New answers tagged

1

From the open system (control volume) version of the first law of thermodynamics, between cross sections x and y, $$\dot{Q}-\dot{W}_s-\dot{m}\Delta h=0$$where $\dot{Q}$ is the rate of heat addition to the control volume, $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}$ is the mass flow rate, and $\Delta h$ is the change in specific enthalpy between ...


1

For an adiabatic system like a piston where $\delta q = 0$, using the first law of thermodynamics gives you the following expression: $$dU = \delta q + \delta w$$ $$dU = - PdV$$ This expression is pretty much useless however, in that you can't integrate it, since T, V, and P are all constantly changing interdependently in the system. However if you assume ...


1

Using Mathematica, I obtained $T=546 K$, as follows. One of the nice things about using Mathematica for physical calculations is that it has the ability to understand/keep track of/cancel out/convert units. The volume is non-physically small -- about 750,000 x smaller than what we'd expect for a mole of, say, ideal gas at $P=100 Pa, T = 546 K$). One does ...


2

Very late to the party, but $\ce{PSF_3}$ (thiophosphoryl fluoride) burns with a cool flame. As reported in wikipedia (https://en.wikipedia.org/wiki/Thiophosphoryl_fluoride), $\ce{PSF_3}$ was discovered in 1888 by J.W. Rodger and T.E. Thorpe. As stated on the wiki page: "The discoverers were able to have flames around their hands without discomfort, and ...


0

I know I'm really late to the party, but I was interested in this as well. Essentially, it appears that water and ethanol form a solution that has some structure to it, not unlike micellar aggregation. The OH end of the ethanol molecules aggregate with each other and the polar water, which creates a packing structure that has negative excess volume. ...


1

There are actually two school of thought on this. One is that residual entropy (the entropy a substance has as a result of not being a perfectly ordered crystal at $\pu{0 K}$) is a thermodynamic entropy. Accordingly, according to this school, a substance must be a perfectly ordered crystal in order to have zero entropy at $\pu{0 K}$. The other is that ...


6

The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process: $$\mathrm dw_\mathrm{rev} = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV$$ where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work ...


0

I would solve this question rather simply. Let's consider the general equation $\ce{ X-Y_{(s)} \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$ with a $\Delta H = C$. Intuitively, if the reaction is endothermic ($\Delta H > 0$), an increase in temperature should help the dissolution (more heat to be absorbed by the salt while dissolving). You can rationalize this ...


2

As air temperature increases(in the vicinity of the hot shower) air can hold more water molecules, or with more technical terms: the saturation water pressure increases with the temperature. Near the water, the air temperature is high and since there's plenty of hot water that evaporates then the vapor pressure increases. Then because of convective motion, ...


3

The nearest I came to an explanation is an article (Ref. 1) by authors of the database explaining that, in the case of $\ce{H2O (cr,l)}$, the notation (cr,l) refers to the condensed state. Also, a link from the above data page explains that "cr=crystal" and "l=liquid" (thanks MaxW for hinting at the solution). A further update: it finally dawned on me that ...


0

The vapor pressure curve describes the relationship between temperature and pressure for which the pure substance can exist as liquid and vapor, together in equilibrium. So it is not necessarily all liquid or all vapor. Both can be present in different proportions. As far as specific volume is concerned, your tables should list both the vapor specific ...


-1

Every gas at decompressing cools down (adiabatic expansion). See for example: https://physics.stackexchange.com/questions/14140/why-does-the-gas-get-cold-when-i-spray-it


2

Both, since atmospheric pressure does not exist without gravity. Atmospheric pressure is essentially the weight of air above something, which happens because of gravity.


-1

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.


3

A combination of gravity and the suction you apply with your lips ("suck") help empty the cup, assuming as you do that the coffee in the cup leaves a vacuum as it exits. This is similar to the "problem" of emptying a bottle of liquid when you invert it (the mouth facing downward). The challenge is most evident with honey or ketchup, viscous liquids or those ...


10

Don't blame yourself. This is a badly-worded question. Products and reactants don't have "heat content", they have thermal energy. I.e., heat is not a property of substances—they don't 'contain heat.' Instead, heat is a property of processes; specifically, it is the flow of thermal energy across a boundary. So, for instance, in an exothermic reaction in a ...


3

You might want to check your units. The energy should be in J/mol, according to the values you put in, and not in kJ/mol. $e^{-22200/(8.314*298)} = 0.000128397 \approx 1.3*10^{-4}$


1

The enthalpy of combustion (aka the heat of combustion) is the enthalpy change in going stoichiometrically from pure reactants to pure product at constant temperature and pressure. The change in internal energy between these same two states is equal to $$\Delta U=\Delta H-\Delta(PV)=\Delta H-P\Delta V$$For a reaction in the ideal gas region, this reduces to ...


5

Equation (1) is an empirical one that does not consider entropy. If you substitute it for the Eyring equation (considering free energy of activation $\Delta G^\ddagger$) you end up with equation (6) as expected, not equation (5). Wikipedia has a nice section on the relationship of activation energy and Gibbs energy of activation: Although the equations ...


3

The mathematics of equilibrium systems are far simpler than others It isn't strictly true that thermodynamics doesn't apply to systems that are not at equilibrium only that the simple theorems and formulae of equlibrium thermodynamics only apply to systems at equilibrium. The reason why most discussion and most teaching is about those equilibrium systems ...


0

Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes [?] A reversible process is a hypothetical process where the entropy of the system and its surrounding is constant. This is hypothetical because nothing happens unless the entropy increase at least a tiny bit. In other words, if the system is at equilibrium, ...


2

$\ce{P4}$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here). In a molecular solid, molecules containing a discrete number of covalently bonded atoms are arranged in a lattice stabilized by weak interactions such as van der Waals forces or hydrogen bonds between the molecules. In a covalent-network solid, the "...


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