A scuba diver takes a 2.8-L balloon from the surface, where the pressure is \(1.0 \mathrm{~atm}\) and the temperature is \(34^{\circ} \mathrm{C}\), to a depth of \(25 \mathrm{~m}\), where the pressure is \(3.5 \mathrm{~atm}\) and the temperature is \(18^{\circ} \mathrm{C}\). What is the volume of the balloon at this depth?

Short Answer

Expert verified
The volume of the balloon at a depth of 25 m is obtained after converting the temperatures to Kelvin and applying the Combined Gas Law. V2 is calculated using the equation with the appropriate units.

Step by step solution

01

Convert Temperature to Kelvin

Convert the initial and final temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = C + 273.15. For the initial temperature, Ti = 34°C = 34 + 273.15 K. For the final temperature, Tf = 18°C = 18 + 273.15 K.
02

Apply the Combined Gas Law

Use the Combined Gas Law, which relates pressure, volume, and temperature: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Since we need to find the final volume, V2, rearrange the equation to solve for V2: \( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \).
03

Plug in the Known Values

Substitute the known values into the equation: P1 = 1.0 atm, V1 = 2.8 L, T1 = 34 + 273.15 K, P2 = 3.5 atm, T2 = 18 + 273.15 K.
04

Calculate the Final Volume

Compute the final volume V2 using the combined gas law equation with the values converted to their proper units. After doing the math, V2 will give us the volume of the balloon at the depth of 25 m.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Gas Law Calculations in Practical Scenarios
In trying to solve problems related to gases, gas law calculations provide a robust framework. They allow prediction of how a gas will behave when subjected to changes in pressure, temperature, and volume. The Combined Gas Law,
particularly useful, amalgamates Charles's Law, Boyle's Law, and Gay-Lussac's Law into a single equation

  • Charles's Law states that the volume of a gas is directly proportional to its temperature, assuming pressure remains constant.
  • Boyle's Law postulates that the volume of a gas is inversely proportional to its pressure at constant temperature.
  • Gay-Lussac's Law asserts that the pressure of a gas is directly proportional to its temperature at constant volume.

The Combined Gas Law allows you to solve for any one of the state variables (pressure, volume, temperature) if the others are known. This law is crucial in fields such as meteorology, engineering, and even undersea explorations—where understanding the behavior of gases under varying conditions is essential.
Temperature Conversion: Celsius to Kelvin
When dealing with gas law calculations, it's imperative to use an absolute temperature scale, which is why temperatures must be converted to Kelvin. Unlike the Celsius scale, where 0 degrees is based on the freezing point of water, the Kelvin scale starts at absolute zero—the coldest possible temperature where particles have minimum thermal motion.
To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. The formula is a simple yet crucial step in ensuring the accuracy of gas law calculations:

  • Celsius to Kelvin: K = °C + 273.15

Precision in this conversion process is vital, as errors can lead to significant miscalculations in the final result. For instance, in the given exercise, the surface temperature of 34°C is equivalent to 307.15 K, and the underwater temperature of 18°C converts to 291.15 K. For any gas law equation, using Kelvin as the temperature unit ensures consistency and coherence in the study of thermodynamic gas behavior.
Scuba Diving and Pressure: The Physics Beneath the Surface
As a scuba diver descends, the pressure from the weight of the water above increases. This change in pressure must be calculated when considering the effects on the volume of gases, such as the air in a balloon or in the diver's tank. At sea level, the atmospheric pressure is about 1 atm, but for every 10 meters of depth underwater, the pressure increases by approximately 1 atm.

Pressure Effects on Scuba Diving


Scuba divers must manage these changes in pressure to avoid physical harm like decompression sickness (also known as 'the bends'), where dissolved gases come out of solution in bubbles and can cause injury to tissues. The increase in pressure also affects buoyancy and the consumption of air from tanks.
  • At the surface: 1 atm pressure.
  • Every 10 m underwater: +1 atm pressure.

The exercise showcases how dive depth (25 m here, hence a pressure of 3.5 atm) will compress the balloon's volume, adhering to the Combined Gas Law. Scuba diving provides a tangible instance of gases behaving in line with fundamental physical laws, affirming the relevance of gas law calculations in practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free