/*! 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 26 Consider two different container... [FREE SOLUTION] | 91影视

91影视

Consider two different containers, each filled with 2 moles of \(\mathrm{Ne}(\mathrm{g}) .\) One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

Short Answer

Expert verified
When the temperature increases, in the rigid container, the pressure will increase and the density will remain constant. In the flexible container, the pressure will remain constant and the density will decrease as the temperature increases.

Step by step solution

01

Analyze the rigid container

In this case, the volume of the container remains constant as the temperature increases. We can rearrange the Ideal Gas Law to find the relationship between pressure and temperature: \(P = \frac{nRT}{V}\) Since both n and V are constants for this container, we can see that the pressure P is directly proportional to the temperature T.
02

Describe the pressure change in the rigid container

As the temperature of the gas inside the rigid container increases, the pressure will also increase because they are directly proportional, as shown in Step 1.
03

Calculate the density change in the rigid container

Since the volume of the rigid container remains constant even when the temperature increases, the density of the gas will remain constant: 蟻 = (mass of gas) / (constant volume)
04

Analyze the flexible container

In this case, the volume of the container can change to maintain a constant pressure equal to the external pressure. We can rearrange the Ideal Gas Law to find the relationship between volume and temperature: \(V = \frac{nRT}{P}\) Since both n and P are constants for this container, we can see that the volume V is directly proportional to the temperature T.
05

Describe the pressure change in the flexible container

In this flexible container, as the temperature increases, the volume of the container will adjust to keep the pressure constant. Therefore, there is no pressure change in the flexible container.
06

Calculate the density change in the flexible container

Since the volume of the flexible container changes as the temperature increases, we can find the relationship between density and temperature: 蟻 = (mass of gas) / (volume) 蟻 = (mass of gas) / (\( \frac{nRT}{P} \)) As the temperature increases, the mass of the gas remains constant while the volume increases (because they are directly proportional). Therefore, the density of the gas in the flexible container will decrease as the temperature increases. #Summary# - In the rigid container, as the temperature increases, the pressure will increase, and the density will remain constant. - In the flexible container, as the temperature increases, the pressure will remain constant, and the density will decrease.

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 and Temperature Relationship
The interconnection between gas pressure and temperature is fundamental in understanding the behavior of gases and is elegantly described by the Ideal Gas Law. Think of the atomic or molecular hustle within a gas; as the temperature goes up, the particles move more vigorously, bumping against the container's walls more frequently and with greater force. This increase in collision rate and strength leads to a rise in pressure when the volume doesn't have room to expand鈥攚hich is what we observe in a rigid container.

For instance, in a sealed, non-expandable container in our exercise, raising the temperature will inevitably lead to an increase in pressure. This correlation is a direct one, mathematically speaking: if you double the temperature (measured in Kelvin), you also double the pressure, assuming the amount of gas (moles) and the volume remain unchanged. Consequently, if you're plotting a graph of pressure against temperature for a fixed volume of gas, you'd get a straight line shooting up, reflecting this direct proportionality.
Constant Volume Gas Behavior
When we're dealing with a gas at constant volume鈥攍ike in the rigid container of our exercise鈥攖he Ideal Gas Law simplifies the scene. The density of a gas is defined as its mass per unit volume. Since the volume is steadfast, and the mass of gas remains unchanged, the density stays the same, no matter how hot or cold the gas gets.

However, according to the Ideal Gas Law presented as
P = \(\frac{nRT}{V}\), where P stands for pressure, n is the number of moles, R is the universal gas constant, T is the temperature, and V is the volume - we observe that pressure goes hand in hand with temperature under these circumstances. So, imagine turning up the heat under a fixed-size balloon filled with helium鈥攖he pressure inside the balloon increases, but since the balloon can't expand, it feels tighter; however, its density stays put.
Flexible Container Gas Behavior
Switching over from rigid to flexible containers, we're adding a twist to the gas tale. A flexible container, much like a balloon, is free to change its shape and size. What this means in terms of gas laws, is that if you increase the temperature, the volume can increase to maintain the pressure constant if the exterior pressure is unchanging鈥攁n important condition in our exercise scenario.

The equation describing this behavior takes the form
V = \(\frac{nRT}{P}\) - indicating the volume is proportional to temperature (when the number of moles and the pressure are constants). As the gas heats up and expands, the density, which is inversely related to volume, decreases. Therefore, as you fill a balloon with warm air, it doesn't just expand because of the air's gentleness; it's the gas density going down while maintaining the same pressure inside that grants you a bigger, more floaty balloon.

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

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and a pressure of \(2.00 \mathrm{~atm}\). The products of the reaction are \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

An \(11.2-\mathrm{L}\) sample of gas is determined to contain \(0.50 \mathrm{~mol} \mathrm{~N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

Consider the following reaction: $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ It takes \(2.00 \mathrm{~L}\) pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Do all the molecules in a 1 -mol sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K}\) ? Do all molecules in a \(1-\mathrm{mol}\) sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K}\) ? Explain.

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.