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 $cal/g\,solution$.

Enthalpies in the steam tables

 Enthalpy is a thermodynamic concept that can be hard to get a grip on. Perhaps, the reason is that it appears in several contexts so that understanding of it becomes confusing.

In this post the usage of enthalpy in the context of steam is presented. This is an interesting case since steam services are present in almost every chemical process industry and different daily applications, such as the heating of buildings, for example.

On the other hand, the interpretation of enthalpy, among the professionals of the steam, is as  thermal energy (as simple as that). Why? Because, engineers and technician need something easy to use and measurable rather than something that makes their work harder.

Some parameters used in steam tables

Data in steam table are usually order either by pressure or by temperature. Also there are different two types of steam: saturated and superheated. For simplicity, we will only refer to saturated steam tables.

For reference, consult the ASME Steam tables. Compact edition book.

Fig. 01 Sample steam tables by temperature and pressure. These tables were taken from the ASME Steam Tables. Compact edition book.

The steam tables present data for volume, enthalpy and pressure. In this post, we will focus on volume and enthalpy only.

Volume of liquid and vapor

These parameters, appearing in the columns 3 and 4 in tables of Fig. 01, partly indicate how much liquid water is present in the steam. Unless the steam is completley dry the steam will be a two phase fluid made of: vapor and liquid (in the form of small droplets). As you  can see from the data in tables of Fig. 01, the saturated steam is not dry at all.

This volume is presented as [volume of liquid or vapor]/[unit mass of steam]. In other words, and for the present case, this is the $ft^3$ of liquid or vapor per $lb_m$ of steam. Also, the summation $V_L+V_V$ is also called the specific volume.

The enthalpies

In Fig. 01 only two enthalpies ares shown: $h_L$ and $h_V$. However, a third enthalpy parameter usually appear in this kind of tables: $h_{evap}$. Theses parameters can be referred as,

• $h_L$ the enthalpy of the liquid phase,
• $h_V$ the enthalpy of the vapor phase and
• $h_{evap}$ the enthalpy of evaporation,

all in the same units: $Btu/lb_m$ or $kJ/kg$ (for SI units). In fact,

$h_{evap}=h_V - h_L$

For illustration purposes, let us consider the first row, for steam at $32\,F$, in Table 1 of Fig. 01. Thus, $h_L$ is interpreted as the energy required to increase the temperature of $1\,lb_m$ of water (or $1\,kg$ of water, in SI) from its freezing point to its boiling temperature $32\,F$ for the present case. Notice, that $h_L$ will only make sense if the water remains liquid.

For the same row in Table 1 in Fig. 01, $h_{evap}$ is interpreted as the energy required for the change of phase of $1\,lb_m$ (or $1\,kg$ in SI) of liquid water at its boling temperature into vapor. Notice, that either liquid or vapor are at $32\,F$, the boiling temperature. This amount of energy referred by the steam professionals as the one that can be used for heating purposes since the steam can very quickly transfer it.

Finally, $h_V$ is interpreted as the thermal energy carried by the vapor. This is the energy of the liquid at its boiling temperature plus an added energy so that the fluid can be in vapor phase.

One parameter I have no mentiones is the pressure. All above comments are for constant pressure. If you change the pressure, the values of temperature, volume and enthalpies will also change but the interpretation will be the same.

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

Other stuff of interest

• Understanding the law of Boyle
• Measuring atmospheric pressure
• Some examples of temperature instruments
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables
• Composition variables for mixtures

==========
Ildebrando.

a and b constants data for the van der Waals equation

 Here are some data for constants a and b appearing in the van der Waals equation:

Gas	Formula	a (atm L2/mol2)	b (L/mol)
Acetylene	C2H2	4.390	0.05136
Ammonia	NH3	4.170	0.03707
Carbon dioxide	CO2	3.592	0.04267
Ethane	C2H6	5.489	0.0638
Ethylene	C2H4	4.471	0.05714
Helium	He	0.03508	0.0237
Hydrogen	H2	0.244	0.0266
Hydrogen chloride	HCl	3.667	0.04081
Krypton	Kr	2.318	0.03978
Mercury	Hg	8.093	0.01696
Methane	CH4	2.253	0.0428
Neon	Ne	0.2107	0.01709
Nitric oxide	NO	1.340	0.02789
Nitrogen	N2	1.390	0.03913
Nitrogen dioxide	NO2	5.284	0.04424
Oxygen	O2	1.360	0.03183
Propane	C3H8	8.664	0.08445
Sulfur dioxide	SO2	6.714	0.05636
Xenon	Xe	4.194	0.05105
Water	H2O	5.464	0.03049

Please, be aware of the units.

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

Other stuff of interest

• Understanding the law of Boyle
• Measuring atmospheric pressure
• Some examples of temperature instruments
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables
• Composition variables for mixtures

==========
Ildebrando.

Does the state equation matter for Boyle's law?

 First, you need to know that the works on which Boyle's law are based are very old.

Boyle's law is due to the investigations of Robert Boyle by the years 1660! Follow this link if you want read the originial text.

On the other hand, the first equation of state, which is that for the ideal gas law,

$PV=n\text{R}T$

was assembled in 1834 by Emil Clapeyron. Notice, that I said assembled because, as in many fields of science, the ideal gas law is based on the findings of researchers before  1834, Robert Boyle included!

On  the other hand, the second most known equation of state is that of van der Waals,

$\left(P+\dfrac{an^2}{V^2}\right)\left(V-nb\right)=n\text{R}T$

in 1873! Then, what is clear at this point, in answer to the question, is that Boyle's law does not care about the state equation you are using.

Why is it that Boyle's law does not care about the state equation?

The answer is simple. Boyle's law describe the behavior of a system and although it was empirically determined it has been demonstrated its validity in many different systems. However, the state equations are based on Boyle's law (or the findings of Robert Boyle) and worst, the state equations (as the two mentioned early), not always work for every condition.

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

Other stuff of interest

• Understanding the law of Boyle
• Measuring atmospheric pressure
• Some examples of temperature instruments
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

==========
Ildebrando.

Some comments on the ideal gas constant $R$

 The constant $R$ in the ideal gas equation of state may have different values according to the unit system in use. For example

$R$	Units
SI units
8.314	$J\,K^{−1}\,mol^{−1}$
8.314	$m^3\,Pa\,K^{−1}\,mol^{−1}$
8.314	$kg\,m^2\,s^{-2}\,K^{−1}\,mol^{−1}$
Other units
8314.462	$L\,Pa\,K^{−1}\,mol^{−1}$
8.314	$L\,kPa\,K^{−1}\,mol^{−1}$
$8.314\times10^{-2}$	$L\,bar\,K^{−1}\,mol^{−1}$
$8.314\times10^{7}$	$erg\,K^{−1}\,mol^{−1}$
$7.302\times10^{-1}$	$atm\,ft^3\,lb_{mol}^{−1}\,R^{−1}$
10.731	$psi\,ft^3\,lb_{mol}^{−1}\,R^{−1}$
1.986	$BTU\,lb_{mol}^{−1}\,R^{−1}$
297.031	$inH_2O\,ft^3\,lb_{mol}^{−1}\,R^{−1}$
554.984	$torr\,ft^3\,lb_{mol}^{−1}\,R^{−1}$
$8.206\times10^{-2}$	$L\,atm\,K^{−1}\,mol^{−1}$
62.363	$L\,torr\,K^{−1}\,mol^{−1}$
1.987	$cal\,K^{−1}\,mol^{−1}$
$8.206\times10^{-5}$	$m^3\,atm\,K^{−1}\,mol^{−1}$

However, as you may imagine if a gas in question has an idealized behavior then the constant $R$ may apply. A gas behaves as an ideal gas under certain conditions of pressure, $P=0$, and temperature so that the following equation of state applies

$PV=nRT$

One way to check for the constant $R$ for any given gas is in a plot of $PV/nT$ versus $P$ which should be a straight line. Also, when this line cuts the vertical axis ($PV/nT$) at $P=0$ you should read any of the values of the table above (according to the units you were using).

