# Immediate pressure change after compressing a gas

I'm confused with some of the pressure and temperature implications of compressing gases with minute, incremental changes and doing it suddenly. Below is an exercise I've been attempting:

Here is my thought process:

• In the case of (a), this seems to be an isothermal process. I'm assuming the gas is enclosed by a thermal reservoir of constant $T$, an the gas is being compressed incrementally, or reversibly, such that each small change $dV$ in volume there is a subsequent supplying of $dP$, in this case $dV$ is negative and $dP$ is positive. Since $T$ is constant, $P_1V_1 = P_2V_2$, so if $V_2 = 1/2V_1$, $P_2 = 2P_1$.

• In this new case (b), the compression is done quickly, not at step $dV$ then $dP$, but several contributions of $dV$ before a contribution of $dP$. This means there is an imbalance, and since $PV \ne K$ where $K$ is a constant at every turn, $T$ must not be conserved. If $T$ is not conserved, there is a $\Delta U$ s.t. $\Delta U = Q + W$. $W$ is positive and contributing here. Since $\Delta U \propto T \propto P$, then $P$ will increase.

However, I have the following issues with my logic:

• $V \propto T$ by the same equation I used for my deduction in the second bullet point: $PV=nk_bT$. And volume decreases. How am I then justified in saying this?
• How does this imply the pressure will be higher than (a)?

I'm seeking any advice on my issues, and whether anything in my thought process is wrong other than the ones pertaining to my issues.

• I think that in (b) what they are trying to get you to consider is that, if the compression is carried out very quickly, there is not adequate time for heat to transfer to occur between the system and the surroundings. So, in this case, during the compression, the internal energy increases, while Q is close to zero. So at least immediately after the process in imposed, the internal energy and temperature will be higher. This will cause the pressure to be higher. – Chet Miller Oct 24 '17 at 17:19
• Let me see if I interpret you correctly: in (a), for every contribution of $W$ in $\Delta U = Q + W$ there is a contribution $Q$ s.t. $Q = -W$ to ensure $\Delta U = 0$ since there is enough time per $dV$ to allow heat to be added and $U \propto T$ so $\Delta T = 0 \implies \Delta U = 0$? And for (b), you're saying since volume is changing rapidly, there will be too fast a change in volume for heat to transfer in each step, so there will be a net change in $\Delta U$ at least initially, and thus $\Delta U = PdV$, $U \propto T \therefore T \propto P$ implies an increase in $P$. – sangstar Oct 24 '17 at 23:05
• That may just be a new interpretation of mine based on what I'm piecing together. However, if I'm on the right track, I'm not exactly sure how $P$ in (b) is larger than $P$ in (a). – sangstar Oct 24 '17 at 23:07
• From the ideal gas law, if the final volume is the same and the final temperature is higher, the final pressure must be higher. In my judgment, you analyzed this very nicely. – Chet Miller Oct 25 '17 at 0:10

Thermodynamics analysis mostly uses two states: start and final. So your thought of $dV$ and $dP$ confused you.
For (a), start state is assumed to be $P_1$, $V_1$ and $T_1$. The final state is $P_2$, $0.5 \cdot V_1$ and $T_1$. As you said $T_2=T_1$. Using state equation, $\frac{P_1 \cdot V_1}{T_1}=\frac{P_2 \cdot 0.5 \cdot V_1}{T_1}$, you will get $P_2=2 \cdot P_1$.
For (b), start state is the same. The final state is $P_2$, $0.5 \cdot V_1$ and $T_2$. Using the ideal gas law, you will get $P_2=2 \cdot P_1 \cdot \frac{T_2}{T_1}$.
Then look at the first law $dU = \delta Q-P d V$. As it is a rapid process, $\delta Q =0$. Then $dU = -P d V$. This gives $U_2-U_1 = \int_{V_1}^{V_2} (-P) dV$. We know pressure is positive and volume is decreasing. Thus $U_2>U_1$. This shows that $T_2>T_1$. Thus $P_2 > 2 \cdot P_1$.