So there was a throw away question that wanted us to calculate the difference in $q$ for a two step process vs single step. Went basically like this:

Gas in a container with a massless a piston. When an external pressure $P_1$ is applied, the gas compresses from $V_i$ to $V_1$. When the external pressure is increased to $P_2$, the gas further compresses from $V_1$ to $V_f$ . In a separate experiment with the same initial conditions, a pressure of $P_2$ was applied to the gas, decreasing its volume from $V_i$ to $V_f$ in one step. If the final temperature was the same for both processes, calculate the difference between $q$ for the two-step process and $q$ for the one-step process.

That was simple enough, heat and work are path functions, $\Delta q=\Delta w$, and we were given the values for pressure and volume. Answer submitted, autograder coughs up the points.

Except that "If the final temperature was the same for both processes" bit is bugging me. How the heck is that possible?

Internal energy is a state function, so $\Delta U=\Delta U_1+\Delta U_2$ (one step, two step), which means $$nC(\Delta T)+P_2(\Delta V_1+\Delta V_2)=nC\Delta T_2+P_2\Delta V_2+nC\Delta T_1+P_1\Delta V_1.$$

And since $\Delta q=\Delta w$,

$$nC(\Delta T)-(nC\Delta T_2+nC\Delta T_1)=P_2(\Delta V_1+\Delta V_2)-(P_2\Delta V_2+P_1\Delta V_1),$$

which can't be true if the final temperature is the same, since that would make $\Delta T=\Delta T_1+\Delta T_1$. The left side would be zero, the right side wouldn't be zero and that's a problem.

What the heck am I missing? Or is the question just badly worded?


1 Answer 1


Since both processes reach the same final state, specified completely by $p_2,V_2$, final temperature as well as every other thermodynamic property must be the same in that final state.

As you mentioned yourself $\Delta q$ is a path function. So you should not equate $\Delta q$ for both processes. In fact, temperature of final state is the same for both processes tells you that $\Delta q$ cannot be equal.

  • $\begingroup$ Where am I equating Δq for both processes? $\endgroup$ Oct 14, 2016 at 14:21
  • $\begingroup$ What is $nC\Delta T+\Delta (pV)$ equal to? $\endgroup$
    – Deep
    Oct 15, 2016 at 3:32
  • $\begingroup$ the total internal energy change of the system for the single step process. ΔU $\endgroup$ Oct 15, 2016 at 16:26
  • $\begingroup$ To be completely clear. nCΔT+Δ(pV)nCΔT+(pΔV) is the change in internal energy for the single step $\endgroup$ Oct 15, 2016 at 19:08
  • $\begingroup$ So apparently if the text boxes loses focus and you hit enter it just submits the comment and doesn't let me edit. nCΔT+Δ(pV) will be the change in internal energy for the single step process nCΔT1+P1ΔV1 is the change in internal energy for the first step of the two step process. nCΔT2+P2ΔV2 should be the change in internal energy for the second step of the two step process. For the two step the total change in internal energy should be the sum of the change in internal energy for both steps: nCΔT1+P1ΔV1+nCΔT2+P2ΔV2 $\endgroup$ Oct 15, 2016 at 19:13

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