Question

In a constant-volume process, dU = nCV dT. But in a constant-pressure process, it is not true that dU = nCp dT. Why not?

Short answer

At the constant volume, the work done W=0. So$d\underset{~}{O}=dU=n{C}_{v}dT$ . While at constant pressure$d\underset{~}{O}=n{C}_{p}dT$ and work done has a value so dU is not equal$d\underset{~}{O}$ and the equation is

$dU=n{C}_{p}dT-pdV$

Step-by-step solution

Step 1:

The molar heat capacity($C_{v}or{C}_p$) of a substance depends on the conditions under which the heat is added. The first law of thermodynamics is in differential form is

$d\underset{~}{O}=dU+dW$

where

$d\underset{~}{O}=n{C}_{v}dT.............\left(1\right)$

At the constant volume (the molar heat capacity ) there is no work done and W=0. So, dQ = dU, and the equation (1) can also be written as

$dU=n{C}_{v}dT$

While at constant pressure (the molar heat capacity ) the work done is given by dW =pdV and the heat

$d\underset{~}{O}=n{C}_{p}dT$

Therefore, dU is not equal$d\underset{~}{O}$ . So it is not true that$dU=n{C}_{p}dT$ when the pressure is constant. But the truth is that constant pressure

$$\begin{gathered} dU=d\underset{~}{O}-dW \\ dU=n{C}_{p}dT-pdV \end{gathered}$$

Conclusion

Hence, at the constant volume, the work done W=0. So$d\underset{~}{O}=dU=n{C}_{p}dT$ . While at constant pressure$d\underset{~}{O}=n{C}_{p}dT$ and work done has a value so dU is not equal and the equation is$d\underset{~}{O}$

$dU=n{C}_{p}dT-pdV$