Calculating Work in an Adiabatic Process for a Diatomic Gas

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SUMMARY

The discussion focuses on calculating the work done during an adiabatic process involving 5.00 moles of diatomic oxygen (O2) at an initial temperature of 20.0°C and pressure of 1.00 atm, compressed to one-tenth of its original volume. The correct work output is determined to be 46.1 kJ. Key equations utilized include the ideal gas law (PV = nRT) and the work integral W = ∫P dV, with the internal energy change related to work through the equation L = -ΔU. The specific heat capacity for the diatomic gas is identified as C_V = (5/2)R.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of internal energy and work relationships (ΔU = Q + W)
  • Concept of specific heat capacity for diatomic gases (C_V = (5/2)R)
NEXT STEPS
  • Study the derivation and application of the adiabatic process equations
  • Learn how to calculate work done in various thermodynamic processes
  • Explore the implications of the first law of thermodynamics in adiabatic conditions
  • Investigate the behavior of diatomic gases under different thermodynamic conditions
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone involved in the study of gas laws and adiabatic processes.

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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