VOLUMETRIC ASSAYS.
These have been already described as those in which the results are got
by measuring, either--(1) the volume of a reagent required to complete
some reaction, or (2) the volume of the resulting product. For example,
if a permanganate of potash solution be added to a solution containing a
weighed amount of iron, dissolved in sulphuric acid, the strong colour
of the permanganate of potash will be removed until a certain quantity
of it has been added. Repeating the experiment, it will be found that
the same amount of iron decolorises the same volume of the permanganate
solution within certain narrow limits of variation, known as "error of
experiment." This error is due to variation in the method of working and
to slight differences in the weighings and measurings; it is present in
all experimental methods, although the limits of variation are wider in
some than in others. Apart from this error of experiment, however, it is
certain that a given volume of the permanganate of potash solution
corresponds to a definite weight of iron, so that if either is known the
other may be calculated. Similarly, if a known weight of zinc (or of
carbonate of lime) be dissolved in hydrochloric acid, a gas will be
given off which can be measured, and so long as the conditions of the
experiment do not vary, the same weight of zinc (or of carbonate of
lime) gives off the same volume of gas. The weight of the one can be
determined from the volume of the other.
Or, again, the quantity of some substances may be measured by the colour
of their solutions, on the principle that, other things being equal, the
colour of a solution depends upon the quantity of colouring matter
present. So that if two solutions of the same substance are equally
coloured they are of equal strength. In this way an unknown may be
compared with a known strength, and a fairly accurate determination may
be made. These three illustrations serve as types of the three chief
classes of volumetric assays--titrometric, gasometric, and colorimetric.
~Titrometric Assays.~--Within the limits of the error of experiment, a
definite volume of a solution or gas represents a certain weight of
metal or other substance, hence the exact weight may be determined by
experiment. The error of experiment may be reduced to insignificant
dimensions by repeating the experiment, and taking the mean of three or
four determinations. This will at the same time show the amount of
variation. Thus, if 0.5 gram of iron were dissolved and found to require
50.3 cubic centimetres of the solution of permanganate of potash, and if
on repeating, 50.4, 50.2, and 50.3 c.c. were required, the experimenter
would be justified in saying that 50.3 c.c. of the permanganate solution
represent 0.5 gram of iron, and that his results were good within 0.2
c.c. of the permanganate solution. So that if in an unknown solution of
iron, 50.5 c.c. of the permanganate solution were used up, he could
state with confidence that it contained a little more than 0.5 gram of
iron. With a larger experience the confidence would increase, and with
practice the experimental error will diminish.
But supposing that the unknown solution required, say, 100.5 instead of
50.5 c.c., he would not be justified in saying that, since 50.3 c.c. are
equivalent to 0.5 gram, 100.6 c.c. are equivalent to twice that amount;
and that, consequently, the unknown solution contained a little less
than 1 gram of iron; or, at least, he could not say it except he (or
some one else) had determined it by experiment. But if on dissolving 1
gram of iron, he found it to require 100.6 c.c. of the solution, and in
another experiment with 0.8 gram of iron that 80.5 c.c. of the solution
were required, he would be justified in stating that _the volume of
solution required is proportional to the quantity of metal present_.
There are a large number of volumetric assays of which this is true, but
that it is true in any particular case can only be proved by experiment.
Even where true it is well not to rest too much weight upon it, and in
all cases the quantity of metal taken, to determine the strength of the
solution used, should not differ widely from that present in the assay.
There are certain terms which should be explained here. When the
solution of a reagent is applied under such conditions that the volume
added can be correctly determined, the operation is called "titrating,"
the solution of the reagent used the "standard solution," and the
process of determining the strength of the standard solution is
"standardising." The "standard" is the quantity of metal equivalent to
100 c.c. of the standard solution.
~Standard Solutions.~--In making these the salt is accurately weighed
and transferred to a litre flask, or to the graduated cylinder, and
dissolved. The method of dissolving it varies in special cases, and
instructions for these will be found under the respective assays.
Generally it is dissolved in a small quantity of liquid, and then
diluted to the mark. For those substances that require the aid of heat,
the solution is made in a pint flask, cooled, and transferred; after
which the flask is well washed out. After dilution, the liquids in the
measuring vessel must be thoroughly mixed by shaking. This is more
easily and better done in the cylinder than in the litre flask. The
solution is next transferred to a dry "Winchester" bottle and labelled.
The label may be rendered permanent by waxing it.
