar\thinspace mass\) \(mass_{c} = 2.842 \thinspace mol \times 44.09 \thinspace g/mol \) \(mass_{c} = 125.29 \thinspace g\)

d) Find pressure and moles for Scenario D

First, we need to convert the given temperature from Celsius to Kelvin, and the volume from mL to L: \(T_d = 13°C + 273.15 = 286.15 K\) \(V_d = 2244 mL * (1 L / 1000 mL) = 2.244 L\) Now, replacing the values in the ideal gas law formula, and solving for the pressure (P): \(PV = nRT\) \(P(2.244 L) = (47.25 g)(0.0821 atm L/mol K)(286.15 K) / 44.09 g/mol\) \(P_d = \frac{(47.25)(0.0821)(286.15)}{(44.09)(2.244)}\) \(P_d = 1.438 atm\) Next, find moles of propane: \(moles_{d} = \frac{mass_{d}}{molar\thinspace mass}\) \(moles_{d} = \frac{47.25 \thinspace g}{44.09 \thinspace g/mol}\) \(moles_{d} = 1.071 \thinspace mol\)

Key concepts

Propane C3H8 Properties

Propane, represented by the chemical formula C3H8, is a hydrocarbon, more specifically an alkane, that finds its place in various household and industrial applications, predominantly as a fuel source. It is a colorless and normally compressed liquid gas that has a distinct odor. Propane's physical properties include having a boiling point of -42°C (-44°F) and a melting point of -187.7°C (-305.9°F). In chemistry, understanding the properties of a compound like propane is critical, especially when working with gas laws, as these properties affect how the gas behaves under different conditions of temperature and pressure.

When dealing with gases, we often refer to them as 'ideal gases' for the purpose of calculation. This is an important assumption in chemistry which greatly simplifies the application of the gas laws. By considering a gas as 'ideal', we assume that the individual gas particles do not attract or repel each other and have negligible volume. Although no real gas perfectly fits these criteria, propane behaves close enough to an ideal gas under many conditions to use these simplifications for calculations like those demonstrated in the exercise.

Converting Units in Chemistry

Converting units in chemistry is a fundamental skill as it ensures that you're using the correct measurements when following a formula or comparing results. Units like temperature, pressure, volume, and mass are often converted from one system to another – for instance, degrees Celsius to Kelvin, or psi to atmospheres (atm). This process is crucial in gas law calculations, as seen in the given exercise where pressures and temperatures were converted before applying the ideal gas law.

Having a strong handle on unit conversion can prevent errors in experiments and calculations, and understanding the reasoning behind the conversions can provide deeper insight into the problem at hand. Being familiar with the conversion factors, such as knowing that 1 atm equals 760 mmHg or 14.7 psi, and that 0°C corresponds to 273.15 K, is essential for accurate gas calculations. Remember that consistency in units throughout a problem is key to obtaining a correct solution. When dealing with volume, particularly, paying attention to whether you should use liters or milliliters is important, as mixing units without proper conversion can lead to incorrect results.

Gas Laws Calculations

The ideal gas law, represented by the equation PV = nRT, is an equation of state for a hypothetical ideal gas. It is a good approximation to the behavior of many gases under a wide range of conditions. The ideal gas law provides a clear relationship between the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas.

The 'R' in the formula is the ideal gas constant, which can have multiple values depending on the units used for pressure, volume, and temperature. In the exercise, R is given as 0.0821 atm L/mol K, which suits the units atm for pressure, L for volume, and K for temperature. By manipulating the ideal gas law formula, you can solve for any of the four variables provided you have the other three. This flexibility makes it extremely useful in various scenarios as shown in the exercise. The ability to adopt and rearrange the formula to find unknown values is an essential skill in chemistry that allows us to work out conditions of a gas in different environmental settings.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance, essentially relating the mass of a molecular or ionic compound to its amount in moles. The unit typically used for molar mass is grams per mole (g/mol). To calculate the molar mass, you add together the atomic masses of all the atoms present in a molecule, where atomic masses are taken from the periodic table and are usually averaged due to natural isotope variations.

For propane (C3H8), the molar mass calculation involves summing the product of the number of carbon atoms and the atomic mass of carbon (3 times 12.01 g/mol) with the product of the number of hydrogen atoms and the atomic mass of hydrogen (8 times 1.01 g/mol), which results in 44.09 g/mol. Knowing the molar mass allows chemists to convert between moles of a substance and the mass in grams, as was necessary in the exercise for converting the given grams of propane to moles or vice versa, using the formula \(mass = moles \times molar\thinspace mass\). This calculation is integral in nearly all chemical equations and reactions, as well in the context of the ideal gas law when working with the physical behavior of gases.