Immediate pressure change after compressing a gas

Question

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.

Answer 1

Thermodynamics analysis mostly uses two states: start and final. So your thought of $dV$ and $dP$ confused you.

For this case, you need to use two equations: ideal gas law and first law. And again look at two states.

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 $.

This is a elaboration of @Chester Miller's comments.