The density of a mixture

- Two components of equal mass but different volume - 

Knowing the density of such a mixture is a common strange situation in engineering. This demonstration can be extended to more components provided the mass of each component is the same.

The situation

Consider that you have a solution, of a solid with certain density $\rho_S$ dilute in certain solvent of density $\rho_L$, in which both the solute and solvent are mixed in equal quantities. This is, the mass of the solute is equal to the mass of the solvent. What is the density of such a mixture?

An analytical approach

The solution to this case is not knew and can also be found elsewhere. However, in this post, some different flavor shall be given.

The density of the solute is,

$\rho_S=\dfrac{m_S}{V_S}$        Eq. (01)

while the density of the solvent is,

$\rho_L=\dfrac{m_L}{V_L}$        Eq. (02)

but since the mass of solute and solvent are equal, it follows,

$m=m_S=m_L$        Eq. (03)

Also, the density of the mixture should be,

$\rho=\dfrac{m_S+m_L}{V_S+V_L}=\dfrac{2m}{V_S+S_L}$       Eq. (04) 

The volumes, $V_S$ and $V_L$, in Eq .(04) can be determined from Eqs. (01-02) condering Eq. (03) as follows,

$V_S=\dfrac{m}{\rho_S}$        Eq. (05)

$V_L=\dfrac{m}{\rho_L}$        Eq. (06)

Substitution of Eqs. (05-06) into Eq. (04) produces,

$\rho=\dfrac{2m}{\dfrac{m}{\rho_S}+\dfrac{m}{\rho_L}}$

which can be simplified,

$\rho=\dfrac{2m}{\dfrac{\rho_L m+\rho_S m}{\rho_S \rho_L}}=\dfrac{2m}{\dfrac{m\left( \rho_L+\rho_S \right)}{\rho_S \rho_L}}=\dfrac{2\rho_S \rho_L}{\rho_L+\rho_S}$            Eq. (07)

Of course, Eq. (07) applies for the mixture of two liquids as well. Therefore, if you know the density of both substances, you can readily estimate the density of the mixture (provided the mass of the two components is the same).

A case to estimate the density of a mixture

Let us consider a slurry fluid made of the mixture of coal and water. The coal has specific gravity 2.5 while the slurry has composition 50% coal w/w. What is the density of the slurry?

Well, the slurry mixture is made of coal and water mixed in the same mass proportion. This is, the mass of the coal is the same as the mass of the water used to make the slurry. The volume is unknown: it can be the same but, who knows.

Starting with the specific gravity of the coal (solute), its density should be,

$\rho_S=\left(2.5\right) \left( 1000\,kg/m^3 \right)=2500\,kg/m^3$

For the water, let us consider that it is at room temperature (say, 25 °C). Similar temperatures around this one will give a very similar water density, so that you do not have to worry too much. Then,

$\rho_L=997.05\, kg/m^3$

Therefore, the density of such a slurry should be,

$\rho=1425.56\, kg/m3$

This is the end of the post. I hope you find it useful.

Ildebrando.

On the vapor pressure data for different NaCl dilutions at different temperatures

 The data presented in this post were extracted from the International Critical Tables.

As is usual in evaporation operations the boiling temperature elevation (BPE) is a key data for engineering calculations. Then, the vapor pressure for different combinations of solute concentration and temperatures are to be combined with the Duhring approximation (Duhring's lines).

The data was just rewritten from the source previously mentioned.

	Vapor pressure, mm Hg
Wt %	0.0	2.5	5.0	7.5	10.0	12.5	15.0	17.5	20.0	22.5	25.0	27.5
t °C
0	4.579	4.5	4.4	4.4	4.3	4.2	4.1	4.0	3.8	3.7	3.5
10	9.21	9.1	8.9	8.8	8.6	8.4	8.2	8.0	7.7	7.4	7.1
20	17.54	17.3	17.0	16.7	16.4	16.1	15.7	15.3	14.8	14.2	13.6
30	31.83	3.4	30.9	30.4	29.8	29.2	28.5	27.7	26.8	25.8	24.7
40	55..34	54.5	53.6	52.7	51.7	50.7	49.5	48.1	46.6	44.9	43.0
50	92.54	91.2	89.7	88.1	86.4	84.7	82.8	80.5	78.1	75.3	72.2
60	149.46	147.2	144.8	142.3	139.7	136.8	133.7	130.0	126.0	121.7	116.8
70	233.79	230.2	226.4	222.4	218.3	213.9	208.9	203.5	197.5	190.7	183.1
80	355.47	350	344	338	332	325	318	309.5	300.5	290.2	278.9	266
90	526	517	509	500	491	481	470	458	445.1	430	414	395
100	760	748	736	723	710	695	680	665	643	622	599	572
110	1075.4	1057	1040	1022	1003	983	961	936	911	881	849	810
B. P., °C	100	100.44	100.9	101.4	101.93	102.51	103.16	103.89	104.72	105.68	106.78	108.12

On the heat capacity for NaCl solutions

 This post is based on the results published by James C. S. Chou and Allen M. Rowe Jr. in their paper Desalination, 6(1969) 105-115.

