## Ideal Gas Law and Isobaric Processes

Hi,

So let's take the standard example of a gas in a container with a piston at the top. Charles' Law states that at constant pressure, an increase in temperature (kinetic energy of gas molecules) will increase the volume. This makes sense both conceptually and mathematically (per PV = nRT). However, in an isobaric process (pressure is constant again), the kinetic energy of the gas molecules is what is moving the piston so it must have lost some energy in doing so. Therefore, although initially Esystem = q that was added, it's net energy change would be Esystem = q - Pext*V.

So this is my problem, wouldn't this isobaric process be inconsistent with the ideal gas law? That is, using strictly the ideal gas law, woudn't the ending temperature not take into account the work done on the piston? Or are we assuming in using the ideal gas law that the work to keep the pressure constant is from an external source?

Thanks.

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Recognitions:
Gold Member
 Quote by nothing123 Hi, So let's take the standard example of a gas in a container with a piston at the top. Charles' Law states that at constant pressure, an increase in temperature (kinetic energy of gas molecules) will increase the volume. This makes sense both conceptually and mathematically (per PV = nRT). However, in an isobaric process (pressure is constant again), the kinetic energy of the gas molecules is what is moving the piston so it must have lost some energy in doing so. Therefore, although initially Esystem = q that was added, it's net energy change would be Esystem = q - Pext*V. So this is my problem, wouldn't this isobaric process be inconsistent with the ideal gas law? That is, using strictly the ideal gas law, woudn't the ending temperature not take into account the work done on the piston? Or are we assuming in using the ideal gas law that the work to keep the pressure constant is from an external source? Thanks.
You're mixing two different concepts. The ideal gas equation of state shows the relationship between P, V and T at a specified state. An isobaric process (or any process) shows how a substance changed from an initial state to a final state.

Does that help?

CS

 Could you clarify exactly what you mean by states? I mean, since the heat added only changes the kinetic energy of the system (which is proportional to temperature), wouldn't we be able to find the same change in temperature whether we used Echange = q + w or whether we used T = PV/nR? Thanks for your help so far.

Recognitions:
Gold Member

## Ideal Gas Law and Isobaric Processes

 Quote by nothing123 Could you clarify exactly what you mean by states? I mean, since the heat added only changes the kinetic energy of the system (which is proportional to temperature), wouldn't we be able to find the same change in temperature whether we used Echange = q + w or whether we used T = PV/nR? Thanks for your help so far.
By state I mean a set of properties that completely describe the condition of the system. If the system is not changing, it is in equilibrium. If the system undergoes a process, something has changed the system and it is now in an alternate state. The ideal gas law describes the state of an ideal gas. If the ideal gas undergoes a process, the process path from state 1 to state 2 will describe how the system changed. Once at state 2 the ideal gas relation can describe the system at that state.

Also, the properties of an ideal gas at two different states for a fixed mass are related by the ideal gas law as well and is called the combined gas law IIRC. However, you would need to know some of the properties at both states in order to solve for the unknown. Hence, the process path must be known.

BTW,

$$\Delta U = Q_{net,in} - W_{net,out}$$

Shows that for a closed system the internal energy of a substance decreases if it does work on it's surrounds. Hence it's temperature would decrease. So for an isobaric process like you described, the change in internal energy and thus temperature would depend on how much heat was added, and how much work was done by the piston on its surroundings.

This is typically stated using the enthalpy for simplicity:

$$Q - W_{other} = H_2 - H_1$$

Does that help?

CS

 Thank you very much. Clearing up that state definition really helped.

Recognitions:
Homework Help