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$\pu{10 L}$ of a monoatomic ideal gas at $\pu{0 ^\circ C}$ and $\pu{10 atm}$ pressure is suddenly released to $\pu{1 atm}$ pressure and the gas expands adiabatically against this constant pressure. What are the final temperature and volume of the gas? (Answer: $\pu{174.9 K}$, $\pu{64.04 L}$)

There are some points I don't get:

  1. First of all, it is not given whether it is reversible of irreversible expansion.
  2. If it is reversible, I am at wits end deciding whether to use one of the following (without success): $$ pV^\gamma = \mathrm{constant} \quad \mathrm{or} \quad T^\gamma p^{1 - \gamma} = \mathrm{constant}$$
  3. If it is irreversible, I tried using
    $$W_\mathrm{irreversible} = \Delta E$$
    i.e. $$-P_{ext} \left(\frac{nrT_2}{p_2} - \frac{nrT_1}{p_1}\right) = nC_\mathrm{v}\Delta T$$ I am having trouble using $r = \pu{0.0821 L atm mol-1 K-1}$ and $R = \pu{8.314 J mol-1 K-1}$.
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    $\begingroup$ Just a note: IUPAC recommends to use lowercase $p$ for pressure to avoid confusion with power $P$. ACS allows both. I personally prefer lowercase $p$ as the safest option, hence my edit. $\endgroup$ – andselisk Nov 26 '17 at 5:17
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The process should indeed be considered irreversible, as the problem statement gives a rather explicit hint stating that the pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation $pV^\gamma = \text{const}$ is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:

$$W = - \Delta U = nC_\mathrm{v}(T_1 - T_2) \tag{1}$$

For ideal monoatomic gas the following is applied:

$$n = \frac{p_1V_1}{RT_1} \tag{1.1}$$ $$C_\mathrm{v} = \frac{3R}{2} \tag{1.2}$$

so that (1) can be rewritten as such:

$$W = nC_\mathrm{v}(T_1 - T_2) = \frac{3p_1V_1R}{2RT_1}(T_1 - T_2) = \frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} \tag{1.3}$$

At the same time work by gas expansion is

$$W = p_2(V_2 - V_1) = p_2V_2 - p_2V_1 \tag{2}$$

and from the ideal gas law the reduced product is

$$p_2V_2 = nRT_2 \tag{2.1}$$

so that (2) can also be rewritten:

$$W = p_2(V_2 - V_1) = nRT_2 - p_2V_1 \tag{2.2}$$

I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature $T_2$ from there:

$$\frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} = nRT_2 - p_2V_1$$

Here I bring up the final result:

$$T_2 = \frac{3p_1V_1 + 2p_2V_1}{5p_1V_1} T_1 = \frac{3 \cdot \pu{10 atm} \cdot \pu{10 L} + 2 \cdot \pu{1 atm} \cdot \pu{10 L}}{5 \cdot \pu{10 atm} \cdot \pu{10 L}} \cdot \pu{273.15 K} = \pu{174.7 K} \tag{3}$$

Volume after expansion is also to be found using ideal gas law and (1.1):

$$V_2 = \frac{nRT_2}{p_2} = \frac{p_1V_1RT_2}{RT_1p_2} = \frac{p_1T_2}{p_2T_1}V_1 = \frac{\pu{10 atm} \cdot \pu{174.7 K}}{\pu{1 atm} \cdot \pu{273.15 K}} \cdot \pu{10 L} = \pu{64.0 L} \tag{4}$$

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  1. Your book has a printing mistake (or you might have typed it out wrong), as Gert pointed out, such a process doesn't even make sense. The part "against this constant temperature." should be replaced with "against this constant pressure."

    Now, those words "suddenly" and "against this constant pressure" tell us the process has been performed irreversibly, it's like a heavy block was sitting on top of that piston, and now suddenly the block just vanished away, all hell breaks loose for the gas molecules during such an expansion, there's no way we can define stuff like pressure or temperature for the gas during this period. In reversible processes, parameters like P,V,T could have been defined all through out, and that's why we can draw such a nice sweet curve on a PV graph.

  2. So, now that we know the process has been performed irreversibly, we cannot use those formulas involving $\gamma$, you can use them only if the process was reversible. We can go back to the basics like what the first law of thermodynamics says.

  3. You're doing great so far with the equation you wrote. Now try substituting known values in that. I really don't think you should have a problem with R, because on substituting the Cv of an ideal monoatomic gas, you'll see you can just divide R out from both sides of the equation.

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    $\begingroup$ I think you're right, so +1 from me. 'In my day' we solved most of these problem assuming reversibility, maybe that's changed now? $\endgroup$ – Gert Nov 25 '17 at 18:47
  • $\begingroup$ Sorry, typing error. $\endgroup$ – Sujay Mohanty Nov 25 '17 at 20:12
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The gas cannot "expand adiabatically against this constant temperature" (of $\pu{0 {^\circ}C}$) and then end up at $\pu{174.9 K}$, at the same time. That's a contradiction in terminis. Adiabatic expansions cannot be isothermal, in any case. Adiabatic processes mean there is no heat transfer. Fast processes are often adiabatic because heat transfer takes time and fast processes don't allow that time.

The fact that the pressure is lowered "suddenly" makes me believe the process is irreversible because mainly slow, equilibrium processes without friction, turbulence, etc are reversible. But I believe the spirit of the question will allow you to simply use:

$$P_0V_0^{\gamma}=PV^{\gamma}$$

to find final volume $V$. If the process was irreversible that would have been specified but it is common for problems like this to use reversibility as a simplifying approximation, just like most gasses are assumed ideal (when few are).

You could then use the Ideal Gas Law to find final temperature $T$.


As said in the comments, the numbers don't stack up.

For example:

$$V=V_0\Big(\frac{P_0}{P}\Big)^\frac{1}{\gamma}=10\times 10^\frac{3}{5}=39.8\ \mathrm{L}$$

So the question is either badly formulated or the process is irreversible or polytropic.

However, the Ideal Gas Law is respected:

$$\frac{P_0V_0}{PV}=\frac{T_0}{T}=\frac{10\times 10}{1 \times 64.04}=\frac{273.16}{174.9}=1.562$$

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  • $\begingroup$ I am getting $ V_2 $ as $ \pu{39.83 L} $ that way. $\endgroup$ – Sujay Mohanty Nov 25 '17 at 17:20
  • $\begingroup$ Let me try. 5 minutes, plse. $\endgroup$ – Gert Nov 25 '17 at 17:34
  • $\begingroup$ What value for $\gamma$ did you use? $\endgroup$ – Gert Nov 25 '17 at 17:39
  • $\begingroup$ Using $\gamma=\frac{7}{5}$ (diatomic gas) the numbers don't stack up for me either. Nothing works out at all. Either this question is badly formulated or this process is VERY irreversible. Or the expansion may be polytropic: en.wikipedia.org/wiki/Polytropic_process. Quitee common in industrial processes. $\endgroup$ – Gert Nov 25 '17 at 18:27
  • $\begingroup$ Ooopsie: it's a mono atomic gas, so $\gamma=\frac{5}{3}$ but still no joy. $\endgroup$ – Gert Nov 25 '17 at 19:05

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