For enthalply calculations the heat capacity is key. However, this last parameter depends on temperature and pressure, and in the case of dilutions on the concentration of solute.

From the thermodynamical point of view the formal relationship between enthalpy and heat capacity is expressed as,

$h=h_0+\int_{T_0}^Tc_P\,dT+\int_{P_0}^P\left[ v-T\left( \dfrac{\partial v}{\partial T} \right)_P \right]dT$    Eq. (01)

where the subscript $0$ indicates a reference data or condition that must be known in order to estimate the data at another set of conditions (without subscripts). $c_P$ and $v$ are the heat capacity at constant pressure and the specific volume, respectively. Chou and Rowe provide math expressions for these two parameters (fortunately):

$c_P=1.3041791 - 8.1519942x + 16.203997x^2 -\left( 0.19159475\times 10^{-2}\right.$

$\left. -0.029952864x+0.0037589577x^2\right)T+\left( 0.29944976\times 10^{-5} \right.$

$\left. -0.498581\times 10^{-4}x-0.89329066\times 10^{-6}x^2 \right)T^2$        Eq. (02)

where $x$ is the mole fraction and the temperature is given in $K$. The units of $c_P$ are $cal/g\,C$. The specific volume is:

$v=A(T)-P\, B(T)-P^2\, C(T)+w\, D(T)+w^2\, E(T)-wP\,F(T)$

$-w^2P\, G(T)-\dfrac{1}{2}wP^2\, H(T)$        Eq. (03)

where $w$ is the salt weight fraction in the solution and the remperature $T$ must be given in $K$. The $A$ through $H$ temperature functions are defined as,

$A(T)=5.916365-0.010357941T+0.92700482\times 10^{-5}T^2$

$-\dfrac{1127.5221}{T}+\dfrac{100674.1}{T^2}$

$B(T)=0.52049144\times 10^{-2}-0.10482101\times 10^{-4}T+0.83285321\times 10^{-8}T^2$

$-\dfrac{1.1702939}{T}+\dfrac{102.27831}{T^2}$

$C(T)=0.11854697\times 10^{-7}-0.65991434\times 10^{-10}T$

$D(T)=2.5166005+0.011176552T-0.17055209\times 10^{-4}T^2$

$E(T)=2.8485101-0.015430471T+0.22398153\times 10^{-4}T^2$

$F(T)=-0.0013949422+0.77922822\times 10^{-5}T-0.17736045\times 10^{-7}T^2$

$G(T)=0.0024223209-0.13698670\times 10^{-4}T+0.20303356\times 10^{-7}T^2$

$H(T)=0.55541298\times 10^{-6}-0.36241535\times 10^{-8}T+0.60444040\times 10^{-11}T^2$

Finally, for the purpose of a reference situation we may take the data of enthalpy at $25\,C$ and pressure of $1\,atm$. Again, Chou and Rowe provide an expression for it in the range of salt weight fraction $w$: $28.8524\%$ to $0.0006\%$. This is,

$h_0=24.953(1-w)+30.805561w^{1.5}-161.50632w^2$

$+79.059598w^{2.5}+114.83149w^3$        Eq. (04)

where the enthalpy $h_0$ is given in $cal/g\,solution$.

Como calcular la densidad de una solución

 En el caso de una solución acuosa de ácido acético al 9% en peso a una temperatura de 17 °C.

Procedimiento de cálculo

Primero que nada, nótese que por cada 100 gr de solución se tienen 9 gr de C$_2$H$_4$O$_2$. Por lo tanto, debe haber también 91 gr de H$_2$O. Luego, por separado buscamos las densidades del ácido acético y del agua a 17 °C,

Densidad C$_2$H$_4$O$_2$ = 1.05 gr/cm$^3$
Densidad H$_2$O = 0.9987 gr/cm$^3$

Calculamos ahora el volumen de C$_2$H$_4$O$_2$ y de H$_2$O

Volumen C$_2$H$_4$O$_2$ 

$= \dfrac{9 \text{ gr}}{(1.05 \text{ gr/cm}^3)} = 8.58 \text{ cm}^3$ de C$_2$H$_4$O$_2$

Volumen H$_2$O 

$= \dfrac{91 \text{ gr}}{0.9987 \text{ gr/cm}^3} = 91.1118 \text{ cm}^3$ de H$_2$O

La masa total de la solución son 100 gr y el volumen total de la solución es:

 $\left( 8.58+91.1118 \right)$ cm$^3$ = 99.6918 cm$^3$ 

Por la tanto, la densidad de la solución $\rho_{sol}$ es

$\rho_{sol}= \dfrac{100 \text{ gr}}{99.69118 \text{ cm}^3}$

= 1.00309 gr/cm$^3$

= 1003.09 kg/m$^3$

Any question? Write in the comments and I shall try to help.

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• Who was Clausius?
• Units conversion
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• About PID controllers
• Ways to control a process
• About pilot lights
• Solving the Colebrook equation

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Ildebrando.