Standard solutions should not be kept in a place exposed to direct
sunlight. Oxidising and reducing solutions, such as those of
permanganate of potash, ferrous sulphate, iodine, hyposulphite of soda,
&c., gradually weaken in strength; the solutions of other salts are more
stable; while those of potassium bichromate and baric chloride are
almost permanent. Solutions of potassium permanganate may be kept for a
month or so without much change. The solutions of hyposulphite of soda
and of iodine should be examined weekly. Ferrous sulphate solutions, if
acidulated with sulphuric acid, may be depended on for two or three
weeks without fresh standardising. Before filling the burette, the
"Winchester" bottle should be well shaken and a portion of about 50 or
100 c.c. poured into a dry beaker or test-glass. Besides the standard
solutions, which are required for titrating an assay, permanent
solutions of the metal or acid of equivalent strength are very useful.
When the finishing point of a titration has been overstepped (_i.e._,
the assay has been "overdone"), a measured volume, say 5 or 10 c.c., of
a solution containing the same metal may be added. The titration can
then be continued, but more cautiously, and the value in "c.c." for the
quantity added be deducted from the final reading.
~Standardising.~--Suppose the object is to standardise a solution of
permanganate similar to that referred to above. A convenient quantity of
iron (say 0.5 gram) would be weighed out, dissolved in dilute sulphuric
acid, and the solution titrated. Suppose 49.6 c.c. of the permanganate
solution are required, then
49.6 : 0.5 :: 100 : _x_
_x_ = 1.008 gram.
This result, 1.008 gram, is the "_standard_." When a gas is measured,
the standard may be calculated in the same way. For example: with 0.224
gram of zinc, 75.8 c.c. of gas were obtained. Then the quantity of zinc
equivalent to 100 c.c. of the gas is got by the proportion.
75.8 : 0.224 :: 100 : _x_
_x_ = 0.2955 gram.
Using the term "standard" in this sense, the following rules hold
good:--
To find the weight of metal in a given substance:--_Multiply the
standard by the number of c.c. used and divide by 100._ For example: a
piece of zinc was dissolved and the gas evolved measured 73.9 c.c. Then
by the rule, 0.2955×73.9/100 should give the weight of the piece of
zinc. This gives 0.2184 gram.
To find the percentage of metal in a given substance:--_Multiply the
standard by the number of c.c. used and divide by the weight of
substance taken._ For example: if 2 grams of a mineral were taken, and
if on titrating with the permanganate solution (standard 1.008) 60.4
c.c. were required, then (1.008×60.4)/2 = 30.44. This is the
percentage.
If the standard is exactly 1 gram, and 1 gram of ore is always taken,
these calculations become very simple. The "c.c." used give at once the
percentage, or divided by 100 give the weight of metal.
If it is desired to have a solution with a standard exactly 1.0 gram, it
is best first to make one rather stronger than this, and then to
standardise carefully. Divide 1000 by the standard thus obtained and the
result will be the number of c.c. which must be taken and be diluted
with water to 1 litre. For example: suppose the standard is 1.008, then
1000/1.008 gives 992, and if 992 c.c. be taken and diluted with water to
1000 c.c. a solution of the desired strength will be obtained. The
standard of this should be confirmed. A simpler calculation for the same
purpose is to multiply the standard by 1000; this will give the number
of c.c. to which 1 litre of the solution should be diluted. In the above
example a litre should be diluted to 1008 c.c.
It has been assumed in these rules that the titration has yielded
proportional results; but these are not always obtained. There can be no
doubt that in any actual re-action the proportion between any two
re-agents is a fixed one, and that if we double one of these then
exactly twice as much of the other will enter into the re-action; but in
the working it may very well be that no re-action at all will take place
until after a certain excess of one or of both of the re-agents is
present. In titrating lead with a chromate of potash solution, for
example, it is possible that at the end of the titration a small
quantity of the lead may remain unacted on; and it is certain that a
small excess of the chromate is present in the solution. So, too, in
precipitating a solution of silver with a standard solution of common
salt, a point is reached at which a small quantity of each remains in
solution; a further addition either of silver or of salt will cause a
precipitate, and a similar phenomenon has been observed in precipitating
a hydrochloric acid solution of a sulphate with baric chloride. The
excess of one or other of the re-agents may be large or small; or, in
some cases, they may neutralise each other. Considerations like these
emphasise the necessity for uniformity in the mode of working. Whether a
process yields proportional results, or not, will be seen from a series
of standardisings. Having obtained these, the results should be arranged
as in the table, placing the quantities of metal used in the order of
weight in the first column, the volumes measured in the second, and the
standards calculated in the third. If the results are proportional,
these standards will vary more or less, according to the delicacy of the
process, but there will be no apparent order in the variation. The
average of the standards should then be taken.
