Heat Change and Specific Heat by pptfiles

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									Heat Change and Specific Heat
• • Specific heat (Chapter 6): amount of heat required to raise temperature of 1g of substance by 1oC. Calculate heat change (q) in stages: related to phase changes/ heating curve.

Heat Change and Specific Heat
• Measure heat change (q) at each stage:
– –
Fig 11.8

q1, q3 and q5: heat change within a phase (solid, liquid or vapor) q2 and q4: heat change during phase change

q4 q3

q5

q2 q1

Heat Change and Specific Heat
• • Specific heat (Chapter 6): amount of heat required to raise temperature of 1g of substance by 1oC. Calculate heat change (q) in stages: related to phase changes/ heating curve.
q = nH
use for phase changes (temp constant) solid-liquid (Hfus ) or liquid-vapor (H vap ) equilibrium n = # moles of substance

q = mst

use for temp changes within a phase m = mass of substance (g) s = specific heat (J/ g • oC) t = change in temp |tfinal - tinitial| (should be positive value)

Heat Change and Specific Heat
• Calculate heat change (q) in stages: related to phase changes/ heating curve.
q5 = mst q4 = nHvap q3 = mst q2 = nHfus

q1 = mst

Clausius-Clapeyron Equation
• Relationship of vapor pressure (P) to absolute temperature (T).
– Can determine values of P, T or Hvap.
ln P = y= Hvap RT mx + B b
P = vapor pressure (mm Hg) Hvap= heat of vaporization R = gas constant (8.314 J/K mol) T = temperature (K) B = constant

pg 329

Clausius-Clapeyron Equation
• At two different temperatures, can generate the two-point formula.
– – – P1, T1: pressure at temp 1 P2, T2: pressure at temp 2 Hvap is a constant
ln P2 P1 = Hvap R 1 T1 1 T2 or ln P1 P2 =

(on pg. 329) ln P1 P2 =

Clausius-Clapeyron Equation
• The vapor pressure of ethanol is 100 mm Hg at 34.9oC. Calculate the vapor pressure of ethanol at 63.5oC.
– The Hvap of ethanol is 39.3 kJ/ mol

•

What information is given or known?
– – – – P1 and T1 (100 mm Hg and 34.9oC) T2 (63.5oC) Hvap (39.3 kJ/ mol) R (8.314 J/ K mol)

Clausius-Clapeyron Equation
• The vapor pressure of ethanol is 100 mm Hg at 34.9oC. Calculate the vapor pressure of ethanol at 63.5oC.
– The Hvap of ethanol is 39.3 kJ/ mol

•

How to solve?
– – Convert oC to K and kJ to J Use Clausius-Clapeyron and solve for P2
ln P2 P1 = Hvap R 1 T1 1 T2 or ln P1 P2 =

(on pg. 329) ln P1 =

Clausius-Clapeyron Equation
• The vapor pressure of ethanol is 100 mm Hg at 34.9oC. Calculate the vapor pressure of ethanol at 63.5oC.
ln P2 P1 P2 100 mm Hg P2 100 mm Hg = Hvap R 1 T1 1 T2 1 307.9 K 1 336.5 K 39.3 x 103 J/ mol 8.314 J/ K mol

ln

=

ln

= 1.305

Take antilog of both sides of the equation. The right side becomes ex.
P2 100 mm Hg P2 = (e1.305) (100 mm Hg) = (3.688) (100 mm Hg) = 369 mm Hg
1.305 = e


								
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