Reading Quiz

Question 1:

Eq. (5.46) relates the slope of the phase boundary line dP/dT for the gas/liquid phase boundary, and is given as dP/dT = (S_g - S_l)/(V_g - V_l). What would the slope of the phase boundary line dP/dT be for the liquid/solid phase boundary?

Answer:

We can repeat the analysis shown in eq. (5.44) and eq. (5.45), or we can just write dP/dT = (S_l - S_s)/(V_l - V_s).
  1. [S(l)-S(s)]/[V(l)-V(s)]
  2. =L/TDeltaV
  3. dP/dT = (S_l - S_s)/(V_l - V_s)
  4. The slope of the phase boundary line for the liquid/solid boundary would be very steeply positive for a thing like carbon dioxide, or very steeply negative for a thing like water.
  5. (Sl-Ss)/(Vl-Vs)
  6. dP/dT = (S_l - S_s)/(V_l - V_s)
  7. (S_l-S_s)/(V_l-V_s)
  8. dP/dT = (S_l-S_s)/(V_l-V_s)
  9. (S_l - S_s)/(V_l - V_s). This follows from the thermodynamic identitiy as stated in 5.23
  10. dP/dT=L/T*deltaV for any phase boundary line.
  11. (S_l-S_s)/(V_l-V_s)

Question 2:

Use your result from the previous question to explain why the slope of the liquid/solid phase boundary of H2O is negative, as shown in the phase diagram given in Figure 5.11. You'll likely need to use what you know about the volume of ice vs. water.

Answer:

The entropy of the liquid is larger than the entropy of the solid, so S_l - S_s > 0. The volume of the liquid is smaller than the volume of an equal amount of the solid, so V_l - V_s < 0. So their ratio will be negative, so dP/dT is negative.
  1. V(s) > V(l) for water, so the slope, dp/dT is negative
  2. deltaV=Vl-Vs=>Positive T=273K L=Latent heat= negative since the ice is freezing
  3. The volume of ice is larger than that of water, so V_l-V_s is negative. Since S_l-S_s is still positive, dP/dT is negative.
  4. The slope of the liquid/solid phase boundary of water is negative because the volume of ice is slightly bigger than the volume of water. (V_l - V_g) then turns out to be negative.
  5. This is because althought the entropy of water is greater than the entropy of the solid, the volume of ice is more than that of water so this gives a negative value.
  6. The slope comes out to be negative because of the special case that the volume of ice is greater than the volume of liquid water (for the same amount of stuff). So, V_ice > V_water and V_water - V_ice < 0 so the slope dP/dT comes out negative.
  7. The volume of a mole of ice is larger then the volume of a mole of liquid water, so the denominator above is negative. Since the entropy of liquid water is greater then that of ice, the numerator is positive, which makes the slope negative.
  8. The volume of liquid water is smaller than the volume of solid ice. The entropy of liquid water is greater than the entropy of sold ice. So the relation above for dP/dT is a negative quantity, meaning a negative slope on a PT phase diagram.
  9. Well, the volume of ice is > than the volume of water. This means that in the value dP/dT calculated above, V_l-V_s would be negative, and since S_l-S_s is positive, the dP/dT would be negative.
  10. As water freezes, it expands, and as it melts it contracts, this causes a negative delta V to be found in the denominator of this equation when the water goes from solid to liquid. Because the latent heat is positive, and temperature is constant and positive, this causes the negative slope on the phase diagram.
  11. Because the volume of ice is greater than the volume of water so the bottom half of the equation is negative.

Question 3:

The version of the Clausius-Clapeyron relation given in eq. (5.47) is dP/dT = L/(T ΔV). I find the choice of the symbol L to be confusing here. What symbol do we usually use instead of L? More specifically, instead of L/T, what do we usually use?

Answer:

For a phase transition where the temperature is constant, we usually use ΔS = Q/T. I'm not sure why the text uses L instead of Q; in fact, we usually use Q = mL for this situation...
  1. We usually use Q instead of L and S instead of L/T. For vaporization at constant pressure, Q = delta_H and delta_V is approximately Vm(g) or RT/p.
  2. We would usually use Q instead of L and S instead of L/T
  3. We usually use Q instead of L, or S instead of L/T = Q/T.
  4. We usually use the specific heat capacity, c, instead of L, and we usually use the heat capacity C, instead of L/T.
  5. Delta S is usually written as Q/T,so we usually use Q/T instead of L/T (although I think they are different Delta S's..)
  6. L is just what we use for the heat Q involved with changing phase, and Q/T is the change in entropy delta_S.
  7. L=Q/m And I'm not sure what we usually use instead of L/T.
  8. We usually use Q because we say S = Q/T. Here we use L because this a phase transition and it stands for latent heat.
  9. The book replaces S_g-S_l with L/T. This has the more familiar form of Delta*S=Q/T. With L, we mean the total heat needed to convert the material from liquid to gas. (or whatever two phases are being studied).
  10. delta S.
  11. L is the latent heat, I'm not sure what's confusing. Do we use specific heat instead?

Question 4:

What material from the reading (or previous classes) would you like me to go over in more detail?

Answer:

Your responses below.
  2. -
  4. I sent this quiz to you earlier, but the website was very slow and it seems that the internet may have eaten the answer. I hope that this one goes through.
  6. I guess we'll see if using the Clausius-Clapeyron relation is as simple as its derivation was.
  7. See 3.
  9. none
  10. This reading was rather straight-forward. No problems comprehending.
  11. Question 3, i'm confused.