# Tag Info

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You've got the right idea — you want to simplify the problem — but I don't think you're using quite the right simplification. Take a look at the book's set-up of the situation, and ask yourself how the water is stopping the lava. You'll see the idea is that we're using liquid water, not ice, to solidify the lava. So your question should really be: How much ...

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Good question You need to specify the conditions a bit more completely: At 1 °C, the water would never freeze, while the molten lava might in an hour. At -50 °C, not uncommon near Mt. Erebus, an uninformed guess on my part is that water would freeze first... depending: What is the shape of the sample Since water might freeze first on top (since below 4 °C,...

5

The problem is in the word "always". The statement "Work and heat energy are always viewed from the perspective of the system?" is wrong. Please inform your teacher that is true in the chemist's world only- and chemists did not invent thermodynamics. Chemists have adopted a different convention than physicists. All we have to do is to be ...

4

It is the consequence of the Hess's law and that is the consequence of the energy conservation law. Overall it says the enthalpy change does not depend on the path, but only on the start and final point. Comparing to the real reaction path, you can formally get there also via the path of conversion reactants to pure elements in their standard form and then ...

4

The question didn't specify which process the given molar heat capacity was specified for. So we will assume that it is molar heat capacity for the first process. The first process is isobaric so $$q=nC_p\,\Delta T$$ You calculated $\Delta T$ correctly. The only mistake you did was assume that the gas was monoatomic. The expression for $\Delta U$ you wrote ...

4

That first equation is supposed to be at constant T. You can rewrite it as $$RTd\ln{(f/P)}=RT(z-1)\frac{dP}{P}$$or $$\ln{\frac{f}{P}}=bT=\int_0^P{(z-1)\frac{dP'}{P'}}$$So if z behaves like z = 1+aPT, this becomes: $$bT=aPT$$ or b = aP

4

The thermodynamics involved might overwhelm the child, but the question is a good one. It shows the child has taken the information in the book and re-applied it to a new problem (the stated question was about freezing and has nothing to do with steam, contrary to a previous answer). He seemingly already understands that the cold water has cooled and ...

3

The internal energy $U$ is the sum of all the energies stored in the chemical bonds. If a chemical reaction happens, the bonds are losing or getting energy. Heat is getting in or out of the system, and this heat $\Delta Q$ can be measured. $\Delta U$ =$\Delta Q$, if the volume is constant. But if the transformation is carried out in contact with the ...

2

When we say a bonded substance is more stable, it means that it has less energy than a the total of its atoms freely floating around. Part of the total energy from the composite are stored in chemical bonds, which makes the resulting product has lower energy. When a catabolic reaction occurs, the large molecule is broken into smaller molecules with the ...

