Kelvin’s law

The relative pressure domain close unity is subject to experimental limitations. This is because the temperature of the sample may be slightly greater than the temperature of the cryogenic bath owing to the cxothermicity of the adsorption and heat transfer phenomena. There is thus a great deal of uncertainty in the application of Kelvin s law to high relative pressure levels. In practice, application is limited to pores less than 50 nm in size. 

The second hydraulic process that is included in the fracture is the variation of capillary pressure caused by changes in the aperture. According to Kelvin s law the capillary pressure necessary to desaturate a fracture is given by ... 

Man, C.S. Proof that the Clausius-Serrin inequaUty is equivalent to a strengthened Kelvin s law . Rend. Accad. Naz. Lincei 8-69(6), 399-409 (1980)... 

These applications of Kelvin s law merely express the quantity of energy required to transfer a molecule from a macroscopic object (a plane surface) to a droplet, thereby increasing the surface area of the latter. They explain... 

Kelvin s law in electrical engineering states that the economic size of a conductor for transmitting current is that for which the annual investment charges for copper just equal the annual cost of the power lost. The similarity between this law and the concept of impedance matching discussed in Ch. 10, Sec. 12, should be noted. 

Lord Kelvin s close associate, the expert experimentalist J. P. Joule, set about to test the former s theoretical relationship and in 1859 published an extensive paper on the thermoelastic properties of various solids—metals, woods of different kinds, and, most prominent of all, natural rubber. In the half century between Gough and Joule not only was a suitable theoretical formula made available through establishment of the second law of thermodynamics, but as a result of the discovery of vulcanization (Goodyear, 1839) Joule had at his disposal a more perfectly elastic substance, vulcanized rubber, and most of his experiments were carried out on samples which had been vulcanized. He confirmed Gough s first two observations but contested the third. On stretching vulcanized rubber to twice its initial length. Joule ob-... 

The relationship between Kelvin temperature and the volume of a gas is expressed as Charles s law The volume of a confined gas, at a constant pressure, is directly proportional to its Kelvin temperature. Mathematically, Charles s law is ... 

Arts. Boyle s law may be used because the temperature is unchanged. Alternately, the combined gas law may be used, with the Kelvin equivalent of 18°C used for both T[ and 7. ... 

Ans. (a) through (e) Kelvin, (r/) No temperature is used in the calculation for Boyle s law problems since the temperature must be constant. 

A Charles s law states that the volume/temperature ratio is constant (temperature in kelvins). 

There is a direct relationship between the Kelvin temperature and the volume as one increases, the other also increases. Mathematically, Charles s law can be represented as ... 

Gay-Lussac s law describes the relationship between the pressure of a gas and its Kelvin temperature if the volume and amount are held constant. Figure 8.5 represents the process of heating a given amount of gas at a constant volume. 

Charle s law / ley de Charles establece que el volumen de una masa determinada de un gas, bajo condiciones constantes de presidn, es directamente proporcional a su temperatura en grades Kelvin, (pag. 424)... 

When you work with Charles s law, converting Celsius temperatures to the Kelvin scale is crucial. If you don t make the conversion, your answer will be incorrect. 

C. The question provides an initial temperature, an initial volume, and a final volume. You re asked to find the final temperature, T. Apply Charles s law, plugging in the known values and solving for the final temperature. But take Ccire — Chcirles s law requires you to convert all temperatures to kelvins (where... 

C. Charles s law is the method here. The unknown is the final temperature, T. You re given an initial temperature as well as the initial and final volumes. After converting the initial temperature to units of kelvins (20.0°C+273 = 293 K), plug in the known values and solve for final temperature ... 

As defined by (4.19) or (4.21), it is easy to recognize that TK is an absolute (strictly non-negative) quantity. Furthermore, one can see from (4.19) that the highest possible efficiency ( —> 1) is achievable only at the absolute zero of the Kelvin scale (7"cK —> 0). In addition, the lowest efficiency of converting heat to work ( —> 0) occurs when the two reservoirs approach the same temperature (7j —> 7"cK), consistent with the statement of Kelvin s principle in Section 4.4. Such limits on engine efficiency can be used to paraphrase the three laws of thermodynamics in somewhat whimsical form as follows (the ultimate formulation of the no free lunch principle) ... 

Every chemistry textbook describes the basic experiments that relate the volume of a gas sample to its pressure and Kelvin temperature. For a given quantity of gas it may be shown, by combining Bovle s law (PV = k", at constant T) and Charles law (V = k T, at constant/5), at... 

He is remembered for Joule s Law lhat describes the rale at which heal is produced by an electric current. Joule s work showed there were different kinds or energy, which can be changed into each other. He established the mechanical equivalence of heat. His work led to the law of conservation of energy. Alsu, he collaborated with William Thomson (Lord Kelvin) and verified experimentally the Joule-Thomson refrigeration effect. 

When data obtained from similar measurements on different gases and at different pressures are plotted, it is found that the straight lines can be extended to the same point (Fig. 4.12). The extension of a graph beyond the data is called extrapolation, and we say that, in this case, the straight lines extrapolate to the same point. For Charles s law, this unique point corresponds to zero volume and —273.15°C. Because a volume cannot be negative, this temperature must be the lowest possible temperature. It is the value corresponding to 0 on the Kelvin scale. It follows that if we use temperature, T, on the Kelvin scale, then we can write Charles s law as... 

STRATEGY To use Kirchhoff s law, we need to know the molar heat capacities of the reactants and products these can be found in Appendix 2A. Combine them in the same way that enthalpies of formation would be combined to calculate a reaction enthalpy and then substitute in Eq. 32. Note that the temperatures must be expressed in kelvins. 

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