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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.

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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.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{~L} / \mathrm{min}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0\) L/min. If \(5.30 \mathrm{~g}\) methanol is produced per minute, what is the percent yield of the reaction?

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00 \mathrm{~atm}\) and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. Be complete. b. Above what temperature would you heat the balloon so that it would float?

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{~g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to \(650 .\) psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

At STP, \(1.0 \mathrm{~L} \mathrm{Br}_{2}\) reacts completely with \(3.0 \mathrm{~L} \mathrm{~F}_{2}\), producing 2.0 L of a product. What is the formula of the product? (All substances are gases.)

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