/*! 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 24 Oxygen is filled in a closed met... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 24

Oxygen is filled in a closed metal jar of volume \(1 \cdot 0 \times 10^{-3} \mathrm{~m}^{3}\) at a pressure of \(1.5 \times 10^{5} \mathrm{~Pa}\) and temperature \(400 \mathrm{~K}\). The jar has a small leak in it. The atmospheric pressure is \(1.0 \times 10{ }^{5} \mathrm{~Pa}\) and the atmospheric temperature is \(300 \mathrm{~K}\). Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

Short Answer

Expert verified
0.16 grams of oxygen leaks out.

Step by step solution

01

Understand Initial Conditions

We start with a closed metal jar of volume \( V = 1.0 \times 10^{-3} \, \text{m}^3 \) containing oxygen at an initial pressure of \( P_1 = 1.5 \times 10^5 \, \text{Pa} \) and temperature \( T_1 = 400 \, \text{K} \). We need to determine the initial number of moles using the ideal gas law.
02

Calculate Initial Moles of Oxygen

Using the ideal gas equation \( PV = nRT \), where \( R = 8.314 \, \text{J/mol} \cdot \text{K} \), calculate the initial moles, \( n_1 \).\[ n_1 = \frac{P_1 V}{RT_1} = \frac{1.5 \times 10^5 \times 1.0 \times 10^{-3}}{8.314 \times 400} \approx 0.045 \text{moles}.\]
03

Understand Final Conditions

After the leak, the pressure and temperature inside the jar equalize with the surroundings. The final pressure \( P_2 = 1.0 \times 10^5 \, \text{Pa} \) and final temperature \( T_2 = 300 \, \text{K} \).
04

Calculate Final Moles of Oxygen

Using the ideal gas equation again, calculate the final moles, \( n_2 \), of the oxygen left in the jar.\[n_2 = \frac{P_2 V}{RT_2} = \frac{1.0 \times 10^5 \times 1.0 \times 10^{-3}}{8.314 \times 300} \approx 0.040 \text{moles}.\]
05

Determine Moles of Oxygen Leaked

The moles of oxygen that leaked out, \( \Delta n \), is the difference between the initial and final moles.\[\Delta n = n_1 - n_2 = 0.045 - 0.040 = 0.005 \text{moles}.\]
06

Calculate Mass of Leaked Oxygen

Using the molar mass of oxygen \( (32 \text{g/mol}) \), the mass \( \Delta m \) is\[\Delta m = \Delta n \times \text{molar mass} = 0.005 \times 32 = 0.16 \text{grams}.\]
07

Conclusion

The mass of oxygen that leaks out of the jar by the time the pressure and temperature inside equalise with the surrounding is 0.16 grams.

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.

Moles Calculation
In scenarios involving gases, accurately determining the number of moles helps us understand how much substance we are dealing with. When working with an ideal gas, we use the Ideal Gas Law, expressed as:
  • Equation: \(PV = nRT\)
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant (8.314 J/mol路K), and
  • \(T\) is the temperature in Kelvin.
When oxygen was initially placed in the jar, we calculated the moles using the above equation. Having a pressure of \(1.5 \times 10^5 \text{ Pa}\) and a temperature of \(400 \text{ K}\), we found the initial moles to be approximately \(0.045 \text{ moles}\). This precise moles calculation sets the foundation for further analysis in gas-related problems.
Pressure Equalization
Pressure equalization is a significant concept in gas dynamics, especially when dealing with leaks or changes in containers. In our scenario, the pressure in the jar eventually balances with the outside environment. How does it happen?
  • The initial pressure was higher inside the jar than outside (1.5 vs 1.0 x 105 Pa).
  • As the gas leaks out, the internal pressure decreases.
  • Equilibrium is achieved when the internal pressure matches the atmospheric pressure.
The equilibrium concept helps predict how changes will occur in closed systems and plays a key role in calculating the leaked amount as the system stabilizes.
Temperature Effect
Temperature significantly influences the behavior of gases. In our scenario, the temperature of the oxygen gas in the jar initially was 400 K, higher than the atmospheric temperature of 300 K. Temperature change impacts:
  • Energy: Higher temperatures provide more kinetic energy to gas molecules, often increasing pressure.
  • Volume and Moles Calculation: Using the Ideal Gas Law, a decrease in temperature upon equilibrium reduces the moles calculated at equilibrium.
When the temperature equalizes with its surroundings, calculations of final conditions (post-leak) use this lower temperature. This results in fewer moles remaining inside the jar, showing how temperature can deeply impact gas calculations.鈥�
Leakage in Gases
Leakage is an essential topic when discussing gases in containers. It naturally occurs when there is a pressure difference. A higher pressure inside a container will push gas out towards an area of lower pressure, like the atmosphere.
  • Mechanism: When the pressure inside is greater than outside, gases escape through even small openings.
  • Calculation Impact: Leaks need to be accounted for in order to calculate the net moles remaining, as detailed in earlier steps.
  • Mass Loss: Using moles, we can calculate the mass of gas that escapes (0.16 grams in our case) based on molar mass.
Understanding leakage helps with real-life situations like maintaining gas levels in containers or systems malfunctioning due to unexpected pressure changes.

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

\(50 \mathrm{cc}\) of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is \(99 \cdot 4 \mathrm{kPa}\) and the room temperature is \(27^{\circ} \mathrm{C}\). The water level in the jar is same as the level outside. The saturation vapour pressure at \(27^{\circ} \mathrm{C}\) is \(3.4 \mathrm{kPa}\). Calculate the number of moles of oxygen collected in the jar.

A vertical cylinder of height \(100 \mathrm{~cm}\) contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to \(75 \mathrm{~cm}\) of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

The average translational kinetic energy of air molecules is \(0.040 \mathrm{eV} \quad\left(1 \mathrm{eV}=1 \cdot 6 \times 10^{-19} \mathrm{~J}\right) . \quad\) Calculate the temperature of the air. Boltzmann constant \(k=1 \cdot 38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\)

An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals \(76 \mathrm{~cm}\) of mercury. The lengths of the mercury column and the trapped air column are \(20 \mathrm{~cm}\) and \(43 \mathrm{~cm}\) respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of \(60^{\circ}\) ? Assume the temperature to remain constant.

Air is pumped into an automobile tyre's tube up to a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(20^{\circ} \mathrm{C}\). During the day the temperature rises to \(40^{\circ} \mathrm{C}\) and the tube expands by \(2 \%\). Calculate the pressure of the air in the tube at this temperature.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Thermodynamics

Read Explanation

Kinematics Physics

Read Explanation

Work Energy and Power

Read Explanation

Quantum Physics

Read Explanation

Oscillations

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.