Problem 57 - Chemistry Matter and Change

Question

Use Boyle’s, Charles’s, or Gay-Lussac’s law to calculate the missing value in each of the following. a. \(V_{1}=2.0 L, P_{1}=0.82 \mathrm{atm}, V_{2}=1.0 \mathrm{L}, P_{2}=?\) b. \(V_{1}=250 \mathrm{mL}, T_{1}=?, V_{2}=400 \mathrm{mL}, T_{2}=298 \mathrm{K}\) c. \(V_{1}=0.55 L, P_{1}=740 mm \mathrm{Hg}, V_{2}=0.80 \mathrm{L}, P_{2}=?\)

Step-by-Step Solution

Answer

a. Using Boyle's law, we find \(P_2 = 1.64 \mathrm{atm}\). b. Using Charles's law, we find \(T_1 ≈ 187 K\). c. Using Boyle's law, we find \(P_2 ≈ 509 \mathrm{mmHg}\).

Step 1: Write down the Boyle's law formula

Since Boyle's law states that \(PV = constant\), we can write: \[P_1V_1 = P_2V_2\]

Step 2: Plug in the given values and solve for the missing value

We know \(V_1 = 2.0 L\), \(P_1 = 0.82 atm\), and \(V_2 = 1.0 L\). Plug these values into the formula: \[0.82 \cdot 2.0 = P_2 \cdot 1.0\] Now, solve for \(P_2\): \[P_2 = 0.82 \cdot 2.0\] \[P_2 = 1.64 \mathrm{atm}\] In this case: \(P_2 = 1.64 \mathrm{atm}\). b. We are given \(V_1\), \(V_2\), \(T_2\), and need to find \(T_1\). This problem refers to Charles's law, as the volume changes and the pressure is constant.

Step 1: Write down the Charles's law formula

Since Charles's law states that \(V \propto T\), we can write: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]

Step 2: Plug in the given values and solve for the missing value

We know \(V_1 = 250 mL\), \(V_2 = 400 mL\), and \(T_2 = 298 K\). Plug these values into the formula: \[\frac{250}{T_1} = \frac{400}{298}\] Now, solve for \(T_1\): \[T_1 = \frac{250 \cdot 298}{400}\] \[T_1 = 186.75\] In this case: \(T_1 ≈ 187 K\). c. We are given \(V_1\), \(P_1\), \(V_2\), and need to find \(P_2\). This problem refers to Boyle's law, as the volume and pressure change while temperature is constant.

Step 1: Write down the Boyle's law formula

Since Boyle's law states that \(PV = constant\), we can write: \[P_1V_1 = P_2V_2\]

Step 2: Plug in the given values and solve for the missing value

We know \(V_1 = 0.55 L\), \(P_1 = 740 mmHg\), and \(V_2 = 0.80 L\). Plug these values into the formula: \[740 \cdot 0.55 = P_2 \cdot 0.80\] Now, solve for \(P_2\): \[P_2 = \frac{740 \cdot 0.55}{0.80}\] \[P_2 = 508.75\] In this case: \(P_2 ≈ 509 \mathrm{mmHg}\).

Key Concepts

Boyle's Law

Imagine you have a balloon. If you squeeze it, making the volume smaller, you'll notice the pressure inside increases. This is the essence of Boyle's Law, which states that the pressure (\(P\)) of a gas tends to decrease as the volume (\(V\)) increases, provided the temperature remains constant. In essence, \(P_1V_1 = P_2V_2\) where \(P_1\) and \(P_2\) are the initial and final pressures, and \(V_1\) and \(V_2\) are the initial and final volumes.

• If the volume of the gas is halved, the pressure will double, assuming constant temperature.
• Conversely, if the volume doubles, the pressure will halve.

In the exercise, Boyle's Law helped us find how pressure changes when volume changes. It's a practical guide for various real-world applications like engines and syringes. Boyle's Law has been tested and proven reliable for scenarios where temperature remains stable.

Charles's Law

Picture a balloon expanding on a hot day. This phenomenon is explained by Charles's Law. This law states that the volume (\(V\)) of a gas is directly proportional to its temperature (\(T\)) when the pressure is constant. Simply put, \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) where \(T_1\) and \(T_2\) are the initial and final temperatures.

• If the temperature of a gas increases, its volume will increase too, provided the pressure stays the same.
• If the temperature drops, the volume decreases.

Through Charles's Law in our exercise, we calculated how temperature affects volume. It is significant in many fields, such as meteorology and the mechanics of hot air balloons. It's an excellent principle showing the predictable nature of gases related to temperature and volume.

Gay-Lussac’s Law

Imagine a sealed, rigid container with gas inside. As you heat it, the pressure within the container rises. This scenario is described by Gay-Lussac’s Law, which states that the pressure of a gas is directly proportional to its temperature when the volume is constant. Mathematically, it is expressed as \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\).

• If you increase the temperature of the gas, the pressure increases as well.
• Conversely, if the temperature goes down, the pressure decreases.

Gay-Lussac’s Law was not explicitly used in this exercise, but it’s an essential component of gas laws. It is particularly important in understanding how gases behave in closed systems, such as pressure cookers or even our everyday car tires. This law helps predict how pressure changes with temperature variations.