CHAPTER LXII. GAS VOLUMES AND WEIGHTS.

GAS VOLUMES AND WEIGHTS.

343. Oxygen.

Experiment 134.--Weigh accurately, using delicate balances, 5 g.
KClO3, and mix with the crystals 1 or 2 g. of pure powdered MnO2.
Put the mixture into a t.t. with a tight-fitting cork and
delivery-tube, and invert over the water-pan, to collect the gas,
a flask of at least one and a half liters' capacity, filled with
water. Apply heat, and, without rejecting any of the gas, collect
it as long as any will separate.

Then press the flask down into the water till the level in the
flask is the same as that outside, and remove the flask, leaving
in the bottom all the water that is not displaced. Weigh the
flask with the water it contains; then completely fill it with
water and weigh again.

Subtract the first weight from the second, and the result will
evidently be the weight of water that occupies the same volume as
the O collected. This weight, if expressed in grams, represents
approximately the number of cubic centimeters of water,--since 1
cc. of water weighs lg,--or the number of cubic centimeters of O.

At the time the experiment is performed the temperature should be
noted with a centigrade thermometer, and the atmospheric pressure
with a barometer graduated to millimeters.

Suppose that we have obtained 1450 cc. of O, that the temperature
is 27 degrees, and the pressure 758 mm.; we wish to find the
volume and the weight of the gas at 0 degrees and 760 mm.

According to the law of Charles--the volume of a given quantity
of gas at constant pressure varies directly as the absolute
temperature. To reduce from the centigrade to the absolute scale,
we have only to add 273 degrees. Adding the observed temperature,
we have 273 degrees + 27 degrees = 300 degrees. Applying the
above law to O obtained at 300 degrees A, we have the proportion
below. Since the volume of O at 273 degrees will be less than it
will at 300 degrees, the fourth term, or answer will be less than
the third, and the second term must be less than the first. 300 :
273 :: 1450 : x. This would give the result dependent upon
temperature alone.

By the law of Mariotte - Physics, - the volume of a given
quantity of gas at a constant temperature varies inversely as the
pressure. Applying this law to the O obtained at 758mm, we have
the following proportion. The volume at 760mm will be less than
at 758mm; or the fourth term will be less than the third; hence
the second must be less than the first. 760: 758:: 1450: x. This
would give the result dependent on pressure alone.

Combining the two proportions in one:--

300: 273 ):: 1450: x = 1316cc.
760: 758 )

1316cc=1.316 liters. It remains to find the weight of this gas. A liter of
H weighs 0.0896g. The vapor density of O is 16. Hence 1.316 liters of O
will weigh 1.316 X 16 X 0.0896 =1.89g.

(KClO3 = KCl + O3)
From the equation (122.5 48) we make a proportion,
( 5 x)

122.5: 5:: 48: x = 1.95, and obtain, as the weight of O contained in
5g of KClO3, 1.95g. The weight we actually,obtained was 1.89g. This
leaves an error of 0.06g, or a little over 4 per cent of error (0.06 / 1.95
= 0.03 +). The percentage of error, in performing this experiment,
should fall within 10.

Some of the liabilities to error are as follows:--


1. Impure MnO2, which sometimes contains C. CO2 is soluble m H2O.

2. Solubility of O in water.

3. Escape of gas by leakage.

4. Moisture taken up by the gas.

5. Difference between the temperature of the gas and that of the
air in the room.

6. Errors in weighing.

7. Want of accuracy in the weights and scales.

344. Hydrogen.

Experiment 135.--Weigh 5g, or less of sheet or granulated Zn, and
put it into a small flask provided with a thistle-tube and a
delivery-tube. Cover the Zn with water, and introduce through the
thistle-tube measured quantities of HCl, a few cubic centimeters
at a time. Collect the H over water in large flasks, observing
the same directions as in removing O. Weigh the water, compute
the volume of the gas, reduce it to the standard, and obtain the
weight, as before. Should any Zn or other solid substance be
left, pour off the water or filter it, weigh the dry residue, and
deduct its weight from that of the Zn originally taken. Suppose
the residue to weigh 0.5g. Make and solve the proportion from the
equation:-


Zn + 2HCl = ZnCl2 + 2H.
65 2.
4.5 x.


Compute the percentage of errcr, as in the case of O. If the
purity of the HCl be known, i.e. the weight of HCl gas in one
cubic centimeter of the liquid, a proportion can be made between
HCl and H, provided no free HCl is left in the flask. State any
liabilities to error in this experiment.

PROBLEMS.

(1) A gas occupies 2000cc.when the barometer stands
750mm. What volume will it fill at 760mm?

(2) At 750mm my volume of O is 4 1/2 liters. What will it be at
730mm?

(3) At 825mm?

(4) At 200mm?

(5) Compute the volume of a gas at 70 degrees, which at 30
degrees is 150cc.

(6) At 0 degrees I have 3000cc.of O. What volume will it occupy
at 100 degrees?

(7) I fill a flask holding 2 litres with H. The thermometer
indicates 26 degrees, the barometer 762mm. What is the volume of
the gas at 0 degrees and 760mm?

If the volumes of gases vary as above, it is evident that their
vapor densities must vary inversely. A liter of H at 0 degrees
weighs 0.0896. What will a liter of H weigh at 273 degrees? At
273 degrees the one liter has be- come two liters, one of which
weighs 0.0448 (= 0.0896 / 2). The vapor density of a gas is
inversely proportional to the temperature. Also, the vapor
density is directly proportional to the pressure, since a liter
of any gas under a pressure of one atmosphere is reduced to half
a liter under two atmospheres.

PROBLEMS.

(1) Find the weight of a liter of O at 0 degrees; then compute the
weight of a liter at 27 degrees.

(2) Find the weight of 500cc.of N2O at 60 degrees.

(3) Of 200 cc. of CO at -5 degrees.

(4) A given volume of O weighs 0.25g at a pressure of 750mm; find
the weight of a like volume of O at 758mm.