2.1 Why are four values of the properties mass, volume, temperature, and pressure insufficient to describe the state of a nonequilibrium gas ; for example, a turbulent gas ?
2.2 Could n in the ideal gas law have been identified as the number of moles without Avogadro's hypothesis ?
2.3 According to Dalton's law, what is most of the pressure of the atmosphere (that is, air) due to ?
2.4 Why don't all the gas molecules in the atmosphere simply fall to earth ?
2.5 The force on an ion of negative charge -q in a constant electric field E in the z direction is F = -qE. By analogy to the gravitational case, what is the spatial distribution of such ions immersed in a column of gas and subject to a constant vertical field E ? (Ignore the effect of gravity on the ions and on the gas.) P R
jawab;
1. because the ideal gas law, p V = nR T, i s a relation between the four variables that describe the state of any gas. As such, it is an equation of state. The variables in this equation fall into two classes : n and V are extensive variables (extensive properties), while p and T are intensive variables (intensive properties).
2. no, it”s couldn”t. We now inquire about the significance of the characteristic mass M. Avogadro's law
says that equal volumes of different gases under the same conditions of temperature and
pressure contain equal numbers of molecules ; that is, they contain the same amount of
substance. We have compared equal volumes, Vo , under the same temperature and pressure,
To and Po , to obtain the characteristic masses ofthe different gases. According to Avogadro's
law these characteristic masses must contain the same number of molecules. If we choose
Po , To , and Vo so that the number is equal to N A = 6.022 X 102 3, then the amount of
substance in the characteristic mass is one mole and M is the molar mass. Also, M is N A times the mass of the individual molecule, m, or M = Na x m
3. This is Dalton's law of partial pressures, which states that at any specified temperature the
total pressure exerted by a gas mixture is equal to the sum of the partial pressures of the
. constituent gases. The first gas is said to exert a partial pressure PI> the second gas exerts a
partial pressure P2 , and so on. Partial pressures are calculated using Eqs. (2.29).
4.Because, that gas is very light than earth gravitation. suppose that for a certain gas the pressure at ground level is 1 atm and the distribution shows that the pressure decreases to t atm at a height of 10 km. Then for this same gas, the pressure at a height z + 10 km is one-half the value of the pressure at the height z. Thus at any height, the pressure is one-half the value it has at a height 10 km below
5. The differential equation relates tkchange in pressure, dp, to the density ofthe
fluid, the gravitational acceleration, and the increment in height dz. The negative sign
means that if the height increases (dz is + ), the pressure of the fluid will decrease (dp is -).
The effect of change in height on the pressure is proportional to the density of the fluid ; thus
the effect is important for liquids and negligible for gases.
If the density of a fluid is independent of pressure, as is the case for liquids, then
equation may be integrated immediately. Since p and g are constants, they are removed
from the integral and we obtain
r dp = -pg fdZ, Po 0
Tidak ada komentar:
Posting Komentar