Internal energy equals to average kinectic energy ?

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SUMMARY

The discussion clarifies that the molar heat capacity, Cv, equals 3R/2 specifically for ideal monatomic gases. It emphasizes that one cannot equate Cv dT to nkT (3/2) without proper context, as this conflates temperature change with absolute temperature. The integral of nC_v dT from initial to final temperature confirms the relationship between heat capacity and temperature change for monatomic ideal gases.

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  • Basic calculus for integrating heat capacity equations
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Outrageous
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For ideal gas , can I assume Cv dT = nkT (3/2)
, thank you
 
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Outrageous said:
For ideal gas , can I assume Cv dT = nkT (3/2)
, thank you
No. Molar heat capacity, Cv = 3R/2 only for ideal monatomic gases. You are also mixing temperature change with temperature here.

Using molar heat capacity 3R/2 for a monatomic ideal gas:

[itex]\int_{Ti}^{Tf} nC_vdT = \frac{3}{2}nR(T_f - T_i) = \frac{3}{2}nR\Delta T = \frac{3}{2}Nk\Delta T[/itex]
AM
 
Andrew Mason said:
No. Molar heat capacity, Cv = 3R/2 only for ideal monatomic gases. You are also mixing temperature change with temperature here.

Using molar heat capacity 3R/2 for a monatomic ideal gas:

[itex]\int_{Ti}^{Tf} nC_vdT = \frac{3}{2}nR(T_f - T_i) = \frac{3}{2}nR\Delta T = \frac{3}{2}Nk\Delta T[/itex]



AM

Thanks for compact explanation.
 

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