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

[Ruby_338]

How come $adiabatic \,work = C_v(T_2 -T_1)$

 

[Ssnow]

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

 

[Chestermiller]

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.

 

[Ruby_338]

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?

 

[Borek]

Isn't it straightforward? How is $C_v$ defined? What is $nC_v\Delta T$?

 

[Ruby_338]

I don't know XD

 

[Borek]

Then start by checking. Per forum rules you should do your legwork, not ask others to do that for you.

 

[Ruby_338]

Okay. Ill remember that