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I am having some troubles in solving the following problem a professor gave us, and in my opinion I think that either there are insufficient data or the solutions are infinite.

We have a perfect gas, diatomic, which passes through the following cycle:

$AB$ an isothermal transformation; $BC$ an adiabatic transformation; $CD$ a isobaric transformation; $DA$ an isocora (isovolumetric transformation). The ONLY known initial parameters are $T_A$ and $V_A$ (and of course also $C_p$ and $C_v$), and also the text says that the total work from $A$ to $D$ is zero.

The question is: find $V_B$.

Now as I said, either there are infinite solutions or something is missing. Right? The first hypothesis is due to the fact that I can draw many diagrams for the transformation, such like these:

enter image description here

Am I right? Thank you!

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  • $\begingroup$ Although I haven't looked at this in detail, I'm inclined to agree that there is not enough information. Could you share the full text of the question? There might be another piece of information which you aren't recognizing as important. $\endgroup$ – user1476176 May 19 at 18:20
  • $\begingroup$ This is the full text... I believe that she missed something! But it might be, she is new and it's not the first time she misses some data lol $\endgroup$ – Les Adieux May 20 at 13:49
  • $\begingroup$ See my answer. This is definitely a problem with sufficient data to get a unique solution. Do you want me to solve the whole thing for you? $\endgroup$ – Chet Miller May 20 at 13:58
  • $\begingroup$ @ChetMiller By inspection, the null process (all points coincident, $v_B = v_A$) satisfies the given constraints. If there is a unique solution as you suggest then this must be it, no? $\endgroup$ – user1476176 May 20 at 16:50
  • $\begingroup$ @ user1476176 I stand corrected. When I added the heats up for the various steps of the cycle, they gave the exact same equation as adding the works. In. retrospect, I guess I should have expected this. So adding the heats does not provide an independent equation. $\endgroup$ – Chet Miller May 20 at 21:50
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I tried adding the heats in hopes it would give an additional independent equation. But, of course, it did not. So, as pointed out by others, there is no unique solution.

Here are the details of my analysis: I'm going to express everything in terms of $T_A$, $V_A$, $V_B$, $V_C$ (and n, the number of moles) and then write done the Work equation and the Heat equation in terms of these.

STATES:

State A: $P_A=\frac{nRT_A}{V_A}$

State B: $P_B=\frac{nRT_A}{V_B}$, $T_B=T_A$

State C: $P_C=\frac{nRT_A}{V_B}\left(\frac{V_B}{V_C}\right)^{\gamma}$, $T_C=T_A\left(\frac{V_B}{V_C}\right)^{(\gamma-1)}$

State D: $P_D=P_C=\frac{nRT_A}{V_B}\left(\frac{V_B}{V_C}\right)^{\gamma}$, $T_D=\frac{P_CV_A}{nR}=T_A\left(\frac{V_B}{V_C}\right)^{\gamma}\left(\frac{V_A}{V_B}\right)$

WORK AMOUNTS: $$W_{AB}=nRT_A\ln{\left(\frac{V_B}{V_A}\right)}$$ $$W_{BC}=-\frac{nRT_A}{(\gamma-1)}\left[\left(\frac{V_B}{V_C}\right)^{\gamma-1}-1\right]=-nC_vT_A\left[\left(\frac{V_B}{V_C}\right)^{\gamma-1}-1\right]$$ $$W_{CD}=P_C(V_A-V_C)=-nRT_A\left(\frac{V_B}{V_C}\right)^{\gamma-1}\left[1-\frac{V_A}{V_C}\right]$$ $$W_{DA}=0$$

HEAT AMOUNTS: $$Q_{AB}=nRT_A\ln{\left(\frac{V_B}{V_A}\right)}$$ $$Q_{BC}=0$$ $$Q_{CD}=nC_p(T_D-T_C)=-nC_pT_A\left(\frac{V_B}{V_C}\right)^{\gamma-1}\left[1-\frac{V_A}{V_C}\right]$$ $$Q_{DA}=nC_v(T_A-T_D)=-nC_vT_A\left[\left(\frac{V_B}{V_C}\right)^{\gamma-1}\left(\frac{V_A}{V_C}\right)-1\right]$$

From this, it is readily apparent that adding the heat amounts leads to exactly the same result as adding the work amounts. So there is no unique solution.

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  • $\begingroup$ Chet, NICE. I was not on the ball on this one. $\endgroup$ – Bob D May 20 at 22:01
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Pre-calculation

Let's work in terms of $P$ and $v$ (knowing that $T$ can be found from the ideal gas law). Firstly, we can find $v_A$ as $$ v_A = \frac{R_\text{gas} T_A}{P_A} $$ where $R_\text{gas} = R_\text{universal}/M_\text{gas}$ and $M_\text{gas}$ is the molar mass of the diatomic gas. We now know $(P_A, v_A)$.

It's also convenient to pre-calculate the specific heat ratio, $$ \gamma \equiv \frac{c_P}{c_v}. $$

Constraints

  • A-B is isothermal, so \begin{align} P_A v_A &= P_B v_B \\ W^\text{out}_{AB} &= P_A v_A \ln \frac{v_B}{v_A} = - P_A v_A \ln \frac{P_B}{P_A} \end{align}
  • B-C is adiabatic, so \begin{align} P_B v_B^\gamma &= P_C v_C^\gamma \\ W^\text{out}_{BC} &= \frac{1}{1-\gamma} (P_C v_C - P_B v_B) \end{align}
  • C-D is isobaric, so \begin{align} P_C &= P_D \\ W^\text{out}_{CD} &= P_C (v_D - v_C) \end{align}
  • D-A is isochoric, so \begin{align} v_D &= v_A \\ W^\text{out}_{DA} &= 0 \end{align}
  • The net work over A-B-C-D is zero, so $$ W^\text{out}_{CD} + W^\text{out}_{BC} + W^\text{out}_{CD} = 0 $$

By my count, these constraints provide 9 independent equations but introduce 10 unknowns ($P$ and $v$ at each of the three unknown points plus the four works). Applying the Ideal Gaw Law at each of the 3 unknown states would introduce 3 more equations, but also 3 unknowns (the three unknown $T$'s) and is therefore not helpful in constraining the system. Unless there is some extra constraint that we have missed or some clever cancellation, this system has one degree of freedom and there are infinite possible solutions.

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  • $\begingroup$ You omitted the following equation which provides closure on the problem: $$Q_{AB}+Q_{CD}+Q_{DA}=0$$ $\endgroup$ – Chet Miller May 20 at 12:25
  • $\begingroup$ I don't see it - could you illustrate how in your answer? $\endgroup$ – user1476176 May 20 at 16:40
  • $\begingroup$ I was mistaken. Adding the heats give exactly the same equation as adding the works. $\endgroup$ – Chet Miller May 20 at 21:50

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