10

Here's your confusion: You need to consider two different things: The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that ...


9

Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$? According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1: Reference 1 states that: An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{...


6

In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider $N$ monoatomic particles in a cubical box of side $\ell$ (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is $\frac{N}{N_A} = n \ \pu{mol}$ where $N_A$ ...


5

Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics and Physics. They offer both theoretical and experimental evidence that neither pressure ...


5

According to German Wikipedia water potential is defined as $$\psi := \frac{\mu - \mu_0}{\bar{V}}$$ where $\mu_0$ is the standard chemical potential (usually pure water at atmospheric pressure at a specified reference height), and $\bar{V} \approx \pu{18 cm3 mol-1}$ is the the molar volume of pure liquid water. This means that water potential $\psi$ is ...


5

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium: $$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$ Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$: $$p_1 = x_1p\tag{1}$$ To find ...


4

Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units. For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation $$v/(\mathrm{km/h})=3.6\frac {l/\...


4

I couldn't resist. The osmotic pressure equation is $$Π = cRT,$$ where $Π$ is osmotic pressure in atmospheres, $c$ is molar concentration, $R = \pu{0.082057366080960 L atm mol-1 K-1}$ (exactly) and $T$ is temperature in Kelvin. In this problem, $Π = \pu{0.236 torr}$ times exactly $\pu{1 atm}/\pu{760 torr}$ and $T = \pu{(273.15 + 19.0) K}.$ So $c = \pu{1....


4

Chemical potential is a portion of water potential, but not all of it. Factors like gravity and bulk fluid properties also affect the water potential. Water potential is typically used for macroscopic quantities of water, so it is more natural to consider the amount of water by volume rather than the number of molecules. If you convert the number of ...


3

When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you ...


3

This is a summary of the equations to use to calculate phase transitions. The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion. The Clausius-Clapeyron equation describes solid-vapour ...


3

The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$ The additional pressure compresses the liquid and thereby increases its chemical ...


3

For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression: $$\mathrm dU = \delta Q + \delta W$$ $$\mathrm dU = - p\,\mathrm dV$$ 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 ...


3

There's a very nice—and fairly accessible—discussion of your third question in Nature Magazine's News and Views Section[1]. In this article, Bonn discusses the results of a fairly recent work that appeared in Physical Review X[2]. Bonn summarizes the key points as follows: The idea that a thin film of meltwater wets the surface of ice has been accepted ...


2

Diamonds burn, but the temperature at which they burn depends on whether or not the diamonds are in contact with air. The temperature of diamond ignition in pure oxygen is 690º C to 840º C.  In a stream of oxygen gas, diamonds burn at a low red heat initially. They will gradually rise in temperature and reach a white heat. Then, the diamonds will burn ...


2

There are a number of ways to look at this. One is to look at the equations. Another is to plot the pressure p as a function of volume V. The area under the pressure curve on the plot is the total work done on the system. It is very easy to answer your second question by inspection of this figure, but perhaps not the first question without actually ...


2

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 ...


2

Choose either the unit you are asked for (as in hwk or exam) or what is useful. For instance if you want to compare pressures use the same units. If the pressures are large you might use kPa, if small, mTorr, etc. You may seek to avoid dealing with large multipliers (ie 3 bar rather than $3 \times 10^5$ Pa). Some equations in physics imply use of a ...


2

Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given) If no information is given, you should just give the three amounts names and treat them as unknowns. ZnO is exposed to pure CO at 1300 K This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental ...


2

When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture. And when the pressure is extremely low, the volume increases but not as much as calculated with the ...


2

Kevin is right. In that reaction, the number of moles does not change. So carrying it at constant pressure or at constant volume does not make any difference. And $q_v$ = $q_p$


2

Just like any sweeping statements made by the General Chemistry textbooks, this statement is also not completely true. When you work at pressure extremes, as in modern day chromatography, such as 1000 or higher bar, large molecules can change their shape, in that case, one can see a change in retention factors which sort of indicates a change in equilibrium ...


2

At the very least your pressure has to be consistent with a whole number of molecules or atoms in your vapor phase. Suppose you have a pressure of $7×10^{-34}$ atmosphere or (roughly) $7×10^{-29}$ Pa at $300$ Kelvins. The average volume per atom/molecule is computed by the molecular level version of the Ideal Gas Law, in which the gas constant $R$ per mole ...


2

The root of your confusion, as I understand, is that you are not much familiar with other kinds of work done. Although, $p\,\mathrm{d}V$ is everyone's favourite but it's not the only one in the league. Always remember, the $p\,\mathrm{d}V$ is work done during an expansion/compression. You can change the pressure or concentration to get electric work.


2

This is an approach to Question 2. The reaction equilibrium condition for the system is: $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$. Then we have $$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$. For a small change in ...


1

When we apply pressure on a substance, its melting point increases or decreases depending upon the change of volume which occurs during the phase change of that substance. To be specific, if the volume of the substance's liquid phase is greater than the volume of the solid phase, its melting point will increase upon the increase of volume. Again, if the ...


1

Why should decrease when it does decrease ? The gas with liquid vapour, previously in equilibrium with liquid, with suddenly increased pressure: Vapour gets over-saturated and condenses until the saturated vapour pressure is reached again. Gas starts dissolving and gas partial pressure decreases, until the new equilibrium between its partial pressure and ...


1

1) If you induce a sudden chance in pressure you will indeed take the system out of equilibrium. Lets consider a closed recipient containing some liquid and some gas above. If you suddenly inject more gas into the recipient, the gas pressure is initially going to raise fast and then going to decrease slowly while part of the gas is dissolving, until the ...


1

If $\alpha$ is the degree of dissociation, at equilibrium $1-\alpha$ of XY is present and $\alpha$ of X and $2\alpha$ of Y making a total of $1+2\alpha$. The partial pressure of XY is $\displaystyle p_{XY}=\frac{1-\alpha}{1+2\alpha}\frac{P}{P^\mathrm{o}}$. The $P^\mathrm{o}$ is the standard pressure used to make the equilibrium constant dimensionless. In our ...


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