Isobaric process; change in internal energy

Click For Summary

Homework Help Overview

The discussion revolves around an ideal monatomic gas undergoing an isobaric expansion at a fixed pressure of 10^5 Pa, transitioning from a volume of 1m^3 to 3m^3. Participants are tasked with finding the work done on the gas, the heat added, and the change in internal energy.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the use of the ideal gas law to determine temperature changes and question how to calculate heat without knowing internal energy or temperature. There is discussion about the signs of work done and the arithmetic involved in calculating work.

Discussion Status

Some participants have provided guidance on using the ideal gas law and the relationship between heat, work, and internal energy. There is an ongoing exploration of assumptions regarding the number of moles and the implications of constant pressure on heat calculations.

Contextual Notes

Participants express frustration over the lack of certain information, such as temperature and the number of moles, which complicates their ability to apply relevant equations. There is also mention of syllabus constraints regarding calorimetry.

RJWills
Messages
16
Reaction score
0
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:
 
Physics news on Phys.org
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.
 
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?
 
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
 
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
 
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
 
  • Like
Likes   Reactions: gracy

Similar threads

Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K