Isobaric process; change in internal energy

In summary, the problem is to find the work done, heat added, and change in internal energy for a reversible expansion of an ideal monatomic gas at a constant pressure of 10^5 Pa from 1m^3 to 3m^3. The equations used are the first law of thermodynamics, ΔU=ΔQ+ΔW, ΔW=-pΔV, and Q=3/2nRT. The initial attempt to solve the problem involved finding ΔW and then finding ΔQ using the ideal gas law, but it is easier to use the fact that at constant pressure, ΔQ=ΔH=CpΔT. Finally, the solution involves calculating the change in internal
  • #1
RJWills
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I know this question has been done to death and I have had a look through past questions on this forum and others but I can't find anything that is helpful to this particular question!

Homework Statement



An ideal monatomic gas expands reversibly at a fixed pressure of 10^5 Pa from a volume of 1m^3 to 3m^3. Find the work done on the gas, the heat added to the gas and the change in internal energy


Homework Equations


ΔU=ΔQ+ΔW
ΔW=-pΔV
Q=3/2nRT (because it's monatomic)

The Attempt at a Solution



ΔW=-pΔV = -20^5J

How do I find ΔQ? I don't have the internal energy, I don't have the temperature so can't use Q=3/2nRT so how do I do it? I really don't get how I can find this without using Calorimetry which isn't part of our syllabus. I'm really frustrated and don't know what to do next. As soon as I have Q, U will simply pop out :frown:
 
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  • #2
RJWills said:
I know this question has been done to death and I have had a look through past questions on this forum and others but I can't find anything that is helpful to this particular question!

Homework Statement



An ideal monatomic gas expands reversibly at a fixed pressure of 10^5 Pa from a volume of 1m^3 to 3m^3. Find the work done on the gas, the heat added to the gas and the change in internal energy


Homework Equations


ΔU=ΔQ+ΔW
ΔW=-pΔV
Q=3/2nRT (because it's monatomic)

The Attempt at a Solution



ΔW=-pΔV = -20^5J

How do I find ΔQ? I don't have the internal energy, I don't have the temperature so can't use Q=3/2nRT so how do I do it? I really don't get how I can find this without using Calorimetry which isn't part of our syllabus. I'm really frustrated and don't know what to do next. As soon as I have Q, U will simply pop out :frown:

Start out by using the ideal gas law to get ΔT. Incidentally, check your multiplication to get ΔW.
 
  • #3
Chestermiller said:
Start out by using the ideal gas law to get ΔT. Incidentally, check your multiplication to get ΔW.

Okay so pV=nRΔT but this is only useful if I assume a value for n such as 1 right? And should I use the final volume after expansion?

With my multiplication should it be +20^5J because the gas is expanding therefore there is positive work being done on the surroundings?
 
  • #4
RJWills said:
I

Homework Equations


ΔU=ΔQ+ΔW
ΔW=-pΔV
Q=3/2nRT (because it's monatomic)

The Attempt at a Solution



ΔW=-pΔV = -20^5J

How do I find ΔQ?
You can use the first law and determine the change in temperature to find the change in internal energy and add the work done, but there is an easier way, since P is constant.

AM
 
  • #5
RJWills said:
With my multiplication should it be +20^5J because the gas is expanding therefore there is positive work being done on the surroundings?
The sign is right since the question asks for the work done ON the gas. But your arithmetic is wrong: ΔV = 2 m^3; P = 10^5 Pa.AM
 
  • #6
RJWills said:
Okay so pV=nRΔT but this is only useful if I assume a value for n such as 1 right? And should I use the final volume after expansion?

With my multiplication should it be +20^5J because the gas is expanding therefore there is positive work being done on the surroundings?

It should be pΔV=nRΔT, since p is constant and V is changing.

In this problem, you don't need to know n and ΔT separately. All you need to know is the product of n and ΔT in order to determine the change in internal energy. I suggest you do this problem algebraically before substituting in the numbers. Once you know the change in internal energy and the work, you can determine the Q.
As Andrew Mason alluded to, you don't need to first calculate the change in internal energy to get the amount of heat if you remember that, at constant pressure, the heat added is equal to the change in enthalpy, that the change in enthaply is equal to the heat capacity at constant pressure times the temperature change, and that the heat capacity at constant pressure is equal to the heat capacity at constant volume plus the gas constant R.

Chet
 
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1. What is an isobaric process?

An isobaric process is a thermodynamic process in which the pressure of a system remains constant while the volume and temperature may change. This means that the work done on or by the system is equal to the product of the constant pressure and the change in volume.

2. How is the internal energy affected in an isobaric process?

In an isobaric process, the internal energy of a system may change due to the transfer of heat. If heat is added to the system, the internal energy will increase, and if heat is removed, the internal energy will decrease. The change in internal energy is equal to the heat added or removed from the system minus the work done by the system.

3. What is the equation for calculating the change in internal energy in an isobaric process?

The equation for calculating the change in internal energy in an isobaric process is ΔU = Q - PΔV, where ΔU is the change in internal energy, Q is the heat added or removed from the system, and PΔV is the work done by the system.

4. How does an isobaric process differ from an isochoric process?

An isochoric process is a thermodynamic process in which the volume of a system remains constant while the pressure and temperature may change. In contrast, an isobaric process is a thermodynamic process in which the pressure remains constant while the volume and temperature may change.

5. What are some real-life examples of an isobaric process?

An example of an isobaric process is the heating of water in an open container. The pressure remains constant as the water is heated, but the volume and temperature will change. Another example is a gas-filled balloon at a constant pressure being heated, causing the volume of the balloon to increase, but the pressure remaining constant.

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