Problem 52 - Chemistry: An Atoms-Focused Approach

Question

The pressure on a sample of an ideal gas is increased from \(715 \mathrm{mmHg}\) to 3.55 atm at constant temperature. If the initial volume of the gas is \(485 \mathrm{mL},\) what is the final volume of the gas?

Step-by-Step Solution

Answer

Answer: The final volume of the gas is approximately 128.63 mL.

Step 1: Convert the given pressures to the same units

First, we need to convert 3.55 atm to mmHg (millimeters of mercury). We know that 1 atm = 760 mmHg, so: $$ 3.55\,\text{atm} \times \frac {760\,\text{mmHg}}{1\,\text{atm}} = 2698\,\text{mmHg} $$ Now both pressures are in mmHg: Initial pressure (P1) = 715 mmHg Final pressure (P2) = 2698 mmHg

Step 2: Apply the Boyle's Law Formula

Now we'll apply the Boyle's Law formula, P1V1 = P2V2, and solve for the final volume (V2). We have: Initial pressure (P1) = 715 mmHg Initial volume (V1) = 485 mL Final pressure (P2) = 2698 mmHg We need to find: Final volume (V2)

Step 3: Solve for the Final Volume (V2)

Using the Boyle's Law formula, we can set up the equation and solve for V2: $$ P1V1 = P2V2 \\ (715\,\text{mmHg})(485\,\text{mL}) = (2698\,\text{mmHg})(V2) $$ Now we can solve for V2: $$ V2 = \frac{(715\,\text{mmHg})(485\,\text{mL})}{2698\,\text{mmHg}} $$

Step 4: Calculate the Final Volume

Calculating the final volume: $$ V2 = \frac{(715\,\text{mmHg})(485\,\text{mL})}{2698\,\text{mmHg}} = 128.634\,\text{mL} $$ So, the final volume of the gas is approximately \(128.63\,\text{mL}\).

Key Concepts

Ideal Gas

Ideal gases are a key area in chemistry and physics. They are hypothetical gases that help to simplify computations by assuming that gas molecules exhibit perfectly elastic collisions and that they have no volume. This means that ideal gases follow the ideal gas law equations without any deviations caused by interactions between gas molecules or by the volume of the molecules themselves.

Understanding ideal gas behavior allows us to use simplified relationships, such as Boyle's Law, where we only focus on the pressure and volume of the gas at a given temperature. In essence, even though real gases might not behave perfectly like ideal gases, this assumption allows us to make useful predictions about how gases behave under different conditions without getting bogged down in complex calculations.

Pressure Conversion

Pressure conversion is an essential skill when working with gases. Often, pressures are given in different units and need to be consistent for calculations. In the problem we're analyzing, pressure was initially given in millimeters of mercury (mmHg) and later in atmospheres (atm).

To convert from atm to mmHg, we use the relationship 1 atmosphere = 760 mmHg. Hence, to convert 3.55 atm to mmHg:

• Multiply 3.55 atm by 760 mmHg / 1 atm.
• This conversion results in 2698 mmHg.

This consistency allows the use of the Boyle's Law formula efficiently, which necessitates pressure in a uniform unit for direct calculations.

Volume Calculation

Volume calculation in scenarios involving gases often relies on Boyle's Law. This law states that for a given mass of a gas at constant temperature, the volume of the gas is inversely proportional to its pressure. Mathematically, it is expressed as \( P_1V_1 = P_2V_2 \).

In the original exercise, the initial conditions are: pressure \( P_1 = 715 \, \text{mmHg} \) and volume \( V_1 = 485 \, \text{mL} \). The final pressure recorded was \( P_2 = 2698 \, \text{mmHg} \). Using these values, we can calculate the final volume \( V_2 \), by rearranging the equation:

• Substitute the values into the equation: \( (715 \, \text{mmHg})(485 \, \text{mL}) = (2698 \, \text{mmHg})(V_2) \).
• Solve for the final volume: \( V_2 = \frac{(715 \, \text{mmHg})(485 \, \text{mL})}{2698 \, \text{mmHg}} \).
• The final computation gives \( V_2 \approx 128.63 \, \text{mL} \).

This calculation reveals how the volume decreases as the pressure increases, which is a direct application of Boyle's Law.