Internal energy of an ideal gas as a function of pressure?

In summary, after a calorimetry experiment, the change in internal energy (ΔU) can be related to the change in enthalpy (ΔH) at a 1:1 ratio. However, to calculate the "exact" ΔU, a correction factor must be introduced to account for the increase in sensible heat of the products. This correction factor takes into account the change in temperature and pressure, with pressure assumed to be a function of moles of gas. The calculation for the "exact" ΔU involves correcting for the increase in temperature of the products, the change in volume due to the increase in moles of gas, and the heat capacity of the calorimeter.
  • #1
Jacob Dale
Assuming all gases in the combustion reaction of benzoic acid (C6H5COOH) behave ideally, what is the "exact" change in internal energy?

The context in which this question is being asked is after a calorimetry experiment. For all the intents and purposes of calorimetry, the change in internal energy (ΔU) can be related to the change in enthalpy (ΔH) at a 1:1 ratio because the volume of the calorimeter remains constant.

dU = dq + dw simplifies to dU = dq

However, this question asks for the "exact" ΔU. Therefore, a correction factor needs to be introduced to account for the variable thermodynamic values.

The only variables at play here are temperature and pressure. The temperature changed around 2 degrees Celsius for each trial of the combustion of benzoic acid so it's contribution to U is negligible. Pressure, however, changes drastically. Pressure can be assumed to be function of moles of gas:

2 C6H5COOH + 15 O2 => 6 H2O + 14 CO2

There is 3:4 ratio for moles of gas. So the pressure of the system increases by a factor of 4/3.

But now I am stuck. I don't know how to calculate the "exact" ΔU. Does anyone know what my lab instructor is asking for in this situation?

Thank you for your help!
 
Physics news on Phys.org
  • #2
It's somewhat unclear what is being asked. Since ## dH=dU+PdV ##, at constant volume, ## \Delta H=\Delta U ##. The ## \Delta H ## for such a reaction is normally found in a chemistry handbook. Perhaps @Chestermiller can provide an input. Is the reaction being carried out at constant volume? Or is the pressure constant so that the ## P \Delta V ## term can be computed, and ## \Delta H=\Delta U+P \Delta V ##? It would appear they may want you to do the latter. If you have ## \Delta H ##, (usually how it is found in the literature), you can compute ## \Delta U ##. ## \\ ## Editing note: This contains a couple of errors. My thermodynamics is a little rusty. See posts by @Chestermiller below.
 
Last edited:
  • #3
Jacob Dale said:
Assuming all gases in the combustion reaction of benzoic acid (C6H5COOH) behave ideally, what is the "exact" change in internal energy?

The context in which this question is being asked is after a calorimetry experiment. For all the intents and purposes of calorimetry, the change in internal energy (ΔU) can be related to the change in enthalpy (ΔH) at a 1:1 ratio because the volume of the calorimeter remains constant.

dU = dq + dw simplifies to dU = dq

However, this question asks for the "exact" ΔU. Therefore, a correction factor needs to be introduced to account for the variable thermodynamic values.

The only variables at play here are temperature and pressure. The temperature changed around 2 degrees Celsius for each trial of the combustion of benzoic acid so it's contribution to U is negligible. Pressure, however, changes drastically. Pressure can be assumed to be function of moles of gas:

2 C6H5COOH + 15 O2 => 6 H2O + 14 CO2

There is 3:4 ratio for moles of gas. So the pressure of the system increases by a factor of 4/3.

But now I am stuck. I don't know how to calculate the "exact" ΔU. Does anyone know what my lab instructor is asking for in this situation?

Thank you for your help!
The change in internal energy for a reaction assumes that the temperature and volume of the products is the same as that of the reactants. But, in this calorimeter experiment, the temperature of the products as 2 C higher than that of the reactants. So, a small correction is needed to account for the increase is sensible heat of the products. This will give the "exact" ##\Delta U## (within the framework of ideal gas behavior).
 
  • #4
Charles Link said:
It's somewhat unclear what is being asked. Since ## dH=dU+PdV ##, at constant volume, ## \Delta H=\Delta U ##. The ## \Delta H ## for such a reaction is normally found in a chemistry handbook.
This is not correct. ##\Delta H=\Delta U+\Delta(PV)##. At constant volume, ##\Delta H=\Delta U+V\Delta P##. In the case of this reaction, the number of moles of gas has changed. Therefore, ##\Delta (PV)=\Delta n RT##, where ##\Delta n## is the change in the number of moles of gas (this neglects the initial volume of the benzoic acid). So, if one wanted to know the "heat of combustion" of the benzoic acid, one would use ##\Delta H=\Delta U+\Delta nRT##, after ##\Delta U## has been corrected for the increase in T, and where now T is the initial temperature of the reactants.
 
  • Like
Likes Charles Link
  • #5
My thermodynamics is very rusty. I haven't done any calculations like this for quite a while. Thank you @Chestermiller for the corrections. ## \\ ## One question I have that seems to be missing from this is what is the heat capacity of the calorimeter that absorbs all of this energy so that the rise in temperature is only 2 degrees Centigrade? If I have this correct, this heat in the calorimeter, along with the ## V \Delta P ##, and along with the small correction for the heat capacity of the ## CO_2 ## as its temperature rose by 2 degrees, will give you the ## \Delta H ## for this reaction. Is there very little heat given off during this process? Is there no heat sink in this particular calorimeter?
 
  • #6
Charles Link said:
My thermodynamics is very rusty. I haven't done any calculations like this for quite a while. Thank you @Chestermiller for the corrections. ## \\ ## One question I have that seems to be missing from this is what is the heat capacity of the calorimeter that absorbs all of this energy so that the rise in temperature is only 2 degrees Centigrade? If I have this correct, this heat in the calorimeter, along with the ## V \Delta P ##, and along with the small correction for the heat capacity of the ## CO_2 ## as its temperature rose by 2 degrees, will give you the ## \Delta H ## for this reaction. Is there very little heat given off during this process? Is there no heat sink in this particular calorimeter?
Typically, the mass times heat capacity of the calorimeter is much higher than the mass times heat capacity of the gas. So the calorimeter is the sink for most of the heat of reaction.

The sensible heat of the products must also include the sensible heat of the water.
 
  • Like
Likes Charles Link

Related to Internal energy of an ideal gas as a function of pressure?

1. What is the internal energy of an ideal gas?

The internal energy of an ideal gas refers to the total energy within the system that is a result of the motion and interactions of its individual particles.

2. How is the internal energy of an ideal gas related to pressure?

The internal energy of an ideal gas is directly proportional to its pressure. This means that as the pressure of the gas increases, so does its internal energy.

3. Can the internal energy of an ideal gas change without a change in pressure?

Yes, the internal energy of an ideal gas can change without a change in pressure. This can happen through changes in temperature, volume, or the addition/removal of heat.

4. What is the equation for calculating the internal energy of an ideal gas as a function of pressure?

The equation is: U = nRT, where U is the internal energy, n is the number of moles of gas, R is the universal gas constant, and T is the temperature of the gas in Kelvin.

5. How does the internal energy of an ideal gas change with temperature?

The internal energy of an ideal gas is directly proportional to its temperature. This means that as the temperature of the gas increases, so does its internal energy.

Similar threads

Replies
5
Views
436
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
923
  • Introductory Physics Homework Help
Replies
12
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Replies
1
Views
991
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top