Problem 50 - Chemistry The Science in Context

Question

The behavior of \(485 \mathrm{mL}\) of an ideal gas in response to pressure is studied in a vessel with a movable piston. What is the final volume of the gas if the pressure on the sample is a. increased from \(715 \mathrm{mmHg}\) to 3.55 atm at constant temperature? b. decreased from 1.15 atm to \(520 \mathrm{mmHg}\) at constant temperature? c. increased by \(26 \%\) at constant temperature?

Step-by-Step Solution

Answer

Answer: The final volumes of the ideal gas in each scenario are: a. 164.45 mL (when pressure is increased from 715 mmHg to 3.55 atm) b. 677.29 mL (when pressure is decreased from 1.15 atm to 520 mmHg) c. 384.92 mL (when pressure is increased by 26%)

Step 1: Convert all pressures to the same units (atm)

Before applying Boyle's law, we need to ensure that all pressures are in the same units. In this case, let's convert all pressures to atm. Conversion factors: 1 atm = 760 mmHg. Scenario a: Initial pressure, P1 = 715 mmHg = (715/760) atm Final pressure, P2 = 3.55 atm Scenario b: Initial pressure, P1 = 1.15 atm Final pressure, P2 = 520 mmHg = (520/760) atm Scenario c: Initial pressure, P1 = unknown Final pressure, P2 = P1 + 26% of P1

Step 2: Use Boyle's Law to calculate the final volume for each scenario

Boyle's Law: P1V1 = P2V2 Scenario a: P1V1 = P2V2 (715/760)V1 = 3.55V2 --> V2 = ((715/760)*485) / 3.55 Scenario b: P1V1 = P2V2 (1.15)V1 = (520/760)V2 --> V2 = (1.15*485) / (520/760) Scenario c: P1V1 = P2V2 P1V1 = (P1 + 0.26*P1)V2 --> V2 = V1 / (1 + 0.26)

Step 3: Calculate final volumes for each scenario

Now, let's calculate the final volumes for each scenario using the expressions obtained in Step 2. Scenario a: V2 = ((715/760)*485) / 3.55 V2 = 164.45 mL (approx.) Scenario b: V2 = (1.15*485) / (520/760) V2 = 677.29 mL (approx.) Scenario c: V2 = 485 / (1 + 0.26) V2 = 384.92 mL (approx.) In conclusion, the final volumes of the ideal gas in each scenario are: a. 164.45 mL (when pressure is increased from 715 mmHg to 3.55 atm) b. 677.29 mL (when pressure is decreased from 1.15 atm to 520 mmHg) c. 384.92 mL (when pressure is increased by 26%)

Key Concepts

Understanding the Ideal Gas

The ideal gas law is a cornerstone of chemistry and physics, providing insight into how gases behave under different conditions. Imagine gases as tiny particles bouncing around in a container. These particles rarely interact, leading to the term 'ideal gas'. This concept simplifies calculations by assuming no attractive or repulsive forces between the particles. This assumption is quite practical for many gases under typical conditions.Boyle's Law, building on the ideal gas concept, posits that the pressure of a gas inversely correlates with its volume when temperature and number of moles remain constant. This means if you squeeze the gas into a smaller volume, its pressure tends to increase, and vice versa. Boyle's Law, therefore, provides a simple relationship: \[ P_1V_1 = P_2V_2 \] Here, \(P\) represents pressure, and \(V\) denotes volume. This equation allows us to predict how changing the pressure of a gas will impact its volume.

Pressure Conversion Necessities

Pressure is a measure of how much force a gas exerts on the sides of its container. In scenarios involving Boyle’s Law, pressure often needs to be converted into consistent units. Typical units for pressure include atmospheres (atm) and millimeters of mercury (mmHg), among others. To ensure consistency in calculations, converting all pressure measurements into the same unit is crucial. For instance, 1 atmosphere is equivalent to 760 mmHg. This conversion factor is critical when calculations in Boyle’s Law are involved:

• Example: Convert 715 mmHg to atm: \[ 715 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.941 \text{ atm} \]
• Convert 520 mmHg to atm similarly:\[ 520 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.684 \text{ atm} \]

This uniformity in units ensures that Boyle’s Law remains applicable and accurate during calculations.

Final Volume Calculation Using Boyle's Law

Once pressure is converted to consistent units, calculating the final volume of a gas sample is straightforward using Boyle's Law. After establishing the initial and final pressures, and the initial volume, you can rearrange the equation to solve for the final volume. The process involves these steps:

• Identify initial and final pressures \(P_1\) and \(P_2\).
• Use the equation \( P_1V_1 = P_2V_2 \).
• Rearrange to: \[ V_2 = \frac{P_1 \times V_1}{P_2} \]

Take scenario (a) as an example:

• Initial pressure \(P_1 = 0.941 \text{ atm}\) (after conversion from mmHg)
• Final pressure \(P_2 = 3.55 \text{ atm}\)
• Initial volume \(V_1 = 485 \text{ mL}\)
• Therefore, \[ V_2 = \frac{0.941 \text{ atm} \times 485 \text{ mL}}{3.55 \text{ atm}} \approx 128.55 \text{ mL} \]

This calculation shows the gas volume decreases as pressure increases, a typical result predicted by Boyle’s Law.

The Significance of Constant Temperature

Boyle's Law is valid only when the temperature of the gas remains constant. Temperature influences the kinetic energy of gas particles. When the temperature remains unchanged, the energy level of the particles stays steady, allowing pressure and volume changes to depend solely on each other. This principle of constant temperature, known as isothermal conditions, ensures that the relationship between pressure and volume is predictable as dictated by Boyle’s Law. Hence, in our exercises:

• Pressure changes selectively alter the volume.
• The temperature was held constant, making calculations using Boyle's Law feasible without additional adjustments.

So, in any exercise or experiment related to Boyle's Law, check if the temperature is constant. If not, more complex gas laws like the combined gas law might be necessary to account for additional variables.