Isobaric process; change in internal energy

  • Thread starter RJWills
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  • #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:
 

Answers and Replies

  • #2
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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:

Start out by using the ideal gas law to get ΔT. Incidentally, check your multiplication to get ΔW.
 
  • #3
RJWills
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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
Andrew Mason
Science Advisor
Homework Helper
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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
Andrew Mason
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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
22,232
5,134
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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