Calculating Work in an Adiabatic Process for a Diatomic Gas

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Homework Help Overview

The discussion revolves around calculating the work done during an adiabatic process involving a diatomic gas (O2) under specific conditions of temperature and pressure. The original poster is attempting to compress the gas and is seeking guidance on the appropriate calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster mentions using the integral of pressure with respect to volume (Pdv) but expresses confusion about the process. Other participants suggest using the ideal gas law and inquire about the missing variable needed for calculations. There are also discussions about the relationship between internal energy and work in an adiabatic process.

Discussion Status

Participants are actively engaging with the problem, offering various approaches and clarifications. Some guidance has been provided regarding the use of the ideal gas law and the relationship between work and internal energy, but no consensus has been reached on the method to solve the problem.

Contextual Notes

The original poster is working under the constraints of an adiabatic process where heat transfer (Q) is zero, and there is a specific focus on the properties of a diatomic gas. The discussion includes assumptions about temperature and pressure conditions relevant to the calculations.

sisigsarap
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I have 5.00 mol of O2(Oxygen) at 20.0 Celsius and 1.00atm. I will compress this to 1/10 the original volume.

Find the work? The correct answer is 46.1 Kj

I am having a terrible time with this.

This is what I know: Q = 0 for an adiabatic process, and the change in internal energy is equal to work.

I think I use integrate Pdv, but I am getting very confused. I am really lost and need a push in the right direction!
 
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Ideal gas law? You know pressure, temperature and the number of atoms and 1 atm is much less than the critical pressure.
 
PV = nRT
I just don't see how to plug it in. Pleaseee help its driving me crazy!
 
sisigsarap said:
PV = nRT
I just don't see how to plug it in. Pleaseee help its driving me crazy!

What variable do you not have a value for? There's only one. Solve for the unknown plug in your known values. This will yield your initial condition. Apply the condition in the question. Then use:

W=\int_{V1}^{V2}PdV
 
In an adiabatic process Q=0 then \Delta U+L=0 and L=-\Delta U. You can find now
L=-\nu C_V \Delta T.
(Because you deal with a diatomic gas, you'll have C_V=\frac{5}{2}R).

The final temperature is taken from TV^{\gamma-1}=const (here T is in Kelvin, as you probably know).

OBS. In my opinion, the work made by the gas must be negative!
 
Last edited:

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