Imagine some helium in a cylinder with an initial volume of $1$ liter ant initial pressure of $1$ atm. Somehow the helium is made to expand to a final volume of $3$ liters, in such a way that its pressure rises in direct proportion to its volume. Calculate the work done on the gas during this process, assuming there are no other types of work being done.

I know the following:

  • $\text{J}=\text{Pa}\cdot\text{m}^3$
  • $1 \ \text{atm} = 1.013\cdot10^5 \ \text{Pa}$

  • $1 \ \text{litre} = (0.1 \ \text{m})^3$

I want to compute work done during quasistatic compression. I do get the correct integral, where the formula is

$$W=-\int\limits_{V_\text{initial}}^{V_{\text{final}}}P(V) \ dV,$$

where $P(V)=V$ in my case. But the answer should be $-400 \ \text{J}$ and not $-0.4 \ \text{J}$ like I'm getting. I assume I'm doing something wrong with the units here:

$$W=-\int\limits_{1 \ \text{litre}}^{3 \ \text{litres}} V \ \text{atm} \ dV=-\int_\limits{0.001\text{m}^3}^{0.003\text{m}^3}V\cdot1.013\cdot10^5\ \text{Pa} \ dV=-0.4052 \ \text{J}.$$

Can someone show me how to get the correct answer?

  • $\begingroup$ You start with SI units, but when you say $P(V)=V$, you are no longer working in SI units. $\endgroup$ – Sayan Mandal Sep 10 '19 at 8:55
  • $\begingroup$ How can P(V)=V? There must be some kind of proportionality factor for this equation to be correct. $\endgroup$ – DomDoe Sep 10 '19 at 9:00
  • $\begingroup$ @SayanMandal I've added the problem statement now. I thought that since the pressure is rising in direct proporion to its volume, the relation should be $P(v)=V$, I might be wrong here. $\endgroup$ – Parseval Sep 10 '19 at 9:25
  • $\begingroup$ @DomDoe - See the comment above please. $\endgroup$ – Parseval Sep 10 '19 at 9:25
  • $\begingroup$ @DomDoe: One can always choose (or invent) units where you can do this, but I think this is beside the point in this post. $\endgroup$ – Sayan Mandal Sep 10 '19 at 9:26

When you specify units of L for volume and atm for pressure, then when you say $P(V)=V$, what you're really saying is $P=kV$ where $k=1$ atm/L. Let's keep track of this constant as the calculation progresses.

$$W=-\int_{1}^3P(V)dV=-\int_1^3kVdV=-\frac{1}{2}kV^2\bigg\vert_{1}^3\\=-\frac{1}{2}(1\text{ atm/L})(8\text{ L}^2)=-4\text{ atm}\cdot\text{L}$$

Now it's clear that you need to convert from atm$\cdot$L to J. Based on your information,

$$1\text{ atm}\cdot\text{L}=(1\text{ atm})(1\text{ L})=(1.013\times 10^5\text{ N/m}^2)(.001\text{ m}^3)=101.3\text{ N}\cdot\text{m}=101.3\text{ J}$$

which gives you your correct answer.

  • $\begingroup$ Why the negative sign for work? Are you using the chemist's version of first law, i.e., $\Delta U=Q+W$, as opposed to the engineering/physics definition, i.e, $\Delta U=Q-W$? Since the gas is expanding, it is doing work. $\endgroup$ – Bob D Sep 11 '19 at 11:29
  • $\begingroup$ @BobD I simply used the convention that the OP used. $\endgroup$ – probably_someone Sep 11 '19 at 11:30
  • $\begingroup$ Then it must be the chemist's version. I will check with the OP. $\endgroup$ – Bob D Sep 11 '19 at 11:31
  • $\begingroup$ I checked the book the OP is using and yes, it does use the chemist version of the first law. $\endgroup$ – Bob D Sep 11 '19 at 11:42

I think the issue of units has already been addressed. There are, however, issues with the problem statement itself and how to determine the work involved.

The problem involves helium, which we can assume behaves like an ideal gas. The question states the helium is made to expand "in such a way that its pressure rises in direct proportion to its volume". This implies a linear relationship between pressure and volume which is not possible for an ideal gas. For an ideal gas, the relationship between pressure and volume is given by the ideal gas law


or, expressing pressure as a function of volume,


meaning, for an ideal gas, the pressure varies inversely with volume if the temperature is constant (an isothermal process). It can not vary directly (linearly) with volume.

But the problem statement does say "somehow" the helium is made to to expand to its final pressure and volume. "Somehow" leaves open the possibility of a sequence of processes to take the gas from its initial to final state such that the average value of the work done is 400 J.

Refer to the PV diagrams below. Path 1-1a-2 shows a constant volume path followed by a constant pressure path, for which the work done would be 6 L-atm, or 600 J. Path 1-1b-2 shows a constant volume path followed by a constant pressure path, for which the work would be 2 L-atm, or 200 J. Neither path produces 400 J, though the average value is 400 J.

It is possible, on the other hand, to have a series of such constant volume and pressure paths that are small enough to follow the linear relationship. An example is the "staircase" of processes shown in the diagram. The average work done is 4OO J since the area enclosed by the constant volume processes above the line equals that below the line. The smaller the steps become, the closer the sequence approximates a straight line.

Hope this helps.

enter image description here


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