# Tag Info

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Many volatile liquids are not combustible Dichloromethane (DCM) is a widely used solvent by chemists. It boils at around 40°C (the same as diethyl ether) but is not remotely combustible or flammable. Ether is both very volatile and very flammable, so much so that most labs would prefer not to have it used anywhere where flames or sparks could be present. ...

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Volatility ( even if by thermal decomposition ) is the necessary, but not sufficient condition for liquids to be combusted, forming a flame. Liquid helium is the most volatile liquid ever, but there is no way to burn it ( chemically ). As other answers mention, there is correlation, as flammable liquids are generally less polar and more volatile than polar ...

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Why thermodynamical equations are just for gases? They are not. The equation $\Delta U = q + P\Delta V$ applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work. Gases are (1) an easy way to introduce thermodynamics concepts because ...

7

Not always true. Tetrachloroethylene ("perc", as it is sometimes called in the dry cleaning business) is not inflammable, but quite volatile. Carbon tetrachloride, which was also a common solvent some decades ago, is yet another halocarbon solvent that is volatile, but not inflammable (hence its former use in fire extinguishers). In general, volatility and ...

3

Chemical energy of system has been converted to thermal energy of both the system and the surrounding water. Additionally, water did work on system, as the system shrinked down. $$\Delta E_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}-p \cdot \Delta V_\mathrm{system}$$ For $H=E + p \cdot V$, $\Delta H=\Delta E + p \cdot \Delta V + V \... 3 Use the ebullioscopic equation (the first equation) in this Wikipedia article, $$\Delta T = K_\mathrm{b} m$$ First solve for$m_\mathrm{init}$, the initial molality of urea. Second figure out at what molality$m_\mathrm{fin}$the boiling point is elevated by$\Delta T = \pu{1.5 °C}.$Since the amount of urea is constant, the % change in the mass of water ... 3 I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different? The total kinetic energy is different. The average kinetic energy per particle is the same. Through this question, I want to ask whether temperature ... 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 ... 2 I have worked on a computational project where I had to study the sites of deprotonation in this molecule: When deciding between mechanisms I had to consider both thermodynamic and kinetic. For example, I had this case: deprotonations of two different hydrogens(bond with N Blu atoms in the picture)were similarly thermodynamically favored (similar free Gibbs ... 2 The short answer: If 2 samples of water of the same temperature is considered,the average kinetic energy of water molecules of samples A nad B are the same. But, within the same water sample, water molecules have different kinetic energy with a particular statistical distribution. Even if I had set by some magic wand the same energy for all water ... 2 This answer is correct. After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature ... 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 The first law of thermodynamics tells us that, for this process at constant pressure, $$\Delta E_\text{system}=q_\text{system}-p\,\Delta V\tag{1}$$and $$MC_\mathrm s\,\Delta T_\mathrm w=-q_\text{system}\tag{2}$$where$\Delta V$is the change in volume of the system. Eqn. 1 and 2, in terms of system enthalpy, can also be combined and written as $$\Delta H_\... 2 Dissolving substances in water decreases water chemical potential$$\mu=\frac {\mathrm{d}G} {\mathrm{d}n}$$respectively activity a :$$\mu = \mu_9 + RT \ln {a}$$As consequence, saturated water vapour pressure above the solution is lower than the saturate vapour pressure over pure water at the same temperature. This leads to the higher boiling ... 2 1) You are correct that the thermodynamic temperature is a measure only of the translational kinetic energy. Intramolecular vibrations do not contribute to the temperature. 2) You are also correct that rotational and vibrational KE can be converted to translational KE in a collision (as long as total momentum and kinetic energy of the system are conserved).... 1 Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid? No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the ... 1 Since the change in internal energy for the cycle is zero, if you are using the form of the 1st law written as \Delta U=Q+W (where work done on the system is regarded as positive), the heat for the entire cycle must equal minus the work for the entire cycle. The work done on the system in AB must be -(1)(1.5-1)=-0.5 L-atm. Therefore, the work done on the ... 1 If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using dG=VdP=RTd\ln P. Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0. And this is also ... 1 Some people define a "closed system" as one that can exchange neither heat nor work not mass with the broader surroundings. Others, like us engineers, define a "closed system" as on that can exchange both heat and work with its surroundings, but not mass. We engineers call a system that can exchange neither heat nor work with the broader surroundings an "... 1 Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."? No, based on the Gibbs free energy equation$$\Delta G = \Delta H -T \Delta S$$and the condition that$\Delta G<0$for spontaneous dissolution, if you assume$\Delta S\$ is constant then the statement is true. The point of that other ...

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The temperature at which NOx is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean NOx exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right? Yes, the vapor and liquid can coexist, provided P and T are on the coexistence (!) line (as ...

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@Raphaël did the hard part. ΔH is larger than zero. [OP] Since the the solubility increases as the temperature is raised (ΔG∘ becomes more negative), I know that ΔS∘ is positive This argument linking solubility and Gibbs energy is incorrect. Le Chatelier or the Van't Hoff equation connects the temperature-dependence to the reaction enthalpy, not to the ...

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The pressure inside the can/bottle is higher than the atmospheric pressure and due to this higher pressure, the freezing point of the beer increases. When you opened the can/bottle, the pressure of the beer equals the atmospheric pressure, decreasing the freezing point of the beer and hence, starting to freeze. Cheers!

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