9

There's not much to do, as you nearly spotted the difference by yourself: $U$ is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with $Q$ and $W$, it's different: these are path functions (or process functions) ...


8

Everything you've written is correct (for a free expansion, $\Delta U = q = w = 0$), but it applies only if the system is an ideal gas, or has ideal-gas-like properties, namely that there are no attractive or repulsive interactions between the particles. It also requires that the system is closed (no matter can flow into or out of the system). Internal ...


7

The full phrase should be thermodynamic stability with respect to ____, where the dash indicates a process, or a chemical reaction. A mixture of hydrogen and oxygen is thermodynamically unstable with respect to water formation. Similarly, a diamond is not forever (which may not please De Beers and ladies). It is thermodynamically unstable with respect to ...


7

How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential? In the example at hand it means that we divide the ways we can change the energy of the system into exactly two bins. The first we call heat, and all the other ones we call work. Heat is an energy transfer that occurs when two entities ...


7

G and H are like altitude. They are not absolute; rather their values are always relative to whatever you set as a reference. So they can be either positive or negative. There is no general constraint on the sign of their values. For this reason, we usually talk about $\Delta G$ or $\Delta H$, since it's the changes that are meaningful; giving them ...


4

Because you are comparing apples with donuts. If I eat enough apples they will add more calories than that donut I gobbled for breakfast. To compare food calories properly you need to match serving sizes. Similarly, to compare reaction energies fairly you need to use a common basis for the extent of reaction. For oxidation of metals the reactivity of the ...


4

Old question, but still seems to attract some interest. In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule. I recommend calculating the thermal correction for $G$ at the temperature of interest and the default ...


4

The Gibbs free energy of a multicomponent mixture is a function not only of temperature and pressure, but also of the number of moles of the various species in the mixture: $$G=G(T, P, n_1, n_2. m_3. ...)$$So, $$dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial n_1}dn_1+\frac{\partial G}{\partial n_2}dn_2+\frac{\...


4

Imagine that you propose to climb Mount Everest. From your desk you can compute exactly how much energy you will need to reach the top of the mountain starting from base camp. That would simply be the gravitational energy difference between the two locations. With that information at hand you load up on supplies - dividing the gravitational potential energy ...


4

The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0)


3

Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose) $$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$ And since $K_p$ is related to $K_c$ by the relation $$K_p = K_c(RT)^{\Delta n_\...


3

I'm not going to do your research for you (at least not for now—perhaps if the mood takes me I'll do the Mathematica programming at some later point), but I will give you guidance that will help you find the answer: It's hard to find data on melting points at different pressures for a wide range of compounds (and you want to search through a wide range of ...


3

I would like to give an alternative derivation for theorist's answer. You can get the same answer with statistical mechanics using a lattice model. Let us discretize the container into $M_1$ respective $M_2$ cells. The volume of a cell is $V_{cell}$. Each gas particle fits exactly into one cell and can hop from cell to cell in a discrete manner. Since we are ...


3

Yes, there is. The following is a plot of the hydration enthalpy versus row in the Periodic Table. Data was extracted from the Wikipedia, not the primary source (Ref 1). Your thinking is right. The hydration entropy is evidently negative due to structuring of water (including loss of rotational degrees of freedom), becoming less negative for the larger ions (...


2

The accepted answer is incorrect. It has nothing to do with reversibility. The underlining issue is that your $dG$ is incomplete: $$dG = dH - TdS - SdT$$ You are missing $-SdT$, so at constant pressure you don't end up with zero. As "ado sar" has commented in the accepted answer: $$dH = TdS + Vdp + \sum_i \mu_i \cdot dn_i$$ Hence, you end up with: ...


2

I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample. This ...


2

Is the total entropy change of all isothermal processes 0? No. Only for reversible processes. However your equation is correct. For an irreversible process in which an ideal gas expands isothermally against a constant external pressure $$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$ becomes $$\Delta S_\mathrm{total}=\...


2

The equation $$\ce{2A <=> B + C}$$ implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be $$\ce{(x + y) A <=> x B + y C}$$ Here is an example that would be well-described by reaction (1): $$\ce{2 H2O <=> H3O+ + OH-}$$ One water molecule cannot make a ...


2

At low pressures the activity of a gas can usually be related in a straightforward way to its partial pressure as $$a=p/p^{\circ}$$ Since the activity of a substance present in multiple phases is the same in all of the phases, for a volatile compound it suffices to determine the partial pressure in the gas phase to know its activity in condensed phases. This ...


2

When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function. The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings. Since the entropy change for ...


2

I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings. When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\mathrm{\Delta S_{surr}...


2

For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$


2

The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$ it refers to the reaction $ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$ with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K. A more negative value of $\...


2

$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$ converts to gaseous $\ I_2$ at a constant temperature. So the answer should be reaction (b).


1

You determined the $\Delta H$ in going from products to reactants at constant temperature and volume. But, you are dealing with an ideal gas mixture, so the enthalpy of the product mixture is a function only of temperature, and not pressure. So the enthalpy change of the product mixture in going from the final pressure in the reactor to the initial ...


1

We define a state function enthalpy as $$H=U+PV$$ Since we cannot measure the absolute value of enthalpy, but the change in it, we modify the equation to $$\Delta H= \Delta U +\Delta(PV)$$ Which is the correct relation. Now, if you assume pressure is constant, then you can take the $P$ out to get $$\Delta H=\Delta U+P\Delta V$$ Now $PV$ work (or the ...


1

The modern definition of thermodynamic stability is the state of maximum entropy. Some background information is necessary to make sense of this. I hope you will find the following helpful! Phenomenologically, thermodynamic stability is the absence of visible change. This is the 'original' definition, employed by experimentalists during the 18th and 19th ...


1

The way that you describe it is exactly correct. This is the essence of the Clausius Inequality.


1

Gardeners often use a tennis ball for this purpose - it preserves a gap in the ice.


1

There are oils sold for swimming pools to slow or stop evaporative cooling . That would be a problem for air breathers like mosquito larva. I don't know anything about it because I went with 1" thick Styrofoam panels. That is what I currently have on my pond ( 10' X 5 '), works very well .I cheat a little and have 150 watt aquarium heater in the 700 ...


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