Other stuff of interest

• LE01 - AC and DC voltage measurement and continuity test
• LE 02 - Start and stop push button installation 24V DC
• What is an Equation of State
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

==========
Ildebrando.

What is an Equation of State

A relationship between the pressure $P$, the temperature $T$, the volume $V$ and the moles of its components $n$.

If a system is heterogeneous (involving several phases or components) then each phase will have its own equation of state.

Two examples of equation of state are:

The equation of the ideal gas

$PV=nRT$

The equation of van der Waals

$\left( P+\dfrac{an2}{V2} \right)\left(V-nb\right)=nRT$

where $a$ and $b$ are constants different for every gas. Constant $a$ represent a measure of the strength of the intermolecular atraction while $b$ represents the molecular size.

Important comments

• The laws of thermodynamics are so general that cannot be used to determine equations of state for a given system
• Equations of state must be determined experimentally

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

Other stuff of interest

• LE01 - AC and DC voltage measurement and continuity test
• LE 02 - Start and stop push button installation 24V DC
• LE 03 - Turn on/off an 24V DC pilot light with a push button
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

==========
Ildebrando.

Cómo calcular la capacidad calorífica $C_P$ a diferentes temperaturas

 Este es una pregunta de fácil respuesta!

¿Cómo encontrar los datos del calor específico $C_P$?

Todo lo que necesitas es una tabla en la que aparezcan diferentes compuestos y la dependencia de la temperatura (que usualmente es un polinomio). Aquí te muestro una serie de datos de 44 compuestos diferentes en estado gaseoso!

Recuerda que deberás usar la siguiente fórmula:

Fórmula para la capacidad calorífica como función de la temperatura

y que la temperatura $T$ deberá estar en las unidades señaladas en la tercera columna. Las unidades de $C_P$ serán como sigue:

• en $[kJ/mol ^\circ C]$ si usaste los valores de a, b, c y d correspondientes a la fila de unidades de temperatura en $^\circ C$
• y en $[kJ/mol K]$ si usaste los valores de a, b, c y d correspondientes a la fila de unidades de temperatura en $K$

