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Chapter 5: Q 5.69 (page 199)

What happens when you spread salt crystals over an icy sidewalk? Why is this procedure rarely used in very cold climates?

Short Answer

Expert verified

The transition temperature between the phases will be lower than, implying that the ice on the street will not freeze until the temperature drops to the new phase temperature.

Step by step solution

01

Given information

Salt crystals are spread over an icy sidewalk.

02

Explanation

We know that water freezes at 0°C, thus sprinkling salt on the icy sidewalk will lower the phase temperature. As a result, the transition temperature between the phases will be lower than 0°C, implying that the ice on the street will not freeze until the temperature drops to the new phase temperature.

This approach will not work in very cold nations since the temperature is below the phase temperature of the water and salt mixture, and the mixture also causes concrete damage, so it is only used in very cold countries on rare occasions.

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Most popular questions from this chapter

Consider the production of ammonia from nitrogen and hydrogen,

N2 + 3H2 →2NH3
at 298 K and 1 bar. From the values of △Hand S tabulated at the back of this book, compute △Gfor this reaction and check that it is consistent with the value given in the table.

Consider an ideal mixture of just 100 molecules, varying in com- position from pure A to pure B. Use a computer to calculate the mixing entropy as a function of NA, and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B; by how much does the entropy increase? By how much does the entropy increase when you convert a second molecule, and then a third, from A to B? Discuss.

Use a Maxwell relation from the previous problem and the third law of thermodynamics to prove that the thermal expansion coefficient (defined in Problem 1.7) must be zero at T=0.

Use the result of the previous problem and the approximate values of a and b to find the value of Tc, Pc, Vc/N for N2, H2O and He.

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between CPandCV.

(a) With the heat capacity expressions from Problem 3.33 in mind, first considerSto be a function of TandV.Expand dSin terms of the partial derivatives (∂S/∂T)Vand (∂S/∂V)T. Note that one of these derivatives is related toCV

(b) To bring in CP, considerlocalid="1648430264419" Vto be a function ofTand P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP=0and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP-CV

(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

CP=CV+TVβ2κT

(d) Check that this formula gives the correct value of CP-CVfor an ideal gas.

(e) Use this formula to argue that CPcannot be less than CV.

(f) Use the data in Problem 1.46 to evaluateCP-CVfor water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of CPfor three elemental solids, compared to predicted values of CV. It turns out that a graph of βvs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CPand CVagree at low temperatures but diverge in the way they do at higher temperatures.
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