Problem 25 - Chemistry: The Central Science

Question

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant. ll be produced?

Step-by-Step Solution

Answer

Pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

Step 1: Action (a): Doubling the volume while keeping the temperature constant.

For this action, the temperature and the number of moles of gas remain constant. We only need to consider the volume (V) and the pressure (P): \(PV = \text{constant}\). If the volume is doubled and the product PV remains constant, the pressure will be halved (P/2), and not doubled. So action (a) will not double the gas pressure.

Step 2: Action (b): Doubling the temperature while keeping the volume constant.

For this action, the volume and the number of moles of gas remain constant. We only need to consider the temperature (T) and the pressure (P): \(\frac{P}{T} = \text{constant}\). When the temperature is increased from 25°C to 50°C, we must first convert it to Kelvin: 25°C + 273.15 = 298.15K and 50°C + 273.15 = 323.15K. To check the effect on the pressure, let's write the relation for the initial and final states: \[ \frac{P_i}{T_i} = \frac{P_f}{T_f}. \] With the given temperature values: \[ \frac{P_i}{298.15} = \frac{P_f}{323.15}. \] Solving for the final pressure (\(P_f\)): \[ P_f = P_i \times \frac{323.15}{298.15}. \] Since \(323.15 / 298.15 \approx 1.084\), the pressure will not be doubled (it would be increased by about 8.4% only). So action (b) will not double the gas pressure.

Step 3: Action (c): Halving the volume while keeping the temperature constant.

For this action, the temperature and the number of moles of gas remain constant. We only need to consider the volume (V) and the pressure (P): \(PV = \text{constant}\). If the volume is halved and the product PV remains constant, the pressure will be doubled (2P), which is our objective. So action (c) will double the gas pressure. To sum up the results: 1. Action (a) will not double the gas pressure. 2. Action (b) will not double the gas pressure. 3. Action (c) will double the gas pressure. Therefore, pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

Key Concepts

Boyle's Law

Understanding gas pressure in chemistry often begins with Boyle's Law, a principle which deals with the inversely proportional relationship between the pressure and volume of a gas at constant temperature. This means that if you increase the volume of gas in a container, its pressure decreases, and conversely, if you decrease the volume, the pressure increases.

From our exercise, when pushing down on a piston to halve the volume (action c), Boyle’s Law can be applied to predict the outcome on the pressure of the gas. Mathematically, Boyle’s Law is represented as \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively. Halving the volume while maintaining the same amount of gas and constant temperature will therefore double the pressure, as we've seen in the solution to action (c).

Charles's Law

Another fundamental aspect of gas behavior can be captured by Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature, assuming the number of gas particles and the pressure remain constant.

Applying Charles's Law to action (b) from our exercise, where the temperature increases from \(25^\circ \mathrm{C}\) to \(50^\circ \mathrm{C}\) while the volume remains constant, results in an increase in pressure. This is due to the direct relationship between temperature and pressure when volume does not change. Charles's Law is mathematically expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( T \) must be in Kelvin. In practice, though, action (b) doesn't double the pressure, as the temperature increase to \( 50^\circ \mathrm{C} \) results in a rise in pressure of only about 8.4%, not the 100% needed to double it.

Ideal Gas Law

The Ideal Gas Law is a more encompassing equation that relates pressure (P), volume (V), temperature (T), and the number of moles of gas (n) to describe the state of a hypothetical 'ideal' gas. The law is commonly expressed as \( PV = nRT \), where R is the ideal gas constant. This equation meshes the principles of Boyle's Law, Charles's Law, and others into one formula.

With respect to our exercise, the Ideal Gas Law can be used to predict the effect of changing more than one condition of a gas at a time. However, since in our step-by-step solution we're changing one condition at a time—volume in action (a) and (c), and temperature in action (b)—the individual laws are sufficient for prediction.

Temperature and Pressure Relationship

The temperature and pressure relationship of gases is another fundamental concept to grasp when studying gas law in chemistry. According to Gay-Lussac's Law, which is a component of the Ideal Gas Law, the pressure of a gas is directly proportional to its temperature when the volume is kept constant.

In our text exercise, action (b) attempts to double the pressure by increasing temperature while keeping volume constant. This is an application of Gay-Lussac's Law, and while pressure does increase with temperature, the specific increase from \(25^\circ \mathrm{C}\) to \(50^\circ \mathrm{C}\) is not sufficient to double the pressure. This helps illustrate the significance of understanding this temperature-pressure relationship to correctly predict the behavior of gases under various conditions.