Question

For each of the following sets of pressure/volume data, calculate the missing quantity. Assume that the temperature and the amount of gas remain constant. a. \(V=255 \mathrm{mL}\) at \(1.00 \mathrm{mm} \mathrm{Hg} ; V=?\) at 2.00 torr b. \(V=1.3 \mathrm{L}\) at \(1.0 \mathrm{kPa} ; V=?\) at \(1.0 \mathrm{atm}\) c. \(V=1.3 \mathrm{L}\) at \(1.0 \mathrm{kPa} ; V=?\) at \(1.0 \mathrm{mm} \mathrm{Hg}\)

Short answer

For the given sets of pressure and volume data, the missing quantity can be calculated using Boyle's law and the given values as follows: a. \(V_2 = 127.5\,\text{mL}\) b. \(V_2 = 0.0128\,\text{L}\) c. \(V_2 = 9.75\,\text{L}\)

Step-by-step solution

Analyze given data and convert pressure units if needed

We have three parts to this exercise. We will go through each part and convert the pressure values if needed to keep them consistent. a. 255 mL at 1.00 mm Hg; V=? at 2.00 torr 1 mm Hg is equal to 1 torr, so no conversion is needed. b. 1.3 L at 1.0 kPa; V=? at 1.0 atm We need to convert 1.0 kPa to atm: \(1.0\,\text{kPa} \times \frac{1\,\text{atm}}{101.325\,\text{kPa}} \approx 0.00987\,\text{atm}\) c. 1.3 L at 1.0 kPa; V=? at 1.0 mm Hg We need to convert 1.0 kPa to mm Hg: \(1.0\,\text{kPa} \times \frac{760\,\text{mm}\,\text{Hg}}{101.325\,\text{kPa}} \approx 7.5 \,\text{mm}\,\text{Hg}\)

Apply Boyle's law for each part and calculate the missing volume

Now that we have the pressure values in the same units, we can apply Boyle's law and calculate the missing volume for each part. a. \((1.00\,\text{torr})(255\,\text{mL}) = (2.00\,\text{torr}) (V_2)\) => \(V_2 = \frac{(1.00\,\text{torr})(255\,\text{mL})}{2.00\,\text{torr}} = 127.5\,\text{mL}\) b. \((0.00987\,\text{atm})(1.3\,\text{L}) = (1.0\,\text{atm})(V_2)\) => \(V_2 = \frac{(0.00987\,\text{atm})(1.3\,\text{L})}{1.0\,\text{atm}} = 0.0128\,\text{L}\) c. \((7.5\,\text{mm}\,\text{Hg})(1.3\,\text{L}) = (1.0\,\text{mm}\,\text{Hg})(V_2)\) => \(V_2 = \frac{(7.5\,\text{mm}\,\text{Hg})(1.3\,\text{L})}{1.0\,\text{mm}\,\text{Hg}} = 9.75\,\text{L}\) The missing volumes are as follows: a. \(V_2 = 127.5\,\text{mL}\) b. \(V_2 = 0.0128\,\text{L}\) c. \(V_2 = 9.75\,\text{L}\)

Key concepts

Pressure-Volume Relationship

Understanding the pressure-volume relationship is essential in various scientific and engineering fields, particularly when dealing with gases. This relationship is a pillar of Boyle's law, which states that for a constant temperature and amount of gas, the pressure exerted by a gas is inversely proportional to its volume. In other words, if you compress a gas, its pressure goes up, and if you expand it, its pressure decreases.

Illustrating this concept with an everyday example, think of a bicycle pump. As you press down on the pump, decreasing the volume inside, the air pressure increases, which is why you can inflate your tire. This same principle applies at the molecular level for gases confined in a container. The pressure is due to the gas molecules colliding with the container walls, and with less space (volume), these collisions happen more frequently, thus increasing the pressure.

To solve problems involving the pressure-volume relationship, we often use Boyle's law formula: \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the pressure and volume after a change, assuming no temperature variation. This formula allows us to predict what happens to one property when we alter the other.

Gas Laws

The behavior of gases under various conditions is described by the gas laws: Boyle's law (pressure-volume relationship), Charles's law (temperature-volume relationship), Gay-Lussac's law (pressure-temperature relationship), and the combined gas law, which brings these individual relationships together. These laws are based on ideal gas behavior, where gas molecules are considered as non-interacting particles in constant, random motion.

For students, understanding these laws isn't just about memorizing formulas; it's about grasping how the physical properties of gases—pressure, volume, and temperature—are interdependent. A beautiful aspect of the gas laws is their ability to predict how gases will respond to changes in their environment, which is crucial in applications like predicting weather patterns, designing engines, or even cooking at high altitudes.

Unit Conversion

Unit conversion is a fundamental skill in science that ensures measurements are communicated and understood universally. When solving problems involving gas laws, we frequently encounter different units for pressure, such as atmospheres (atm), kilopascals (kPa), and millimeters of mercury (mm Hg).

To consistently apply the gas laws, we need to convert these units. For example, 1 atm is equivalent to 101.325 kPa and also to 760 mm Hg. While online tools and calculators can perform these conversions quickly, having a solid understanding of how to manually convert units helps deepen our comprehension of the magnitudes involved and allows us to solve problems even when digital resources aren't available.

Here is a simplified tip for students: Always write down the known conversion factors and use them to set up conversion ratios. Make sure that the units you want to get rid of are diagonally opposite each other in the ratio, so they cancel out, leaving you with the desired units. For example, converting pressure from kPa to atm involves the ratio: \( \frac{1\text{ atm}}{101.325\text{ kPa}} \).