/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A compressed-air cylinder is kno... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 83

A compressed-air cylinder is known to fail if the pressure exceeds 110 atm. A cylinder that was filled to 25 atm at \(20^{\circ} \mathrm{C}\) is stored in a warehouse. Unfortunately, the warehouse catches fire and the temperature reaches \(950^{\circ} \mathrm{C} .\) Does the cylinder explode?

Short Answer

Expert verified
Using the provided information and applying Gay-Lussac's Law, we can calculate the final pressure in the cylinder under the given conditions. Depending on whether this pressure exceeds the failure threshold of 110 atm, we can conclude whether the cylinder would explode or not.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert Celsius temperatures to Kelvin because the gas laws use absolute temperatures (Kelvin scale). The Kelvin temperature can be found by adding 273.15 to the Celsius temperature. Initial temperature (\(T_1\)) would be 20掳C + 273.15 = 293.15 K, and final temperature (\(T_2\)) would be 950掳C + 273.15 = 1223.15 K.
02

Apply Gay-Lussac's Law

The formula for Gay-Lussac's Law is \(P_1 / T_1 = P_2 / T_2\), where \(P_1\) and \(T_1\) are initial pressure and temperature, and \(P_2\) and \(T_2\) are the final pressure and temperature. We already know \(P_1\) (25 atm), \(T_1\) (293.15 K), and \(T_2\) (1223.15 K). We can solve for \(P_2\): \(P_2 = (P_1 * T_2) / T_1\).
03

Calculate Final Pressure and Compare to Threshold

Substituting the known values into the equation, we calculate \(P_2 = (25 * 1223.15) / 293.15\). This will give us the final pressure in the cylinder when it is exposed to the high temperature. We compare this value to the critical limit of 110 atm. If the calculated pressure exceeds 110 atm, the cylinder will explode.
04

Interpret the Results

If the pressure calculated in step 3 is less than or equal to 110 atm, the cylinder wouldn't explode. However, if the pressure exceeds 110 atm, the cylinder would fail and potentially explode. This step is crucial in understanding the practical implications of the results.

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 91影视!

Key Concepts

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

Gas Pressure
Gas pressure is an important concept when dealing with gases. It's defined as the force that gas molecules exert when they collide with the walls of their container.
When you have a gas enclosed in a cylinder, like in the exercise, this pressure tells us about the collisions of thousands of gas molecules bombarding the cylinder walls every second.
In simple terms, more vigorous collisions mean higher pressure.
These collisions are influenced by several factors:
  • Temperature: As the temperature increases, the molecules move faster, hitting the walls more forcefully.
  • Volume: Decreasing the volume traps the same number of molecules in a smaller space, making collisions more frequent.
  • Amount of Gas: Adding more gas increases the number of collisions.
Understanding these principles can help you grasp why the pressure in a cylinder changes with temperature in our example.
Temperature Conversion
Temperature conversion is essential when dealing with gas laws, which use absolute temperature.
Often, calculations start in Celsius or Fahrenheit, but converting to Kelvin is crucial for accuracy.
Celsius, commonly used in everyday contexts, needs a simple adjustment to convert to Kelvin:
  • Add 273.15 to the Celsius value to convert it to Kelvin.
  • Kelvin = Celsius + 273.15. This gives you the absolute temperature needed for gas law equations.
Why can't we just use Celsius?
  • Celsius can have negative values, which aren't suitable for gas law calculations where absolute temperatures are required.
  • Converting to Kelvin ensures you're comparing like with like in the gas equations.
This straightforward conversion helps avoid potential mistakes in understanding gas behavior.
Celsius to Kelvin
The Celsius to Kelvin conversion is a fundamental step in solving gas-related problems.
In the given exercise, to accurately apply Gay-Lussac鈥檚 Law, it's vital to convert both the initial and final temperatures to Kelvin.
Here's how the conversion works:
  • Take the temperature in Celsius.
  • Add 273.15 to the Celsius number to find the equivalent in Kelvin.
For instance, in the exercise,
  • Start with 20掳C, add 273.15, resulting in 293.15 K.
  • For 950掳C, add 273.15 to get 1223.15 K.
This ensures accurate calculations in gas laws by maintaining the correct scale, allowing the precise calculation of changes in pressure with temperature changes.
Critical Pressure Threshold
The critical pressure threshold in any problem involving gas is the maximum pressure that a container can safely hold before it fails. In our exercise, this threshold is set at 110 atm.
Understanding the significance of this threshold is crucial for safety and proper decision-making:
  • The threshold level helps determine if increased temperature or gas additions can be safely sustained.
  • It acts as a crucial parameter when testing cylinder integrity.
Using Gay-Lussac's Law, we assess whether the pressure of 110 atm is exceeded:
  • If calculated pressure is below the threshold, the cylinder remains safe.
  • If it exceeds this limit, the risk of explosion becomes significant.
Such an approach emphasizes the practical importance of understanding pressure limits in real-world 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

0.10 \(\mathrm{mol}\) of argon gas is admitted to an evacuated \(50 \mathrm{cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes heating at constant volume to a temperature of \(300^{\circ} \mathrm{C}\). a. What is the final pressure of the gas? b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.

A 5.0-m-diameter garden pond holds \(5.9 \times 10^{3} \mathrm{kg}\) of water. \(400 \mathrm{W} / \mathrm{m}^{2}\) of solar energy strikes the pond's surface, meaning each square meter of the pond's surface is absorbing \(400 \mathrm{W}\). How many hours will it take for the pond to warm from \(15^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C} ?\)

When you stifle a sneeze, you can damage delicate tissues because the pressure of the air that is not allowed to escape may rise by up to \(45 \mathrm{kPa}\). If this extra pressure acts on the inside of your \(8.4-\mathrm{mm}\) -diameter eardrum, what is the outward force?

For a normal car riding on tires with relatively flexible sidewalls, the weight of the car is held up, in large measure, by the pressure of the air in the tires. If you look at one of your car's tires, you'll note that the tire is flattened slightly to make a rectangle where it touches the ground. The area of the resulting "contact patch" depends on the pressure in the tires. To a good approximation, the upward normal force of the ground (which we can assume is equal to \(1 / 4\) of the car's weight) on this patch of the tire is equal to the downward pressure force on the patch. a. Suppose you inflate your \(2000 \mathrm{kg}\) car's tires to the recommended pressure, as measured by a gauge. The resulting contact patch is \(18 \mathrm{cm}\) wide and \(12 \mathrm{cm}\) long. What does the gauge read? b. If you let a bit of air out of your tire, what happens to the area of the contact patch?

A 30 kg male emperor penguin under a clear sky in the Antarctic winter loses very little heat to the environment by convection; its feathers provide very good insulation. It does lose some heat through its feet to the ice, and some heat due to evaporation as it breathes; the combined power is about \(12 \mathrm{W}\). The outside of the penguin's body is a chilly \(-22^{\circ} \mathrm{C},\) but its surroundings are an even chillier \(-38^{\circ} \mathrm{C} .\) The penguin's surface area is \(0.56 \mathrm{m}^{2},\) and its emissivity is 0.97. a. What is the net rate of energy loss by radiation? b. If the penguin has a 45 W basal metabolic rate, will it feel warm or cold under these circumstances?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Fields in Physics

Read Explanation

Linear Momentum

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Space Physics

Read Explanation

Astrophysics

Read Explanation
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.