New answers tagged thermodynamics
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As gas particles expand into a vacuum there's no transfer of their kinetic energy into anything else, they do not collide with anything but with themselves. Since the temperature of a gas is a measure of the average kinetic energy of its particles, the temperature remains constant. (This is different than in the case when the expansion would involve for ...
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Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed.
Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has ...
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This is an adiabatic irreversible expansion so $q=0$. The change in internal energy is $\Delta U=\int_{T_1}^{T_2}C_V dT$ where $C_V$ is the heat capacity at constant volume for the monoatomic gas (=3R/2). The work done is $w=-\int_{V_1}^{V_2} pdV$ at the expense of the internal energy. As the temperature drops as no heat added or lost and then $\Delta U =w$...
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Your trouble comes from the two opposed concepts used in the world to define the work $\pu{p\Delta V}$.
For theoretical scientists, the work is positive when work is done on the system, when the gas is compressed by an external force. And in such a compression, the volume decreases, so $\pu{\Delta V}$ is negative. So in order for the work to be positive, ...
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The link between entropy and heat follows from the balance between entropy change of system and surroundings during a reversible process. During such a process the total change in entropy is zero, and the change in the entropy of the surroundings is given by the second law of thermodynamics as $\Delta S_\mathrm{surr}=-q_\mathrm{rev}/T$ (assuming constant T),...
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Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give ...
1
[OP] One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$
This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.
[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is ...
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Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK ...
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One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit
$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$
A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for ...
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The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:
$$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$
This is the property usually discussed as representing ...
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Your book uses the "chemistry sign convention", that is, $w=-p_\textrm{ex} \Delta V$, where $p_\textrm{ex}$ is the external pressure.
However the enthalpy is not defined in terms of the external pressure but rather the pressure $p$ in the system: $H=U+pV$. From the definition of enthalpy it follows that $\Delta H= \Delta U+ \Delta (pV)$, or, at ...
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All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably!
And without resorting to entropic ...
4
The purpose of roasting a/o calcination is oxidation a/o decomposition of original ore to form metal oxides.
It is desirable to keep the intermediate product solid to obtain porous product. This allows easy access for further reduction by carbon monoxide by coke or charcoal reduction process. If oxides were in form of a solidified melt, the reduction ...
1
They are not equal. The enthalpy of a compound at temperature T is relative to the pure elements at the reference temperature $T^0$. So it is equal to the standard heat of formation at the reference temperature plus $C_p(T-T^0)$.
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Here is how it is properly supposed to play out for the reaction in your video:
Assuming an ideal solution, the enthalpy of the reactor contents can be represented as:
$$H=n_Ah_a+n_Bh_B$$ where the n's are the numbers of moles of the two species and the h's are the enthalpies of the pure species at the same temperature as the reactor mixture:
$$h_A(T)=h_A(T^...
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You omitted the energenics associated with breaking chemical bonds and making new chemical bonds. If this contributes a negative change to the internal energy (exothermic reaction), then the temperature would have to rise to offset this effect so that the overall change in internal energy would be zero. If it contributes a positive change to the internal ...
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why is dH=dE at constant volume?
It isn't, and nowhere in the problem or answer is this implied.
First of all some definitions.
For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume:
$$\Delta U = q_V ~~~~~\text{constant volume} \...
3
Suppose that in a given transformation, ${\Delta U}$ is defined, and cannot be changed. But the same ${\Delta U}$ is the sum of two terms $Q$ and $W$ that can be changed and modified arbitrarily. If one of these terms, here the heat $Q$, is chosen to be zero, the other term $W$ cannot vary and stays equal to $W_{rev}$.
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For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$ is only true for a reversible process, since only then $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore
$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$
This ...
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Because $q=\Delta H$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $\Delta G = \Delta H ...
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Since joule-thomson effect is isoenthalpic process, so everything is done at the cost of internal energy.
Taking $a$ as the term involving intermolecular forces of attraction and $b$ as the excluded volume, we get:
$$T_i = \frac{2a}{Rb}$$
As per above equation inversion temperature is the function of intermolecular forces of attraction. When the temperature ...
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The temperature of the milk increases because of work done on the milk molecules which cause , increase the velocities of the moecules and since velocity is directly proportional to the temperature.
Velocity α Temperature
So the temperature of milk increases due to the shaking process. Although this effect is very small but heat up the molecules.
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Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This ...
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For a system in contact with a constant temperature reservoir at the same temperature as the system in its initial state, and also in contact with a constant pressure surrounding at the same pressure as the system in its initial state, $$\Delta U=Q-W=Q-P\Delta V-W_{npv}$$where $W_{npv}$ is the amount of non-PV work done by the system. In addition, from the ...
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Let
$$moles\ CH_4=1-e_1$$
$$moles\ H_2O=1-(e_1+e_2)$$
$$moles\ CO=e_1-e_2$$
$$moles\ CO_2=e_2$$
$$moles\ H_2=3e_1+e_2$$
where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.
So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\...
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Sea salt is usually sold as coarse crystals because, as others have noted, it is separated by crystal size before packaging. Sea salt, along with Kosher salt, is desirable by chefs and food enthusiasts because it allows one to see how much salt is actually put a piece of food, let's say a steak, without dissolving as smaller crystals would do. And it gives ...
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tl;dr– "Gas" and "vapor" aren't mutually exclusive. Generally:
a gas is any material that'd fill a volume to its boundaries; and
a vapor is a gas-like material that's associated with a condensed-state transition.
It's a bit misleading for a state-diagram to label a region "vapor" in a manner that might imply that a vapor's ...
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Vapor (or steam for H2O) can be used for the gaseous form of a particular substance which at normal T and P are solid or liquid. They are all gases although not always ideal (steam is certainly not ideal below its critical T).
