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I was wondering if there are specific formulae in chemistry for which it is "compulsory" or convenient to use atmosphere (atm) rather than Pascal (or even mmHg) as a unit of measurement.

Of course we can convert one into the other, but the choice of using one of the two also conditions other units, such as in the case of volume.

Can someone provide me with a good picture of the relationship between these units and the others that we may need to use in the same context? (e.g. we use L with Atm; $m^3$ with Pa).

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    $\begingroup$ Chemistry, and other sciences in general, have been trying for years to switch to the International System of Units exclusively. Older books don't adhere to the convention of course. So the older books may use pressure units such as torr, atmospheres, bar and psi. $\endgroup$ – MaxW Nov 22 '19 at 19:02
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    $\begingroup$ I first started studying chemistry in 1968. All heat analysis was done with the calorie. I did so many problems with calorie that I think calories. I can work the problem out in Joules, but I don't really have a "feeling" for joules like I do calories. $\endgroup$ – MaxW Nov 22 '19 at 19:13
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    $\begingroup$ @Shootforthemoon I am struggling to understand your question, mostly because I've been thinking of units my way for about 10 years and am a little stuck in my understanding. Is there a specific situation/problem that has prompted this question? Maybe that will be easier to answer lol $\endgroup$ – J. Ari Nov 22 '19 at 19:41
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    $\begingroup$ Well when I learned pV=nRT, i was taught pressure was in torr. If you look up the ideal gas constant, R, you can see that it has many values depending on the units. Note that even with all the values listed temperature can be measured in grad, time in fortnights, and length in leagues. So there are a lot of other values "missing" in the table. Having countless values is exactly the situation that the International System of Units tries to avoid. $\endgroup$ – MaxW Nov 22 '19 at 20:37
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    $\begingroup$ Are you thinking something like the cigs system? There are lot of inconsistencies between fields, but few has good arguments to do it $\endgroup$ – Greg Nov 23 '19 at 7:03
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Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units.

For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation

$$v/(\mathrm{km/h})=3.6\frac {l/\mathrm m}{t/\mathrm s}$$ or $$\left\{v\right\}_{\mathrm{km/h}}=3.6\frac{\left\{l\right\}_{\mathrm m}}{\left\{t\right\}_{\mathrm s}}$$

you should write “speed $v$ is the quotient of distance $l$ by time $t$”, preferably in form of the quantity equation

$$\displaystyle v=\frac lt$$

Generally, you can work with any units you like; however, this requires the use of conversion factors. The use of conversion factors is never required when only coherent SI units are used; therefore, the use of coherent SI units (the SI base units and the SI coherent derived units, including those with special names) are preferred.

In order to avoid large or small numerical values, decimal multiples and submultiples of units may be formed with the SI prefixes. However, when prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor in the expression for the unit in terms of the base units.

Furthermore, various non-SI units are accepted for use.

In the International System of Quantities, pressure $p$ is defined as $$p=\frac{\mathrm dF}{\mathrm dA}$$ Where $\mathrm dF$ is the force component perpendicular to the surface element of area $\mathrm dA$.

The coherent derived units in the SI for the pressure has the special name pascal and the symbol $\mathrm{Pa}$. According to the definition of pressure, the pascal can be expressed in terms of other SI units as $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. The pascal can also be expressed in terms of SI base units as $1\ \mathrm{Pa}=1\ \mathrm{m^{-1}\ kg\ s^{-2}}$. The last form is especially useful to check whether units in equations and calculations are converted correctly.

The SI prefixes may be used with any of the names and symbols, but when this is done the resulting unit will no longer be coherent.

The bar (symbol $\mathrm{bar}$) is included in the list of non-SI units that are used by special interest groups for a variety of different reasons. Although the use of SI units is to be preferred for reasons already emphasized, authors who see a particular advantage in using these non-SI units should have the freedom to use the units that they consider to be best suited to their purpose. In particular, the bar is included because since 1982 one bar has been used as the standard pressure for tabulating all thermodynamic data. (Prior to 1982 the standard pressure used to be the standard atmosphere, equal to $1.01325\ \mathrm{bar}$, or $101\,325\ \mathrm{Pa}$.)

The following other non-SI units can be used, which requires the use of conversion factors. However, the use of these units is deprecated.

$$\begin{array}{lll} \hline \text{Unit} & \text{Unit} & \text{Conversion} \\ \text{name} & \text{symbol} & \text{factor} \\ \hline \text{standard atmosphere} & \mathrm{atm} & 1\ \mathrm{atm} := 101\,325\ \mathrm{Pa}\\ \text{kilogram-force per square metre} & \mathrm{kgf/m^2} & 1\ \mathrm{kgf/m^2} = 9.80665\ \mathrm{Pa}\\ \text{technical atmosphere} & \mathrm{at} & 1\ \mathrm{at}=1\ \mathrm{kgf/cm^2} = 98\,066.5\ \mathrm{Pa}\\ \text{conventional millimetre of water} & \mathrm{mmH_2O} & 1\ \mathrm{mmH_2O}:=10^{-4}\ \mathrm{at} = 9.806 65\ \mathrm{Pa}\\ \text{conventional millimetre of mercury} & \mathrm{mmHg} & 1\ \mathrm{mmHg} \approx 13.5951\ \mathrm{mmH_2O} \approx 133.3224\ \mathrm{Pa} \\ \text{torr} & \text{Torr} & 1\ \mathrm{Torr}:=\frac1{760}\ \mathrm{atm} \approx 1\ \mathrm{mmHg} \approx 133.3224\ \mathrm{Pa} \\ \text{pound-force per square inch} & \text{psi}, \mathrm{lbf/in^2} & 1\ \mathrm{lbf/in^2}\approx6\,894.757\ \mathrm{Pa} \\ \hline \end{array}$$

Note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted.

For example, write “the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”.

(The symbol $p_\mathrm e$ is recommended for gauge pressure, defined $p-p_\mathrm{amb}$, where $p_\mathrm{amb}$ is the ambient pressure. Thus the gauge pressure is positive or negative depending on whether $p$ is larger or smaller than $p_\mathrm{amb}$.)

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Choose either the unit you are asked for (as in hwk or exam) or what is useful. For instance if you want to compare pressures use the same units. If the pressures are large you might use kPa, if small, mTorr, etc. You may seek to avoid dealing with large multipliers (ie 3 bar rather than $3 \times 10^5$ Pa).

Some equations in physics imply use of a particular system of units e.g. in electromagnetic theory, where equations can change depending on whether SI or Gaussian units are used. However, I am not aware of a similar situation where equations change depending on the selection of particular pressure units (perhaps when using Planck units in cosmology?). Of course the numerical value of the gas constant R depends on the units in which it is expressed, but it remains constant. Also, you'll find that most common pressure units can be related to the SI unit system by fixed constant factors (or ratio). The exception might be mm Hg (I now understand that both 1 Torr and 1 mm Hg are fixed quantities, see eg this post, but not exactly equal).

Basically different units are more convenient in different contexts depending on the magnitude of what you are discussing, but the same rule always applies: be meticulous and consistent. This means simply that you keep track of the units in use, and perform conversions as required when evaluating the result of an expression.

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In the formula Delta G° = - RT lnK, the numerical value of the equilibrium constant K must be in atmospheres.

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    $\begingroup$ I'm afraid this is not correct and addresses the question only tangentially. $\endgroup$ – andselisk Nov 23 '19 at 11:53
  • $\begingroup$ I’m afraid this is not correct although if it were it would be a full answer to the question. $\endgroup$ – Jan Nov 26 '19 at 10:01

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