Consider the two ideal-gas processes shown in FIGURE 19.18. Even though the initial and final states are the same, the heat added during these two processes is not the same. We can use the first law $\text{Δ}{E}_{\text{th}}=W+Q$ to see why.

FIGURE 19.18 Is the heat input along these two paths the same or different?

The thermal energy is a state variable. That is, its value depends on the state of the gas, not the process by which the gas arrived at that state. Thus $\text{Δ}{E}_{\text{th}}=$$E_{\text{th f}}-E_{\text{th i}}$ is the same for both processes. If $\text{Δ}{E}_{\text{th}}$ is the same for processes A and B, then $W_{\text{A}}+Q_{\text{A}}=W_{\text{B}}+Q_{\text{B}}$.

You learned in Section 19.2 that the work done during an ideal-gas process depends on the path in the pV diagram. There’s more area under the process B curve, so $\left|W_{\text{B}}\right|>\left|W_{\text{A}}\right|$. Both values of $W$ are negative because the gas expands, so $W_{\text{B}}$ is more negative than $W_{\text{A}}$. Consequently, $W_{\text{A}}+Q_{\text{A}}$ can equal $W_{\text{B}}+Q_{\text{B}}$ only if $Q_{\text{B}}>Q_{\text{A}}$. The heat added or removed during an ideal-gas process depends on the path followed through the pV diagram.