2

The equation $$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$ does not assume that $\Delta S^{\ddagger}$ is temperature independent. To evaluate its T dependence you might proceed as follows: $$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{... 2 Both the answers above are accurate and touch on my point here. There are two factors that contribute to whether a reaction is spontaneous (if it will happen on its own) such as the one you described (iron with oxygen). The thermodynamic properties of the reaction (like the net heat exchange, exothermic vs. endothermic, etc.) The entropy of the reaction. ... 2 Internal energy of any gas is given by$$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$Last three aren't affected by ordinary heating. And U_\text{intermolecular} for ideal gas is zero. That's why for ideal gas U is only function of temperature. But ... 2 The friction they are talking about is viscous friction of the gas itself. This comes into play when the gas is deformed rapidly, and is roughly proportional to the square of the rate of gas deformation. Regarding entropy, the change is entropy is equal to the integral of dq/T only for a reversible process. In an irreversible process, in addition to this ... 2 Quantities such as heat and work can vary depending on the path (how you get from an initial to a final state). In contrast a defining aspect of a state function is that it is independent of the path. Entropy is a state function, as a corollary therefore a path-independent property. The experimental determination of entropy involves measuring the heat ... 2 You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps . We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our ... 2 Consider a gas which was subjected to two different processes: isochoric and isobaric. Isochoric: If a certain amount of heat dq was given to this gas, then this heat will be completely used up in raising it's temperature since dw is zero. So,$$dq_v = dU$$Also$$dq_v = nC_v dT$$where C_v is constant volume molar heat capacity. Therefore we can say ... 2 I'm afraid that the temperature to remove the odor agent depend on the odor components. In the industry you will have a time/temperature protocol with three steps. First near 100°C (a little more), to remove humidity from the activated carbon. Then a temperature dependant of the adsorbed component. (around 400° commonly). Finally above 700°C to reactivate ... 2 Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings), and ignition circuit connected to the bomb. (*) The whole device known as calorimeter consists of a bomb and a water ... 2 You have differential equations. Differential equations are ALL about the initial conditions, and A PROCESS. So, what’s the initial condition? You have a piston with a given volume of gas. What’s happening then? I apply an additional force with an object. (The atmospheric pressure already was there) And what if you want to know the work made by the ... 2 Your analysis is correct in terms of number density. But let's see how it plays out in terms of molar density. Let n be the number of moles and A be Avagadro's number. Then N=nA. If we substitute this into your first equation, then I get$$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where ... 2 If a body absorbs a quantity of heat q its temperature will normally rise by a value \Delta T. The average heat capacity over this temperature range is defined as C_{av}\equiv q/\Delta T. The instantaneous heat capacity at temperature T is C\equiv dq/dT. This definition is not exact enough, however, until the path of heating is specified. From the ... 2 METHOD 1 Suppose we were to start off with pure reactants in stoichiometric proportions in the standard state of 1 bar, and end up with pure products in corresponding proportions in the standard state of 1 bar. The change in Gibbs free energy for this process is \Delta G^0, as you have answered in your comment to me. To bring about this change, we use ... 1 For an isobaric process you can integrate the expression dS = dq/T = C_p dT/T to compute the entropy change at another temperature. Since the heat capacities are assumed constant you can proceed as follows:$$\begin{align}\Delta S_{\pu{380 K}}^\circ &= \Delta S_{\pu{298.15 K}}^\circ + \sum_i \nu_i\int_{298.15 K}^\pu{380 K} \frac{C_{p,i}}{T}dT\\&= \...

1

Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases. I think the problem is you're thinking of it the wrong way, that is, you first think that $q= nC_p \Delta T$ and for constant pressure process $q_p= \Delta H$ , and hence, $\Delta H = nC_p \Delta T$ In actuality, it is the other way around. ...

1

The total work ( and its value for the constant force case ) for pressure $$p=p_\mathrm{force} + p_\mathrm{atm}$$ $$W_\mathrm{tot} = - \int_{V1}^{V2}{p \cdot \mathrm{d}V}$$ is shared between the source of the explicit mechanical force acting on piston and atmosphere. Atmosphere would do work $$W_\mathrm{atm} = - \int_{V1}^{V2}{p_\mathrm{atm} \cdot \... 1 In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well ... 1 In your first equation, the \Delta U is not the change in internal energy of the gas reaction. It is the change in internal energy of the surroundings (the calorimeter and water), which is the heat transferred from the reaction mixture to the calorimeter and water. Since the calorimeter/water and the reaction chamber are both of constant volume, the \... 1 The heat capacities arise when you define various differential forms of U and H. The differential forms define how the energy responds to changes in thermodynamic variables such as pressure, temperature and volume. For instance, one form for U reads:$$\begin{align}dU = \left(\frac{\partial U}{\partial T} \right)_V dT + \left(\frac{\partial U}{\partial V} \...

1

Principally, definitions cannot be derived. They are chosen. What you have described in the question are heat capacities without the adjective specific, being heat capacities of a whole system. Note that specific heat capacities are related to the substance unit mass amount, mostly 1 kg, so I would expect the mass in denominator, like $$c_V = \frac 1m . \... 1$$\Delta S_\mathrm{sys} =\frac{\Delta H_\mathrm{vap}}{T_b}$$is the entropy change of the system.$$\frac {\Delta G}{T_\mathrm{b}} = \Delta S_\mathrm{tot} = \Delta S_\mathrm{ext} + \Delta S_\mathrm{sys} = \Delta S_\mathrm{ext} + \frac{\Delta H_\mathrm{vap}}{T_b} = 0 is proportional to the total entropy change.

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