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In college, I had a thermodynamics teacher who was awesome. He had a way of explaining things that were accurate and easy to understand. He explained this difference to us this way:
A gas will not condense into a liquid with an isothermal compression. (i.e. an Ideal Gas)
A vapor will form liquid when isothermally compressed.
His example:
When you boil water, ...
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I'm surprised the OED has such a strict definition for gas. I could not find a strict definition in the IUPAC color books (certainly not in the gold book). Presumably these words are in such common use that their definition is assumed understood or easily found. The analytical compendium (orange book) and physical chemistry book (green book) mention vapour (...
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Vapor is a much older word alluding to dampness and it was not coined by scientists. It is in use since the 1300s. The actual meaning of meaning of vapor is "Matter in the form of a steamy or imperceptible exhalation; esp. the form into which liquids are naturally converted by the action of a sufficient degree of heat. This is the original 13th century ...
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The normal use distinguishes "vapour" from permanent gas
At normal lab conditions there is a (fairly obvious) distinction between things that could exist as liquids and things where no liquid phase is possible. Oxygen, for example, is a permanent gas, but dichloromethane is not. But the vapour pressure of dichloromethane is pretty high and there ...
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The statement that it is impossible to calculate the absolute energy at an ambient T seems to be intuitively incorrect, difficult possibly, but not impossible. Since U is a state function the value of U for a given set of conditions is invariable and can be set as an arbitrary zero and changes in U calculated. So all one must do is work in reverse to ...
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Because it is not ground as fine , unless a customer wants finely ground sea salt. It is passed over screens to separate it into sizes after grinding, just like table salt. Fun fact ; table salt that is too fine for traditional marketing ,is sold as "popcorn" salt.
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The method of production of sea salt usually involves bulk evaporation of seawater. This allows the salt to crystallise and due to it's bigger size it is sold without much processing. On the other hand, the table salt has much smaller size and is processed a lot more than we could assume before it is sold.
This could be one reason why sea salt has coarse ...
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Enthalpy is a fundamental physical property of the material(s) comprising a system. It can often be presented for a given material in the literature in tables (e.g., the steam tables for water), relative to some specific datum state (at which it is taken to be zero). This is analogous to potential energy which is expressed relative to some specified ...
1
Enthalpy vs ∆Enthalpy
We know that enthalpy represents the energy of a system. So, now think about it? Measuring the energy of a system? Consider all the changes and influences a system is under. This makes it really difficult (we could say impossible) to calculate the enthalpy of a system.
But we do know one thing, that the enthalpy depends on two factors.
...
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The three laws of thermodynamics (or four, depending on the source) are postulates that scientists have assembled in an attempt to describe the way the universe works. The First Law is a reformulation of the principle of conservation of energy. It says very simply that while energy is indestructible it can be exchanged in various ways. It also implies that a ...
2
At the Joule-Thomson inversion temperature
$$\mu = \left(\frac{\partial T}{\partial p} \right)_H = 0$$
It can be shown that
$$ \left(\frac{\partial H}{\partial p} \right)_T=-\mu C_p \tag{1}$$
(this can be derived from the differential form $dH = \left(\frac{\partial H}{\partial p} \right)_Tdp + \left(\frac{\partial H}{\partial T} \right)_pdT$)
Also, the ...
1
An ideal gas is defined as one of non-interacting point particles and its internal energy depends on temperature alone, thus the slope of a plot of $U$ vs $T$ is a straight line with slope $C_V$.
A real gas (or liquid or solid) is made up of molecules and these have internal energy levels due to a molecule's ability to rotate and vibrate. At low temperatures ...
1
For a multicomponent reactive system you would rather write the enthalpy differential as
$$
d(nH) = \left(\frac{\partial (nH)}{\partial T}\right)_{P,n_i}dT + \left(\frac{\partial (nH)}{\partial P}\right)_{T,n_i} dP + \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i
$$
Where $H$ is the molar enthalpy. At constant ...
2
Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase?
Consider the four paths through the phase diagram.
Comparing (A) and (C), we are below the temperature of the critical point in (A), so an increase in pressure will liquify the gas. In (C), we are above, so increase the pressure does not lead to a gas/liquid ...
2
The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats ...
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My textbook says that critical temperature is the temperature above which a gas cannot be liquified no matter how much pressure we apply on it.
Below the critical temperature, you can compress a gas and a second phase (liquid) will gradually form. In places with gravity, the liquid will collect at the bottom of the container. As you decrease the volume, the ...
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One useful distinction between a liquid and a gas is that while they are in equilibrium with each other, so at the same temperature and pressure but not the same density, molecules of the liquid need to gain energy to leave the liquid phase and enter the gas phase. A surface, with surface energy, exists between the two phases, and allows them to segregate ...
1
An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas.
Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (...
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[OP] But according to the equation $dH=nC_{p}dT$, if temperature $T$ is constant then $dT =0$ and $dH =0$.
Even if this equation would describe a system completely, a constant temperature does not imply that $dH$ is zero. Instead, you can have a temperature bath of infinite size (n approaches infinity) so that the average temperature does not increase. Of ...
1
The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.
The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, ...
0
For most salts the equilibrium constant for ionization is very large and thus the reaction practically proceeds only in the forward direction and so we generally say it is irreversible.
But in reality no chemical reaction is irreversible and some backward reaction always takes place.
So, in a strict sense ionization is a reversible equilibrium process but ...
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It really does liquefy. But it does not do so in exactly the same way as you see below the critical temperature and pressure.
As an example, suppose you heat steam to 400°C and then compress it, isothermally, to 5000 bars pressure*. When you are done, you find that the water has a density and viscosity more or less similar to ordinary liquid water; what ...
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