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Chapter 5: Q. 5.21 (page 163)
Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.
Heat capacity is an extensive property
Specific heat is an intensive property.
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An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1掳C. The system is then placed in thermal contact with a low-temperature reservoir at -1掳C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1掳C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor's reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 -Te/Th
Osmotic pressure measurements can be used to determine the molecular weights of large molecules such as proteins. For a solution of large molecules to qualify as "dilute," its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data for the protein hemoglobin dissolved in water at :
| Concentration (grams/liter) | (cm) |
| 5.6 | 2.0 |
| 16.6 | 6.5 |
| 32.5 | 12.8 |
| 43.4 | 17.6 |
| 54.0 | 22.6 |
The quantity is the equilibrium difference in fluid level between the solution and the pure solvent,. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole).
An experimental arrangement for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level,, is just enough to supply the osmotic pressure.
If expression 5.68 is correct, it must be extensive: Increasing both NA and NB by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression 5.68 has this property. Show that it would not have this property had we not added the term proportional to In NA!.
Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.
(a) Use the vapour pressure equation (Problem 5.35) and the data in Figure 5.11 to plot a graph of the vapour pressure of water from 0掳C to 40掳C. Notice that the vapour pressure approximately doubles for every 10掳 increase in temperature.
(b) Suppose that the temperature on a certain summer day is 30掳 C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?
In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.
(a) Expand the van der Waals equation in a Taylor series in , keeping terms through order . Argue that, for T sufficiently close to Tc, the term quadratic in becomes negligible compared to the others and may be dropped.
(b) The resulting expression for P(V) is antisymmetric about the point V = Ve. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary,
( c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find .8, where (3 is known as a critical exponent. Experiments show that (3 has a universal value of about 1/3, but the van der Waals model predicts a larger value.
(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function.
The shape of the T = Tc isotherm defines another critical exponent, called Calculate 5 in the van der Waals model. (Experimental values of 5 are typically around 4 or 5.)
A third critical exponent describes the temperature dependence of the isothermal compressibility, K=-t This quantity diverges at the critical point, in proportion to a power of (T-Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents 'Y and -y' are defined by the relations
Calculate K on both sides of the critical point in the van der Waals model, and show that 'Y = -y' in this model.
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