Problem 130 - Introductory Chemistry Atoms First

Question

You are on the balcony of the upper floor of a high-rise building in Chicago. Way down below on the street, your friend is engaged in conversation with someone and is drinking a particularly flavorful mixed drink. You decide to take a sip from his glass without his knowing, so you collect together every drinking straw in the apartment and start taping them together to create a very long drinking straw. If the atmospheric pressure is currently \(760 \mathrm{~mm} \mathrm{Hg}\), what is the longest straw you can use to get a sip of your friend's drink? (Assume the density of the drink is the same as pure water, \(1.0 \mathrm{~g} / \mathrm{mL}\).)

Step-by-Step Solution

Answer

The longest straw you can use to get a sip of your friend's drink is approximately \(10.33\ \text{m}\).

Step 1: 1. Write the given variables and conversions

: - Atmospheric pressure: \(760\ \text{mm Hg}\) - Density of the drink: \(\rho = 1.0\ \text{g/mL}\) We'll also need some important constants and formulas during the process: - Acceleration due to gravity: \(g = 9.8\ \text{m/s}^2\) - Conversion from mm Hg to Pa (Pascal): \(1\ \text{mm Hg} \approx 133.3\ \text{Pa}\)

Step 2: 2. Convert atmospheric pressure to Pa

: To use the atmospheric pressure in the formula, we need to convert it to Pa. Let's do that: Atmospheric pressure = \(760\ \text{mm Hg} \times 133.3\ \text{Pa/mm Hg} \approx 101320\ \text{Pa}\)

Step 3: 3. Convert the density of the drink to proper unit

: The density of the drink is given in g/mL. To use it in the formula, we need to convert it to kg/m³: Density \(\rho = 1.0\ \text{g/mL} = 1000\ \text{kg/m}^3\)

Step 4: 4. Write the formula for pressure at a depth in liquid

: Pressure at a depth in a liquid can be calculated using the following formula: \(P = \rho \cdot g \cdot h\) Where: - \(P\) is the pressure at depth - \(\rho\) is the density of the liquid - \(g\) is the acceleration due to gravity - \(h\) is the depth (height) in the liquid

Step 5: 5. Solve for the maximum height in the straw

: Using the formula mentioned above and equating the pressure inside the straw to the atmospheric pressure, we get: \(P_\text{atmosphere} = \rho \cdot g \cdot h\) Solve for \(h\): \(h = \frac{P_\text{atmosphere}}{\rho \cdot g}\) Now plug in the values: \(h = \frac{101320\ \text{Pa}}{1000\ \text{kg/m}^3 \times 9.8\ \text{m/s}^2} \approx 10.33\ \text{m}\)

Step 6: 6. Get the final answer

: So, the longest straw you can use to get a sip of your friend's drink is approximately \(10.33\ \text{m}\).

Key Concepts

Density Conversion

Let's begin by understanding what density conversion means. Density is a measure of mass per unit volume. In many physics or chemistry problems, you are often required to convert density from one set of units to another so the formulae can be easily applied.
For example, in this problem, the density of the drink is given in grams per milliliter (g/mL), but we need it in kilograms per cubic meter (kg/m\(^3\)). This is because the formula for calculating pressure at a depth requires the density in kg/m\(^3\).

• To perform this conversion, note that 1 g/mL is equivalent to 1000 kg/m\(^3\).
• This stems from the fact that 1 milliliter equals 0.001 liters, and since 1 liter is 0.001 cubic meters, it translates as: 1 g/mL = 1 g/(0.001 L) = 1000 g/L = 1000 kg/m\(^3\).

This conversion ensures compatibility with the SI units used in the physical formulas.

Pressure Calculation

Pressure calculation involves determining the force exerted by a fluid per unit area. In this scenario, we are asked to find the maximum height of a straw where atmospheric pressure can lift the drink up to that height.
To achieve this, we use the equation for the pressure at a depth within a liquid, which is constructed based on lines of balance between atmospheric pressure and the pressure exerted by the liquid column.
The equation is:\[ P = \rho \cdot g \cdot h \]
Where:

• \(P\) represents the atmospheric pressure exerted.
• \(\rho\) stands for the density of the liquid.
• \(g\) is the acceleration due to gravity.
• \(h\) denotes the height of the liquid column.

In our exercise, we equate the atmospheric pressure outside to the pressure exerted by a column of the liquid inside the straw to find \(h\). By solving for \(h\), we calculate the maximum height that the atmospheric pressure can support, resulting in an approximate length of 10.33 meters for the straw.

Units of Measurement

Understanding units of measurement is crucial for solving physics problems as it ensures consistency when substituting values into equations.
In this problem, different units are used for pressure and density while calculating pressures and heights.

• Atmospheric pressure is initially given in millimeters of mercury (mm Hg), a common unit of pressure in meteorology and aviation. We convert this to Pascals (Pa), the SI unit of pressure, leveraging the relationship: 1 mm Hg ≈ 133.3 Pa.
• Density is provided in grams per milliliter (g/mL), but converted to kilograms per cubic meter (kg/m\(^3\)) for compatibility in equations that utilize SI units.

The SI (International System of Units) ensures uniformity across global scientific analysis and data interchange by adopting standard units such as meters (m), kilograms (kg), and seconds (s).
Through conversion and proper application of these units, the problem is solved accurately, facilitating understanding and precise communication among scientific communities.