			Peso	Unidades de la	a	b	c	d	Rango de	temperatura
		Compuesto	molecular	temperatura	1.00E+03	1.00E+05	1.00E+08	1.00E+12	desde	hasta
1	Acetona	CH3COCH3	58.08	C	71.96	20.1	-12.78	34.76	0	1200
2	Acetileno	C2H2	26.04	C	42.43	6.053	-5.033	18.2	0	1200
				K	30.6720526	5.281346655	-1.627140334	0	300	1500
3	Aire		29	C	28.94	0.4147	0.3191	-1.965	0	1500.00
4	Amoniaco	NH3	1.03	C	35.15	2.954	0.4421	-6.686	0	1200
				K	2.59E+01	3.300010213	-0.304309332	0	300	1500
5	Benceno	C6H6	78.11	C	74.06	32.95	-25.2	77.57	0	1200
				K	-1.712779299	32.47712243	-11.05823528	0	300	1500
6	Bromuro	Br2	159.808	K	35.24084981	0.407491813	-0.148745736	0	300	1500
7	Isobutano	C4H10	58.12	C	89.46	30.13	-18.91	49.87	0	1200
8	n-butano	C4H10	58.12	C	92.3	27.88	-15.47	34.98	0	1200
9	Isobuteno	C4H8	56.1	C	82.88	25.64	-17.27	50.5	0	1200
10	Dióxido de carbono	CO2	44.01	C	36.11	4.233	-2.887	7.464	0	1500
				K	26.64785269	4.226241349	-1.424267446	0	300	1500
11	Monoxido de carbono	CO	28.01	C	28.95	0.411	0.3548	-2.22	0	1500
				K	2.65E+01	0.768339491	-0.117233923	0	300	1500
12	Cloruro	Cl2	70.91	C	33.6	1.367	-1.607	6.473	0	1200
13	Cumeno (isopropil benceno)	C9H12	120.19	C	139.2	53.76	-39.79	120.5	0	1200
14	Ciclohexano	C6H12	84.16	C	94.14	49.62	-31.9	80.63	0	1200
15	Ciclopentano	C5H10	70.13	C	73.39	39.28	-25.54	68.66	0	1200
16	Ciclopropeno	C3H4	40.06	K	26.49819236	12.96640445	-3.95685276	0	300	1500
17	Etano	C2H6	30.07	C	49.37	13.92	-5.816	7.28	0	1200
				K	9.403657221	15.98372294	-4.622841216	0	300	1500
18	Alcohol etílico (etanol)	C2H5OH	46.07	C	61.34	15.72	-8.749	19.83	0	1200
19	Etileno	C2H4	28.05	C	40.75	11.47	-6.891	17.66	0	1200
				K	11.83979477	11.96700605	-3.650880536	0	300	1500
20	Fomaldeído	CH2O	30.03	C	34.28	4.268	0	-8.694	0	1200
21	Helio	He	4	C	20.8	0	0	0	0	1200
22	n-Hexano	C6H14	86.17	C	137.44	40.85	-23.92	57.66	0	1200
23	Hidrógeno	H2	2.016	C	28.84	0.00765	0.3288	-0.8698	0	1500
				K	29.06569842	-0.083643494	0.201126851	0	300	1500
24	Bromuro de hidrógeno	HBr	80.92	C	29.1	-0.0227	0.9887	-4.858	0	1200
				K	27.52087127	0.399509929	0.066183122	0	300	1500
25	Cloruro de hidrógeno	HCl	36.47	C	29.13	-0.1341	0.9715	-4.335	0	1200
				K	28.16607357	0.180922707	0.154649005	0	300	1500
26	Cianuro de hidrógeno	HCN	27.03	C	35.3	2.908	1.092	0	0	1200
27	Sulfuro de hidrógeno	H2S	34.08	C	33.51	1.547	0.3012	-3.92	0	1500
				K	26.71436839	2.386250771	-0.506350773	0	300	1500
28	Metano	CH4	16.04	C	34.31	5.469	0.3661	-11	0	1200
				K	14.14290091	7.549532057	-1.799249711	0	300	1500
29	Alcohol metílico (metanol)	CH3OH	32.04	C	42.93	8.301	-1.807	-8.03	0	700
30	Metil ciclohexano	C7H14	98.18	C	121.3	56.53	-37.72	100.8	0	1200
31	Metil ciclopentano	C6H12	84.16	C	98.83	45.857	-30.44	83.81	0	1200
32	Óxido nítrico	NO	30.01	C	29.5	0.8188	-0.2925	0.3652	0	3500
				K	29.37167064	-0.154649005	1.065082661	-4.548011052	300	1500
33	Nitrógeno	N2	28.02	C	29	0.2199	0.5723	-2.871	0	1500
				K	26.98375698	0.590992003	-0.033756718	0	300	1500
34	Dióxido de nitrógeno	NO2	46.01	C	36.07	3.97	-2.88	7.87	0	1200
35	Tetraóxido de nitrógeno	N2O4	92.02	C	75.7	12.5	-11.3	0	0	1200
36	Óxido nitroso	N2O	44.02	C	37.66	4.151	-2.694	10.57	0	1200
37	Oxígeno	O2	32	C	29.1	1.158	-0.6076	1.311	0	1500
				K	25.50295119	1.361160675	-0.425534197	0	300	1500
38	n-Pentano	C5H12	72.15	C	114.8	34.09	-18.99	42.26	0	1200
39	Propano	C3H8	44.09	C	68.032	22.59	-13.11	31.71	0	1200
				K	10.08544316	23.93068631	-7.335850368	0	300	1500
40	Propylene	C3H6	42.08	C	59.58	17.71	-10.17	24.6	0	1200
				K	13.61077531	18.87632448	-5.748619454	0	300	1500
41	Dióxido de sulfuro	SO2	64.07	C	38.91	3.904	-3.104	8.606	0	1500
				K	25.71663288	5.792686106	-3.808855325	8.60546881	300	1500
42	Trióxido de sulfuro	SO3	80.07	C	48.5	9.188	-8.54	32.4	0	1000
43	Tolueno	C6H5CH3	92.13	C	94.18	38	-27.86	80.33	0	1200
				K	2.411194159	39.11788373	-13.06534656	0	300	1500
44	Agua	H2O	18.016	C	33.46	0.688	0.7604	-3.593	0	1500
				K	30.20644269	0.993578283	0.111413799	0	300	1500