+-------------+---------------+----------+
| Weight. | Volume found. | Standard |
+-------------+---------------+----------+
| 0.2160 gram | 72.9 c.c. | 0.2963 |
| 0.2185 " | 73.9 " | 0.2957 |
| 0.2365 " | 79.9 " | 0.2959 |
| 0.2440 " | 82.3 " | 0.2964 |
| 0.2555 " | 85.9 " | 0.2974 |
+-------------+---------------+----------+
Any inclination that may be felt for obtaining an appearance of greater
accuracy by ignoring the last result must be resisted. For, although it
would make no practical difference whether the mean standard is taken as
0.2961 or 0.2963, it is well not to ignore the possibility that an error
of 0.4 c.c. may arise. A result should only be ignored when the cause of
its variation is known.
In this series the results are proportional, but the range of weights
(0.216-0.2555 gram) is small. All processes yield fairly proportional
results if the quantities vary within narrow limits.
As to results which are not proportional, it is best to take some
imaginary examples, and then to apply the lesson to an actual one. A
series of titrations of a copper solution by means of a solution of
potassic cyanide gave the following results:--
+---------------+---------------+-----------+
| Copper taken. | Cyanide used. | Standard. |
+---------------+---------------+-----------+
| 0.1 gram | 11.9 c.c. | 0.8403 |
| 0.2 " | 23.7 " | 0.8438 |
| 0.3 " | 35.6 " | 0.8426 |
| 0.4 " | 47.6 " | 0.8403 |
+---------------+---------------+-----------+
These are proportional, but by using a larger quantity of acid and
ammonia in the work preliminary to titration, we might have had to use
1 c.c. of cyanide solution more in each case before the finishing point
was reached. The results would then have been:
+---------------+---------------+-----------+
| Copper taken. | Cyanide used. | Standard. |
+---------------+---------------+-----------|
| 0.1 gram | 12.9 c.c. | 0.7752 |
| 0.2 " | 24.7 " | 0.8097 |
| 0.3 " | 36.6 " | 0.8191 |
| 0.4 " | 48.6 " | 0.8230 |
+---------------+---------------+-----------+
It will be noted that the value of the standard increases with the
weight of metal used; and calculations from the mean standard will be
incorrect.
By subtracting the lowest standardising from the highest, a third result
is got free from any error common to the other two; thus:--
0.4 gram = 48.6 c.c. "cyanide."
0.1 " = 12.9 " "
--- ----
0.3 " = 35.7 " "
And the standard calculated from this corrected result is 0.8404.
Further, if 0.3 gram requires 35.7 c.c., then 0.1 gram should require
11.9 c.c., or 1.0 c.c. less than that actually found.
We may therefore use the following rules for working processes which do
not yield proportional results. Make a series of two or three
titrations, using very different quantities of metal in each. Subtract
the lowest of these from the highest, and calculate the standard with
the remainder. Calculate the volume required by this standard in any
case, and find the excess or deficit, as the case may be. If an excess,
subtract it from the result of each titration; if a deficit, add it; and
use the standard in the usual way. The following table shows an actual
example:--
+--------------+---------------+-----------+
| Chalk taken. | Gas obtained. | Standard. |
+--------------+---------------+-----------+
| 0.0873 gram | 17.8 c.c. | 0.4904 |
| 0.1305 " | 27.3 " | 0.4780 |
| 0.1690 " | 35.8 " | 0.4721 |
| 0.1905 " | 40.4 " | 0.4715 |
| 0.2460 " | 52.5 " | 0.4686 |
| 0.3000 " | 64.0 " | 0.4687 |
+--------------+---------------+-----------+
It will be seen that the standard decreases as the quantity of chalk
increases; this points to a deficiency in the quantity of gas evolved.
Then
0.3000 = 64.0 c.c.
