How much work is done by the gas?

  • Thread starter Thread starter Tehy
  • Start date Start date
  • Tags Tags
    Gas Work
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the work done by a gas during an expansion process. The scenario includes 5 moles of gas at a temperature of 326K, with pressure changing from 1 atm to 3 atm. Participants are exploring how to approach the calculation of work done by the gas under these conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the ideal gas law and the conditions under which work is done, including isochoric conditions and the implications of changing external pressure. Questions arise about the need for additional information regarding how pressure changes and whether heat is added or released during the process.

Discussion Status

There is an ongoing exploration of different interpretations of the problem, with some participants suggesting calculations based on the ideal gas law and others questioning the assumptions made about the process. Guidance has been offered regarding the relationship between pressure, volume, and work, but no consensus has been reached.

Contextual Notes

Participants note that the problem lacks sufficient information to definitively calculate the work done, emphasizing the need to understand the nature of the pressure change and the thermal conditions of the gas.

Tehy
Messages
7
Reaction score
0
Hello! Any ideas how I could solve this exercise:

5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm. How much work is done by the gas?

Detailed instruction would be good, thanks :)
 
Physics news on Phys.org
W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.
 
sicjeff said:
W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.

Yes, it probably should, but I'm very dumb with physics :P I already tried to find some help from Google but I didn't have any luck :/
 
okay, I assume you are in like a freshman/sophmore level physics course right.

Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.
We are assuming that the size of our container is constant. (isochoric condition)
 
sicjeff said:
okay, I assume you are in like a freshman/sophmore level physics course right.

Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.
We are assuming that the size of our container is constant. (isochoric condition)

Oh, yes it's ideal gas. I forget to mention that :)

Do I need to calculate v1= nRT/1atm and then v2= nRT/3atm?
 
Tehy said:
Hello! Any ideas how I could solve this exercise:

5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm. How much work is done by the gas?

Detailed instruction would be good, thanks :)
The question does not have sufficient information. You have to know how the pressure is changed and whether heat is being added or released.

If the external pressure is increased, the volume must decrease, in which case the work done by the gas is negative. If pressure is increased because heat is added, the volume can be held constant, in which case, no work is done. Or it could increase, in which case the work done by the gas is positive.

Work = Pdv not \Delta (PV)

If you assume that the external pressure is gradually changed from 1 atm to 3 atm, and the temperature is kept constant, the work done is:

W = \int_{P_i}^{P_f} PdV = \int_{P_i}^{P_f} nRTdV/V = nRT\ln(\frac{V_f}{V_i}) = -nRT\ln(\frac{V_i}{V_f})

If it is adiabatic (temperature will increase) it is more complicated.

AM
 
Thanks Andrew! That works great! :)
 

Similar threads

Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
11
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 21 ·
Replies
21
Views
3K