Problem 34 - Chemistry for Allied Health

Question

A balloon inflated with three breaths of air has a volume of \(1.7 \mathrm{~L}\). At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?

Step-by-Step Solution

Answer

The volume with eight breaths is 4.53 L.

Step 1: Determine the volume contribution per breath

The balloon initially holds a volume of 1.7 L with three breaths of air. To find the volume contribution per breath, divide the total volume by the number of breaths. Volume per breath = \( \frac{1.7}{3} \) L.

Step 2: Calculate the additional volume from five more breaths

Each breath contributes \( \frac{1.7}{3} \) L of volume. To find the total volume of five more breaths, multiply the volume per breath by five.Additional volume = \( 5 \times \frac{1.7}{3} \) L.

Step 3: Compute the total volume with eight breaths

The total volume of the balloon is the initial volume (1.7 L from the first three breaths) plus the additional volume from the five extra breaths calculated in the previous step. Add these two volumes together.Total volume = 1.7 L + \( 5 \times \frac{1.7}{3} \) L.

Key Concepts

Volume

When discussing gases, volume is an essential concept. Volume refers to the amount of space a substance occupies. In the context of a gas-filled balloon, volume tells us how much space the air is taking inside the balloon. Gas can expand or compress based on the pressure and temperature, so the volume of a given amount of gas can vary under different conditions. For example, if a balloon with three breaths has a volume of 1.7 L and we add more breaths, the volume increases proportionally assuming pressure and temperature remain constant. This idea helps us understand how much more space the gas will occupy when additional air is introduced into the balloon.

Temperature

Temperature plays a crucial role in determining the behavior of gases. It measures the average kinetic energy of the particles. In simpler terms, temperature tells us how "hot" or "cold" something is, measured in degrees (Celsius, Fahrenheit, or Kelvin). At a molecular level, higher temperatures mean particles move more vigorously and energetically. Although in our exercise, we assumed temperature remains constant when adding more air to the balloon, normally, changes in temperature can cause gases to expand or contract. For example, heating a balloon may cause it to expand because the air particles inside move faster and push outwards with greater force.

Pressure

Pressure in gases is defined as the force that the gas exerts on the surfaces of its container, measured in units like Pascals (Pa) or atmospheres (atm). It's an important concept because the behavior of gases is closely linked to changes in pressure. In the current exercise, we assume the pressure is constant when additional air is added to the balloon. However, in general, if the pressure inside the balloon was to increase while keeping the temperature constant, then the gas might occupy a smaller volume. This relationship is Maxwell's Law, which states that pressure is inversely proportional to volume when temperature is constant. Constant pressure during the inflation of the balloon means that the added gas volume is evenly distributed without causing the balloon to burst.

Chemical Calculations

Chemical calculations involving gases use formulas and principles outlined in the ideal gas law. The ideal gas law is expressed in the formula: \[ PV = nRT \]where:

• \(P\) represents pressure.
• \(V\) represents volume.
• \(n\) is the number of moles of gas.
• \(R\) is the universal gas constant.
• \(T\) stands for temperature.

In simpler terms, this equation helps predict how a gas will behave based on its current state. For the balloon exercise, we used chemical logic to calculate how the volume changes based on a fixed number of breath expansions. Though we didn't need the ideal gas law directly since we kept temperature and pressure constant, understanding such equations helps us predict changes when these conditions vary. This knowledge helps one in predicting experimental results and is the backbone of many chemical calculations in chemistry.