0.0873 = 17.8 "
------ = ----
0.2127 = 46.2 "
and 0.2127×100/46.2 = 0.4604. Then, multiplying the weight of chalk
taken by 100, and dividing by 0.4604, we get the calculated results of
the following table:--
+--------------+------------+-----------------+-------------+
| Chalk taken. | Gas found. | Gas calculated. | Difference. |
+--------------+------------+-----------------+-------------+
| 0.0873 gram | 17.8 c.c. | 18.9 c.c. | -1.1 c.c. |
| 0.1305 " | 27.3 " | 28.3 " | -1.0 " |
| 0.1690 " | 35.8 " | 36.7 " | -0.9 " |
| 0.1905 " | 40.4 " | 41.4 " | -1.0 " |
| 0.2460 " | 52.5 " | 53.4 " | -0.9 " |
| 0.3000 " | 64.0 " | 65.1 " | -1.1 " |
+--------------+------------+-----------------+-------------+
By adding 1 c.c. to the quantity of gas obtained, and taking 0.4604 as
the standard, the calculated results will agree with those found with a
variation of 0.1 c.c. When a large number of assays of the same kind are
being made, this method of calculation is convenient; when, however,
only one or two determinations are in question, it is easier to make a
couple of standardisings, taking quantities as nearly as possible the
same as those present in the assays.
Sometimes it is necessary to draw up a table which will show, without
calculation, the weight of substance equivalent to a given volume of gas
or of solution. The substance used for standardising should be, whenever
possible, a pure sample of the substance to be determined--that is, for
copper assays pure copper should be used, for iron assays pure iron, and
so on; but when this cannot be got an impure substance may be used,
provided it contains a known percentage of the metal, and that the
impurities present are not such as will interfere with the accuracy of
the assay. Including compounds with these, the standard may be
calculated by multiplying the standard got in the usual way, by the
percentage of metal in the compound or impure substance, and dividing by
100. If, for example, the standard 1.008 gram was obtained by using a
sample of iron containing 99.7 per cent. of metal, the corrected
standard would be 1.008×99.7/100 = 1.005.
In volumetric analysis the change brought about must be one in which the
end of the reaction is rendered prominent either by a change of colour
or by the presence or absence of a precipitate. If the end of the
reaction or finishing-point is not of itself visible, then it must be
rendered visible by the use of a third reagent called an indicator.
For example, the action of sulphuric acid upon soda results in nothing
which makes the action conspicuous; if, however, litmus or
phenolphthalein be added the change from blue to red in the first case,
or from red to colourless in the second, renders the finishing-point
evident. Some indicators cannot be added to the assay solution without
spoiling the result; in which case portions of the assay solution must
be withdrawn from time to time and tested. This withdrawal of portions
of the assay solution, if rashly done, must result in loss; if, however,
the solution is not concentrated, and if the portions are only withdrawn
towards the end of the titration, the loss is very trifling, and will
not show-up on the result. The usual plan adopted is to have a solution
of the indicator placed in drops at fairly equal intervals distributed
over a clean and dry white porcelain-plate: a drop or two of the
solution to be tested is then brought in contact with one of these and
the effect noted. Another plan is to have thin blotting-paper, moistened
with a solution of the indicator and dried; a drop of the solution to be
tested placed on this shows the characteristic change. When the assay
solution contains a suspended solid which interferes with the test, a
prepared paper covered with an ordinary filter-paper answers very well;
a drop of the solution to be tested is placed on the filter-paper, and,
sinking through, shows its effect on the paper below.
Except when otherwise stated, all titrations should be made at the
ordinary temperature; cooling, if necessary, by holding the flask under
the tap. When a titration is directed to be made in a boiling solution,
it must be remembered that the standard solution is cold, and that every
addition lowers the temperature of the assay.
On running the solution from the burette into the assay, do not let it
run down the side of the flask. If a portion of the assay has to be
withdrawn for testing, shake the flask to ensure mixing, and then take
out a drop with the test-rod; the neglect of these precautions may give
a finishing-point too early. This is generally indicated by a sudden
finish, in which case on shaking the flask and again testing no reaction
is got. Do not remove the drop on the point of the burette with the
test-rod; let it remain where it is or drop it into the solution by
carefully opening the clip.
Generally the methods of working are as follows:--
(1) _When the finishing-point depends on a change of colour in the
solution._--Increase the bulk of the assay up to from 100 to 150 c.c.
with water. Boil or cool, as the case may be. Run in the standard
solution from a burette speedily, until the re-agent appears to have a
slower action, and shake or stir all the time. Then run 1 c.c. or so at
a time, still stirring, and finally add drops until the colour change is
got.
