Internal Energy of an Isovolumetric Process

[Astrogirl93]

1. We have some gas in a container at high pressure. The volume of the container is 469 cm^3. The pressure of the gas is 2.52*10^5 Pa. We allow the gas to expand at a constant temperature until its pressure is equal to the atmospheric pressure, which at the time is .857*10^5 Pa. (a) Find the work (J) done on the gas. (b) Find the change of internal energy (J) of the gas. (c) Find the amount of heat (J) we added to the gas to keep it at constant temperature. Be sure to include the right signs on the answers.



2. E=ΔQ+W
PV=nRT



3. I know that the work done on the system is equal to 0 because the volume of the gas does not change. This means that the change in internal energy is equal to the change in heat of the system. However, I don't think I can use PV=nRT because the system does not have a constant pressure. I am not sure how to find the change in internal energy.

 

[nasu]

3. I know that the work done on the system is equal to 0 because the volume of the gas does not change.

How do you know that the volume is fixed? Where does it say so in the problem?

 

[cepheid]

Welcome to PF,

1. We have some gas in a container at high pressure. The volume of the container is 469 cm^3. The pressure of the gas is 2.52*10^5 Pa. We allow the gas to expand at a constant temperature until its pressure is equal to the atmospheric pressure, which at the time is .857*10^5 Pa. (a) Find the work (J) done on the gas. (b) Find the change of internal energy (J) of the gas. (c) Find the amount of heat (J) we added to the gas to keep it at constant temperature. Be sure to include the right signs on the answers.



2. E=ΔQ+W
PV=nRT



3. I know that the work done on the system is equal to 0 because the volume of the gas does not change. This means that the change in internal energy is equal to the change in heat of the system. However, I don't think I can use PV=nRT because the system does not have a constant pressure. I am not sure how to find the change in internal energy.

This is not an "isovolumetric" process, as shown by the statement in boldface above. The fact that the gas is allowed to expand means that its volume increases. (I assume that it is released from the container, even though this is not stated explicitly). The fact that the temperature remains constant means that this is an isothermal process.

You can of course still use the ideal gas law. It is always applicable (to an ideal gas). If T = const, then what is the relationship between pressure and volume?

What does the internal energy of an ideal gas depend upon? There is an equation for this that you should be able to look up.

 

[Astrogirl93]

For an isothermal process ΔE=0 and Q=-W. If T is constant than P=nRT/V. Since the pressure and the volume are changing, would it be correct to use the equation P/V=P/V where P1 is the starting pressure, V1 is the unknown, P2 is equal to atmospheric pressure, and V2 is the volume of the container? (I don't think I can assume they are letting the air out of the container, but I might be able to assume that the ending volume is that of the container)

 

[cepheid]

For an isothermal process ΔE=0 and Q=-W.

Correct.

If T is constant than P=nRT/V. Since the pressure and the volume are changing, would it be correct to use the equation P/V=P/V where P1 is the starting pressure, V1 is the unknown, P2 is equal to atmospheric pressure, and V2 is the volume of the container?

To be honest, I'm not sure why you are asking this. You are correct that P = nRT/V = C/V where C is a constant. Therefore, it automatically follows that P₁/V₁ = P₂/V₂, since this ratio is constant (equal to C).

(I don't think I can assume they are letting the air out of the container, but I might be able to assume that the ending volume is that of the container)

Stop and actually think about it for a second. A gas always expands to fill its container. Which means that it is initially taking up the entire volume of the container. So, for this question to make sense, the gas cannot remain in its container, otherwise it would have nowhere to expand to.