¿Cómo calcular el $C_P$ en función de la temperatura?

Hagamos un ejemplo. Si quisiéramos dar el $C_P$ del propano en $[kJ/mol K]$, lo que haríamos sería buscar en la tabla de arriba los datos del compuesto 39; pero usando los datos de a, b, c y d que corresponden a unidades de temperatura en K. En este caso, tenemos los siguientes valores de las constantes $a$, $b$, $c$ y $c$:

a$=10.08544316 \times 10^{-3}$

b$=23.93068631 \times 10^{-5}$

c$=-7.335850368 \times 10^{-8}$

d$= 0$

De esta manera podemos escribir la capacidad calorífica del propano como función de la temperatura como sigue:

$C_P^{(Propano)} = 10.08544316 \times 10^{-3} + 23.93068631 \times 10^{-5}T - 7.335850368 \times 10^{-8}T^2$        Ec. (2)

Por último, basta decir que esta expresión puede proporcionar la capacidad calorífica del propano para cualquier temperatura entre 300 $K$ y 1500 $K$. Es decir, que si deseas conocer el $C_P$ del propano en $[kJ/mol K]$ a una temperatura de 65.3$^\circ C$, harías como sigue:

• convierte la temperatura a $K$. Esto es: $65.3 ^\circ C=338.45K$
• sustituye en la Ec. 2 de arriba. Esto da:

8.27 $\times 10^{-2}$ $[kJ/mol K]$.

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

Other stuff of interest

• Some basic concepts on thermodynamics
• 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

==========
Ildebrando.

Composition variables for mixtures

 When composition of a mixture of several gases is to be considered it can be challenging to use all concepts to represent the right quantities. This is a brief explanation.

About mole $n$ and volume $V$

The number of moles for a pure substance is usually represented by $n$. However, for a mixture with several components the moles of each of these components are to be expressed as follows:

$n_1$, $n_2$, $n_3$,...

where the subscripts 1, 2, 3 indicate the component in the mixture. 

Important note: You should remember that moles are extensive variables which is not recommended for composition calculations purposes. You may go around this difficulty dividing $n$ by an intensive variable, which results in a new intensive variable.

On the other hand, you may also have volumetric concentrations [concentración volumétrica] $\bar{c}$ defined as:

$\bar{c}_i=\dfrac{n_i}{V}$

where $n_i$ stands for the mole of some component and $V$ for the volume of the mixture. When $\bar{c}$ is given in units such as mole/l or mole/dm$^3$ the volumetric concentration is also called molar concentration [molaridad].

Important note: volumetric concentration is recommended for liquid or solid mixtures since these change very little with temperature and pressure. However, the use of $\bar{c}_i$ is not advised for gas mixtures. 

Mole ratio $r_i$ and molal concentration $m_i$

This is another form for referring to composition in terms of moles of components in a mixture. Picking up the moles of component 1 as reference we may define the corresponding ratios $r_i$ for all others as:

$r_i=\dfrac{n_i}{n_1}$

On the hand, molal concentration $m_i$ is in fact a variation of the mass concentration (how it is expressed) of the single component  gas $m$. Remember that the mass $m$ can be defined as:

$m=nM$

where $M$ is the molar mass (molalidad) given in [mole/g]. However, the mass of a component in a gas mixture is defined as:

$m_i=\dfrac{n_i}{n_1M_1}=\dfrac{r_i}{M_1}$

In other words, the mass $m_i$ of a mixture component must be given in terms of the mass and moles of the other components.