(2) _When an outside-indicator is used._--Pour the standard solution
from a burette into the assay until 5 or 6 c.c. from the
finishing-point; then run in 1 c.c. at a time (stirring and testing on
the plate between each) until the indicator shows the change wanted, and
deduct 0.5 c.c. for excess. When greater accuracy is sought for a
duplicate assay is made. In this case the standard solution is run in
close up to the end, and the operation is finished off with a few drops
at a time.
(3) _Where the finishing-point depends upon the absence of a precipitate
and no outside-indicator is used._--As in the last case, run in the
standard solution up to within a few c.c. of the end, then run in 1 c.c.
at a time until a precipitate is no longer formed, but here 1.5 c.c.
must be deducted for excess, since it is evident that the whole of the
last "c.c." must have been, and a portion of the previous one may have
been, in excess.
~Indirect Titration.~--The action of permanganate of potash upon a
ferrous solution is one of oxidation, hence it is evident that if any
other oxidising agent is present it will count as permanganate. In such
a case the titration can be used (indirectly) to estimate the quantity
of such oxidising agent, by determining how much less of the
permanganate is used. For example, suppose that 1 gram of iron dissolved
in sulphuric acid requires 100 c.c. of standard permanganate to fully
oxidise it, but that the same amount of iron only requires 35.6 c.c. of
the same standard permanganate if it has been previously heated with 0.5
gram of black oxide of manganese. Here it is evident that 0.5 gram of
black oxide does the work of 64.4 c.c.[4] of the permanganate solution,
and that these quantities are equivalent; moreover, if 64.4 c.c.
correspond with 0.5 gram, then 100 c.c. correspond with 0.7764 which is
the standard. On theoretical grounds, and by a method of calculation
which will be explained further on (under the heading "Calculations from
Formulæ"), it can be found that if the standard for iron is 1 gram, that
for the black oxide will be 0.7764 gram.
The principles of these indirect titrations become clearer when
expressed in a condensed form. Thus, in the example selected, and using
the formulæ Fe = Iron, KMnO_{4} = permanganate of potash, and MnO_{2} =
oxide of manganese, we have:--
(1) 1 gram Fe = 100 c.c. KMnO_{4}
(2) 1 gram Fe = 35.6 c.c. KMnO_{4} + 0.5 gram MnO_{2}
.'. 100 c.c. KMnO_{4} = 35.6 c.c. KMnO_{4} + 0.5 gram MnO_{2}
(100-35.6) c.c. KMnO_{4} = 0.5 gram MnO_{2}
64.4 c.c. KMnO_{4} = 0.5 gram MnO_{2}
The iron does not enter into the calculation if the same quantity is
present in the two experiments.
An indirect titration thus requires three determinations, but if more
than one assay is to be carried on, two of these need not be repeated.
The standard is calculated in the usual way.
~Colorimetric Assays.~--These are assays in which the colour imparted to
a solution by some compound of the metal to be determined is taken
advantage of; the depth of colour depending on the quantity of metal
present. They are generally used for the determination of such small
quantities as are too minute to be weighed. The method of working is as
follows:--A measured portion of the assay solution (generally 2/3, 1/2,
1/3, or 1/4 of the whole), coloured by the substance to be estimated, is
placed in a white glass cylinder standing on a sheet of white paper or
glazed porcelain. Into an exactly similar cylinder is placed the same
amount of re-agents, &c., as the portion of the assay solution contains,
and then water is added until the solutions are of nearly equal bulk.
Next, a standard solution of the metal being estimated is run in from a
burette, the mixture being stirred after each addition until the colour
approaches that of the assay. The bulk of the two solutions is equalised
by adding water. Then more standard solution is added until the tints
are very nearly alike. Next, the amount added is read off from the
burette, still more is poured in until the colour is slightly darker
than that of the assay, and the burette read off again. The mean of the
readings is taken, and gives the quantity of metal added. It equals the
quantity of metal in the portion of the assay. If this portion was
one-half of the whole, multiply by two; if one-third, multiply by three,
and so on. When the quantity of metal in very dilute solutions is to be
determined, it is sometimes necessary to concentrate the solutions by
boiling them down before applying the re-agent which produces the
coloured compound. Such concentration does not affect the calculations.
