How do I find the work done in an adiabatic process?

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SUMMARY

The discussion centers on the calculation of work done in an adiabatic process, specifically using the formula ##W = n \cdot C_{v} (T_{1} - T_{2})##, where ##C_{v}## is the specific heat at constant volume and ##n## represents the number of moles. Participants clarify that in an adiabatic transformation, the internal energy change ##\Delta U## is equal to the negative work done, as heat transfer ##Q## is zero. The confusion regarding the absence of the mole term in the initial formula is addressed, emphasizing the importance of including it for accurate calculations.

PREREQUISITES
  • Understanding of the first and second laws of thermodynamics
  • Familiarity with the concepts of internal energy and specific heat capacity
  • Knowledge of adiabatic processes in thermodynamics
  • Basic algebra for manipulating thermodynamic equations
NEXT STEPS
  • Study the derivation of the adiabatic process equations in thermodynamics
  • Learn about the relationship between internal energy, work, and heat transfer
  • Explore the concept of specific heat at constant volume (##C_{v}##) in detail
  • Investigate real-world applications of adiabatic processes in engineering
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Students of thermodynamics, engineers working with heat engines, and anyone seeking to deepen their understanding of adiabatic processes and their implications in physical systems.

Ruby_338
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How come ##adiabatic \,work = C_v(T_2 -T_1)##
 
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If the transformation is adiabatic you have that ##\Delta U=-W## for the second law of thermodynamics because ##Q=0##. So ##W=-\Delta U = n\cdot C_{v}(T_{1} -T_{2})## where ## C_{v}## is the specific heat for constant volume and ##n## represents the mole. It is strange that in your formula there isn't ##n##...

Ssnow
 
Ssnow said:
If the transformation is adiabatic you have that ##\Delta U=-W## for the second law of thermodynamics because ##Q=0##. So ##W=-\Delta U = n\cdot C_{v}(T_{1} -T_{2})## where ## C_{v}## is the specific heat for constant volume and ##n## represents the mole. It is strange that in your formula there isn't ##n##...

Ssnow
Maybe he means "per mole" or "per unit mass.". Also, irrespective of the 2nd law, Q is equal to zero for an adiabatic process.
 
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I forgot to add the n. How did we get ## - \Delta U = n C_v (T_2-T_1) ## ?

Can you show me step by step?
 
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Isn't it straightforward? How is C_v defined? What is nC_v\Delta T?
 
I don't know XD
 
Then start by checking. Per forum rules you should do your legwork, not ask others to do that for you.
 
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Okay. Ill remember that
 

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