What is the work done during an adiabatic process?

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SUMMARY

During an adiabatic process involving 5.50 moles of a monatomic ideal gas, the temperature drops from 495°C to 215°C. The work done by the gas is calculated using the formula Q = (3/2)nRΔT, resulting in 19196 Joules. The heat exchanged with the surroundings is zero, consistent with the definition of an adiabatic process. The change in internal energy is equal in magnitude but opposite in sign to the work done.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of thermodynamic principles, specifically adiabatic processes
  • Familiarity with molar heat capacities (C_v and C_p) for monatomic gases
  • Basic calculus for understanding changes in internal energy (ΔU = C_v n ΔT)
NEXT STEPS
  • Study the derivation of the molar heat capacity (C_v) for monatomic gases
  • Learn about the relationship between work done and internal energy in thermodynamic processes
  • Explore the concept of gamma (γ) in adiabatic processes and its significance
  • Investigate real-world applications of adiabatic processes in engines and refrigeration
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Students studying thermodynamics, physics enthusiasts, and professionals in engineering fields focusing on heat transfer and energy systems.

TheBeesKnees
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Homework Statement


During an adiabatic process, the temperature of 5.50 moles of a monatomic idea gas drops from 495 C to 215 C. For this gas, find (a) the work it does, (b) the heat it exchanges with its surroundings, and (c) the change in its internal energy


Homework Equations


I know work is P(delta V)
PV1=nRT1 and
PV2=nRT2 so if I subtract to get the change,
P(deltaV)=nR(delta T)
P(deltaV)=5.50(8.31)(280)
P(deltaV)= 12797.4 Joules.
But my textbook says this isn't the answer.
Where did I go wrong?

I know the answer for b is zero, since that's the definition of an adiabatic process
 
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Oh yeah, and the answer to part c must be the opposite sign but same magnitude of the answer to a.
 
What is the work done during an ADIABATIC process... ?

Is it really (P deltaV) ?

Have a look...

adiab.gif
 
Last edited:
TheBeesKnees said:

Homework Statement


During an adiabatic process, the temperature of 5.50 moles of a monatomic idea gas drops from 495 C to 215 C. For this gas, find (a) the work it does, (b) the heat it exchanges with its surroundings, and (c) the change in its internal energy
Do c) first. You are given the initial and final temperatures so you can determine the change in internal energy. How is the work done related to the change in internal energy if dQ = 0?

You are assuming pressure is constant. That is where you went wrong.

AM
 
I see now that I was wrong to assume pressure is constant, but I'm afraid I'm still a little lost with the symbols and stuff.

What does PV^gamma mean? Is gamma an exponent? And if so, do I need to figure out what both P and V are on their own?

And what constant do they equal? My textbook says (5/3) in the chapter about adiabatic-ness.

In the equation beside the graph, how do I find Vfinal and Vinitial without being given a pressure?
 
You don't need to find pressure or volume to solve this problem. You do need to know that there are only two ways to add energy to a closed system: do work on it or heat it.
 
And in this problem, since it's adiabatic, it does not exchange heat, so if the temperature is going to go down, it must do positive work. I understand this.

I just don't understand how to apply this stuff about Cv and Cp. I am at a loss for what I should do mathematically and what all those symbols mean, and where they came from.
 
Hmmm... I've been searching, and I found the formula
Q= 3/2 nRdT
Which gives me 1.5*5.5*8.31*280= 19196 Joules, which my textbook says is the correct answer.

Where did the 3/2 come from?
 
The energy of an ideal gas depends only on its temperature, by the relationship

\Delta U=c_Vn\Delta T

The molar heat capacity c_V is \frac{3}{2}R for a monatomic gas. The 3/2 factor can be derived (from statistical mechanics), but the derivation is very complex compared to what you're studying now.
 

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