# Elastic Balloon Filled with Ideal Gas

• lluke9
In summary, the conversation discusses the equation PV = nRT and its application to an ideal gas inside an elastic balloon. The question is raised about whether the volume of the balloon would also double if the temperature is increased, taking into account the elasticity and tightness of the balloon. It is mentioned that for a balloon that stretches ideally according to Hooke's law, the volume and pressure would both increase by a factor of √2. However, without more information about the properties of the container, it is not possible to determine how the other variables (P, N, T) would change if one is doubled.
lluke9
This is the equation I'm talking about:
PV = nRT

So say I had a bit of Ideal Gas in a completely insulated elastic balloon, and I raised a temperature just a little bit.
PV = nR(2T)

Because this is an elastic balloon, would the volume also double?
P(2V) = nR(2T)

But what if the balloon is really tight? Would the pressure increase because the volume couldn't expand as much? But to what extent? How much would the balloon expand?
(?P)(?V) = nR(2T)
There has to be a sort of limit, depending on the stretchiness of the balloon, right?

For a balloon that stretches ideally according to Hooke's law, I believe it resists with a constant pressure, meaning you would see the volume doubling.

That is, because the surface increases with the square of the radius, the stretching force also increases with the square of the radius.
But the resulting pressure, which is the quotient of the two, remains constant.

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Say the volume of the balloon increased by √2, then the pressure inside of the balloon would have to increase by √2.

Look at it this way because the balloon is a closed system (xP)(yV) = nR(zT)

x · y = z

if you double z, then x and y have to change so that their product also doubles.

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You have one equation, PV = nRT - or, the version I think with, P = NT - and three variables: P, V or N, and T. Specifying a change to one does not specify how the other two change, you need another equation. Something about the properties of the container (Hooke's law spring), something about heat flow, whatever - but without more information you cannot say what happens to V if you double T.

I can provide an explanation for the behavior of an elastic balloon filled with an ideal gas using the ideal gas law, PV = nRT. This equation relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of an ideal gas.

When the temperature of the gas inside the balloon is increased, the volume will also increase according to the equation PV = nR(2T). This is because as the temperature increases, the gas molecules will have more kinetic energy and will collide more frequently with the walls of the balloon, causing it to expand.

However, the extent to which the volume will increase depends on the elasticity of the balloon. If the balloon is very tight, it will not be able to expand as much, and therefore the pressure will increase to compensate for the smaller volume. This can be represented in the equation as P(2V) = nR(2T), where the pressure has to double to maintain the same number of moles and temperature while the volume is halved.

There is indeed a limit to how much the balloon can expand, and it depends on the stretchiness of the material. If the balloon is made of a very elastic material, it can expand significantly without a large increase in pressure. However, if the material is not very stretchy, the pressure will increase more quickly as the volume expands.

In conclusion, the behavior of an elastic balloon filled with an ideal gas can be explained using the ideal gas law. The volume will increase with temperature, but the extent of the increase will depend on the elasticity of the balloon. The pressure will also increase to compensate for a smaller volume, and this increase will be greater for less stretchy balloons.

## 1. What is an elastic balloon filled with ideal gas?

An elastic balloon filled with ideal gas is a scientific model used to study the behavior of gases under different conditions. The balloon represents a closed system, and the ideal gas is a theoretical gas that follows the ideal gas law, which describes the relationship between pressure, volume, temperature, and amount of gas.

## 2. How does an elastic balloon filled with ideal gas behave?

The behavior of an elastic balloon filled with ideal gas depends on the conditions of the gas inside. If the temperature or amount of gas increases, the volume of the balloon will also increase. Similarly, if the temperature or amount of gas decreases, the volume of the balloon will decrease. This is because the ideal gas law states that the volume of a gas is directly proportional to its temperature and amount, and inversely proportional to its pressure.

## 3. What factors affect the behavior of an elastic balloon filled with ideal gas?

The behavior of an elastic balloon filled with ideal gas is affected by several factors, including temperature, pressure, volume, and amount of gas. Changes in any of these factors can cause the balloon to expand or contract.

## 4. What is the purpose of using an elastic balloon filled with ideal gas in experiments?

An elastic balloon filled with ideal gas is often used in experiments to demonstrate the principles of the ideal gas law and to study the behavior of gases in different conditions. It allows scientists to observe the effects of changing one variable on the other variables, and to make predictions about the behavior of real gases.

## 5. How is an elastic balloon filled with ideal gas different from a real gas?

An elastic balloon filled with ideal gas is a simplified model that assumes the gas particles have no volume and do not interact with each other. In reality, real gases do have volume and interact with each other, which can affect their behavior. However, the ideal gas model is still useful for understanding the general behavior of gases under different conditions.

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