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Assignment 3. Solutions. Problems. February 22. 1. Find a vector of magnitude 3 in the direction opposite to the direction of v = 1 2 i 1 2 j 1 2 k. Solution. The vector we are looking for is 3 v jvj We have jvj= r 1 4 1 4 1 4 = p 3 2 Therefore, 3 v jvj = 2 p 3v = p 3i p 3 1 2 j p 3k 2. Given P 1(1;4;5) and P

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May 05, 2015p * V = constant This relationship between pressure and volume is called Boyle's Law in his honor. For example, suppose we have a theoretical gas confined in a jar with a piston at the top. The initial state of the gas has a volume equal to 4.0 cubic meters and the pressure is 1.0 kilopascal. With the temperature and number of moles held

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O comportamento termodinmico mais simples verificado nos gases, e, por esse motivo, facilmente verifica-se que, para fluidos homogneos em geral, um estado de equilbrio termodinmico inteiramente caracterizado por qualquer par entre as variveis P, V e T (respectivamente, presso, volume e temperatura).

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At room temp we have pV=nRT and T=273K then dip it liquid nitrogen p'V'=nRT with T= 78K, p'V' should be smaller by a factor of 273/78=3.5 compared to pV So why does the balloon get a lot smaller? Since air boils at 90 K, air is liquid at 78 K and is no longer a gas so the ideal gas law does not apply.)

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3.3.32 Let Sbe the tetrahedron in R3 with vertices at the vectors 0, e 1, e 2, and e 3. Let S0be the tetrahedron with vertices at vectors 0, v 1, v 2, and v 3.See the gure in the book on page 210. (a) Describe a linear transformation that maps Sonto S0. Suppose we call this transformation T.

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V k T = According to this law, the volume of the gas increases as the temperature increases, but the ratio of volume to temperature, V/T, is always constant( k'). Like the Boyle's law, the Charles's law can also be applied to two sets of conditions at a constant pressure. The equation applicable to this situation is

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