due to the neutral gas. This result, which may be extended to any number of neutral gases, is simply an expression of Dalton's Law.

We now proceed to the comparison of the formulæ, especially of equation (6), with the results of experiment.

Table I.—Peroxide of Nitrogen.
Experiments at Atmospheric Pressure.
Mitscherlich,—R. Müller,—Deville and Troost.

Temperature		Pressure		Density calculated by eq. (10).		Density observed.					Excess of observed density.
							Deville & Troost					Deville & Troost
						M—h.	${\displaystyle \scriptstyle {\overbrace {\quad \quad \quad \quad \quad } }}$				M—h.	${\displaystyle \scriptstyle {\overbrace {\quad \quad \quad \quad \quad \quad } }}$
						M—r.	I.	II.	III.	M—r.	I.	II.	III.
183.2		(760)		1.592					1.57					-.022
154.0		(760)		1.597					1.58					-.017
151.8		(760)		1.598			1.50					-.10
135.0		(760)		1.607				1.60						-.007
121.8		(760)		1.622			1.64					+.02
121.5		(760)		1.622					1.62					-.002
111.3		(760)		1.641					1.65					+.009
100.25		760		1.677		1.72					-.04
100.1		(760)		1.676					1.68					+.004
100.0		(760)		1.677			1.71					+.03
90.0		(760)		1.728					1.72					-.008
84.4		(760)		1.768			1.83					+.06
80.6		(760)		1.801					1.80					-.001
79		748		1.814		1.84					+.03
77.4		(760)		1.833				1.85					+.02
70.0		(760)		1.920					1.92					.000
70		754.5		1.919		1.95					+.03
68.8		(760)		1.937			1.99					+.05
66.0		(760)		1.976				2.03					+.05
60.2		(760)		2.067					2.06					-.013
55.0		(760)		2.157				2.20					+.04
52		757		2.211		2.26					+.05
49.7		(760)		2.255			2.34					+.09
49.6		(760)		2.256					2.27					+.014
45.1		(760)		2.342				2.40					+.06
39.8		(760)		2.443					2.46					+.017
35.4		(760)		2.524					2.53					+.006
35.2		(760)		2.528			2.66					+.13
34.6		(760)		2.539				2.62					+.08
32		748		2.582		2.65					+.07
28.7		(760)		2.642			2.80					+.16
28		751		2.652		2.70					+.05
27.6		(760)		2.661				2.70					+.04
26.7		(760)		2.676					2.65					+.026

Peroxide of nitrogen.—If we take the constants of the equation for this substances from the paper already cited,^[1] we have

${\displaystyle \log {\frac {15.89({\text{D}}-1.589)}{(3.178-{\text{D}})^{2}}}={\frac {3118.6}{t_{\text{C}}+273}}+\log p-12.451,}$

(10)

${\displaystyle t_{\text{C}}}$ denoting the temperature on the centigrade scale. The numbers 3.178 and 1.589 represent the theoretical densities of N₂O₄ and NO₂