~Gasometric Assays.~--Gasometric methods are not much used by assayers,
and, therefore, those students who wish to study them more fully than
the limits of this work will permit, are recommended to consult Winkler
and Lunge's text-book on the subject. The methods are without doubt
capable of a more extended application. In measuring liquids, ordinary
variations of temperature have but little effect, and variations of
atmospheric pressure have none at all, whereas with gases it is
different. Thus, 100 c.c. of an ordinary aqueous solution would, if
heated from 10° C. to 20° C., expand to about 100.15 c.c. 100 c.c. of a
gas similarly warmed would expand to about 103.5 c.c., and a fall of one
inch in the barometer would have a very similar effect. And in
measuring gases we have not only to take into account variations in
volume due to changes in temperature and atmospheric pressure, but also
that which is observed when a gas is measured wet and dry. Water gives
off vapour at all temperatures, but the amount of vapour is larger as
the temperature increases.
By ignoring these considerations, errors of 3 or 4 per cent. are easily
made; but, fortunately, the corrections are simple, and it is easy to
construct a piece of apparatus by means of which they may be reduced to
a simple calculation by the rule of three.
The volume of a gas is, in practice, usually reduced to that which it
would be at a temperature of 0° C., when the column of mercury in the
barometer is 760 mm. high. But, although convenient, this practice is
not always necessary. The only thing required is some way of checking
the variations in volume, and of calculating what the corrected volume
would be under certain fixed conditions.
Suppose that at the time a series of standardisings is being made, 100
c.c. of air were confined in a graduated tube over moist mercury. These
100 c.c. would vary in volume from day to day, but it would always be
true of them that they would measure 100 c.c. under the same conditions
as those under which the standardisings were made. If, then, in making
an actual assay, 35.4 c.c. of gas were obtained, and the air in the tube
measured 105 c.c., we should be justified in saying, that if the
conditions had been those of the standardising, the 105 c.c. would have
measured 100 c.c., and the 35.4 c.c. would have been 33.7; for 105:
100:: 35.4: 33.7. The rule for using such a piece of apparatus for
correcting volumes is:--_Multiply the c.c. of gas obtained by 100, and
divide by the number of c.c. of air in the apparatus._
If it is desired to calculate the volumes under standard conditions
(that is, the gas dry, at 0° C. and 760 mm. barometric pressure) the
calculations are easily performed, but the temperature and pressure must
be known.
_Correction for Moisture._--The "vapour tension" of water has been
accurately determined for various temperatures, and it may be looked
upon as counteracting the barometric pressure. For example, at 15° C.
the vapour tension equals 12.7 millimetres of mercury; if the barometer
stood at 750 mm., the correction for moisture would be made by
subtracting 12.7 from 750, and taking 737.3 mm. to be the true
barometric pressure.
The vapour tensions for temperatures from 0° C. to 20° C. are as
follows:--
-------+----------++-------+----------++-------+----------
Temp. | Tension. || Temp. | Tension. || Temp. | Tension.
-------+----------++-------+----------++-------+----------
0° | 4.6 mm. || 7° | 7.5 mm. || 14° | 11.9 mm.
1° | 4.9 mm. || 8° | 8.0 mm. || 15° | 12.7 mm.
2° | 5.3 mm. || 9° | 8.6 mm. || 16° | 13.5 mm.
3° | 5.7 mm. || 10° | 9.2 mm. || 17° | 14.4 mm.
4° | 6.1 mm. || 11° | 9.8 mm. || 18° | 15.3 mm.
5° | 6.5 mm. || 12° | 10.5 mm. || 19° | 16.3 mm.
6° | 7.0 mm. || 13° | 11.2 mm. || 20° | 17.4 mm.
-------+----------++-------+----------++-------+-----------
The _correction for pressure_ is:--Multiply the volume by the actual
pressure and divide by 760.
The _correction for temperature_:--Multiply the volume by 273 and divide
by the temperature (in degrees Centigrade) added to 273.
For all three corrections the following rules hold good. _To reduce to
0° C. and 760 mm. dry._
Volume × 0.3592 × (Pressure-tension)
Corrected volume = --------------------------------------
Temperature + 273
To find the volume, which a given volume under standard conditions would
assume, if those conditions are altered.
Volume × 2.784 × (Temperature + 273)
Resulting volume = ------------------------------------
Pressure - tension
As an example, we will suppose that it is desired to enclose in the
apparatus referred to on p. 45, a volume of air, which, when dry (at 0°
C. and 760 mm.), shall measure 100 c.c., whilst the actual temperature
is 15° C., and the pressure 750 mm.
The second formula is the one to be used, and we get 108.7 c.c.
100 c.c.×2.784×288
Required volume = ----------------------
750-12.7
80179.2
= -------
737.3
= 108.7 c.c.
FOOTNOTES:
[4] 100-35.6 = 64.4.