Since mole and molality ratios are temperature and poressure independent, these are preferable for any physicochemical calcuation.

Mole fractions $x_i$

These are obtained dividing each of the number of moles ($n_1$, $n_2$,...), of each component, by the total number of moles $n_t$ (of the whole substance) which is defined as:

$n_t=n_1+n_2+n_3+...$

The mole fraction is then expressed as,

$x_i=\dfrac{n_i}{n_t}$

Also, the summation of the mole fractions is always equal to 1:

$x_1+x_2+x_3+...=1$

Important note: The composition of a mixture is determined when all mole fractions are given or can be determined. Since mole fractions are temperature and pressure independent, these are suitable, and possibly the most used, to describe the composition of any mixture.

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

Other stuff of interest

• Some basic concepts on thermodynamics
• 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

==========
Ildebrando.

Some key basic concepts on thermodynamics

Thermodynamicists tend to use words for technical aspects of this subject. This words are used for other subjects but with some restrictions or further details that help to describe what they are talking about.

Not knowing the meaaning of these concepts would be as trying to comunicate with someone from another country using a different language. This case is not that extreme but difficulties could arise and the worst, you will wste time.

Then, consider the following concepts, presented in a colloquial manner.

Frst, body and system come as words referring to physical things that may be the same for certain circumstances. Perhaps, a difference we can make a first difference between body and system: a system may involve more than one body. In a body and in a system physical and chemical changes, that can be measured, may occur. Next, these measurements, also understood as data, help to describe, in detail, the body or system, or we should say characterized it. Also, the features or parameters to be measured in a body or system may be important due to its changes in time or relation to other features are known as properties.

On the other hand, a system has another feature: it denotes a region or space which is restricted by boundaries with different features such as: thermal conductivity, porosity, etcetera. Measurements of the properties of the system allow to define the system in a given set of conditions, which is called a state. Those properties measured to define the state are also called variables of state. 

From the mathematical point of view, there are independent and dependent variables. Next, state properties can be expressed as independent variables.

Main components of a function.

On thermodynamics, it is known, empirically, that in order to estimate the intensive variables of a system, only two intensive variables need to be known. An intensive variable is one that does not depend on the amount of matter in the system. If the value of the variable is proportional to the quantity of matter, then it would be an extensive variable.

The equations inoliving independent and dependent variables of state are called state equations. 

A thermodynamical process implies changes in time of one or more properties. these changes in time can be referred as state changes as well.

Two types of systems can be found. The first ones are closed systems in which mass entrance or leaving is not allowed or never happens. If mass enters or leaves the system, then we would say that the system is open. Similarly, an adiabatic system is a closed system too since no heat is exchanged with the surroundings.

Thermal equilibrium would mean that state variables remain constant boundaries are allowed to change.

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

==========

Ildebrando.

Who was Clausius?

 He full name was Rudolph Julius Emannuel Clausius and perhaps is well known for his contributions to thermodynamics.

This is how Clausius looks like

Clausius was born in 1822 and died in August, 1888. He only live 66 years!

He was born in Cöslin, Pomerania, located in Poland at the presennt day. His studies were oon physics in universities of Germany were he also worked as a professor.

The most important thing, perhaps, about Clausius is related to his synthesis of the second law of thermodynamics.This means that the formulation of the second laaw of thermodynamics would be, about, 164 year old this 2023!

Read the article

Proceedings of the American Academy of Arts and Sciences, Vol. 24 (May, 1888 - May, 1889), pp. 458-465


if you would like to know more about the life or Rudoph Clausius. It is free!

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

==========
Ildebrando.