# Heat exchange in adiabatic process

I know my teacher told me that

In adiabatic process there is no exchange for heat between system and surrounding.

Then he came to point. From first law of thermodynamics

$$∆U = q + W \tag{1}$$

Since in adiabatic process there is no heat exchange then,

$$∆U = W \tag{2}$$

And then he said as we know

$$∆U = nC_v∆T \tag{3}$$

So,

$$W = nC_v∆T \tag{4}$$

I don't get it. Above equation tells us that the work done when a system is given heat at constant volume equals to $$nC_v∆T.$$ But how can this be applied in adiabatic process, since there is no exchange of heat?

• The above equation does not say that the work done when a system is given heat at constant volume equals $nC_v\Delta T$. The relationship is not equal to the heat transferred. It is equal to the change in internal energy. Did you think that the only way the temperature of a gas can change is by adding or removing heat? Oct 9 '19 at 3:07
• Adiabatic work is done in expense of thermal energy and vice versa. E.g air lifted by 100m adiabatically expands, does the expansion work and cools itself by 0.98 K. Oct 9 '19 at 3:26
• Please note that heat and temperature are completely different concepts. Do not confuse them.
– user23638
Oct 9 '19 at 9:13

## 1 Answer

The problem is that we were taught incorrectly in freshman physics. They told us that $$Q=nC\Delta T$$, where C is called the heat capacity. However, they neglected to tell us that this only applies if no work is done (i.e., the volume is essentially constant, as for a solid or liquid). Otherwise, it gives the wrong answer for the heat Q.

"Heat capacity" is a bit of a misnomer for C. From what we then learn in thermodynamics, a better term would be the "internal energy capacity," because that is how Cv is more properly defined. For an ideal gas, $$nC_v=\frac{\Delta U}{\Delta T}$$ or $$\Delta U=nC_v\Delta T$$This has nothing to do with the heat added unless the volume is constant and no work is done. In that case, $$Q=\Delta U=nC_v\Delta T$$So the freshman form of the equation is recovered only if no work is done.

• Could you please tell me work done in Adiabatic process. Or just write it down on paper and send me it's image Oct 9 '19 at 14:37
• For an ideal gas, the equation for the work in your original post is correct. However, you can't really use it unless you know the final pressure and volume. The final conditions depend on the details of how the process is carried out. Oct 9 '19 at 15:01