please solve these questions. I need only short answer.

Heating Curves and Vapor PressureIntroduction – When studying Thermochemistry, you learned about relating the energy for
systems and surrounding and how to determine the amount of thermal energy exchange
between them. However, in those systems, you were not asked to consider the thermal energy
requirements for a system to change its physical state. In this activity you will explore the
thermal energy requirements for a substance to undergo any state (phase) change and will then
further explore the liquid-gas phase equilibrium as a function of intermolecular forces.
Prerequisite knowledge – This activity requires that you are comfortable with the following
skills and understanding. If you need to review this material, the page numbers provided
indicate where a topic is covered within the OpenStax (2e) text book.
Prerequisite knowledge
Quantifying thermal energy exchange
Reaction enthalpy
Relations involving change in enthalpy
Intermolecular forces
OpenStax
230-237
251-255
260-265
518-530
Changing temperature versus changing state – Most people are familiar with the temperatures
at which pure water changes states. They know that water melts/freezes at 0 C and
boils/condenses at 100 C. In studying Thermochemistry, you dealt with determining the
energy requirements to reach one of these state-change points (temperatures). However,
simply heating a sample of water to one of these temperatures does not instantly change its
state, additional energy is required to weaken or overcome the intermolecular forces between
water molecules – thereby allowing them to transition to the next state.
——————————————————————————————————————————Model 1 – Melting an ice cube
Thermal Energy
The graph below (called a heating curve) shows that as you add thermal energy to melt an ice
cube, the energy is used to both raise the temperature of the sample as well as to convert the
sample from the solid state to the liquid state.
Model 1 continues on next page.
Model 1 continued.
Energy requirements to convert an ice cube at -10.0 C to liquid water at +10.0 C
Energy required to raise
temperature of solid:
𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻
Energy required to convert
from solid to liquid:
𝒒𝟐 = 𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏
Energy required to raise
temperature of liquid:
𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻
Note: you should be familiar with all of the variables and units in these equations from the last
chapter of material on Thermochemistry (Tro, chapter 6). The only new variable is Hfusion, which is the
enthalpy change for the melting (fusion) of a solid.
——————————————————————————————————————————Key Questions – Use the information provided in Model 1 to answer these questions.
KQ 1) According to the heating curve, at what temperature does the ice cube melt?
KQ 2) Did it require more or less thermal energy to raise the temperature of the ice cube to its
melting point or to convert it from a solid to liquid?
KQ 3) Write the equation used to determine the amount of thermal energy (q) required to raise
the temperature of a solid and define all of the variables (including their units) in the equation.
2
KQ 4) Write the equation used to determine the amount of thermal energy (q) required to
convert from a solid to a liquid and define all of the variables (including their units) in the
equation.
KQ 5) How is the equation for raising the temperature of a liquid different from that for raising
the temperature of a solid?
Exercises – These questions will help you develop your understanding of the relationships
presented in Model 1. For each Exercise, report numerical answers with the correct number of
significant digits.
EX 1) Determine the amount of thermal energy (q) required to raise the temperature of 10.0 g
𝐽
solid water from -10.0 C to its melting point. 𝐶𝑠,𝑠𝑜𝑙𝑖𝑑 = 2.09 𝑔∙℃
EX 2) Determine the amount of thermal energy (q) required to convert 10.0 g solid water to
𝑘𝐽
liquid water at its melting point. ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 = 6.02 𝑚𝑜𝑙
EX 3) Determine the amount of thermal energy (q) required to raise the temperature of 10.0 g
𝐽
liquid water from its melting point to +10.0 C. 𝐶𝑠,𝑙𝑖𝑞𝑢𝑖𝑑 = 4.18 𝑔∙℃
EX 4) Determine the total amount of thermal energy required to raise 10.0 g of ice at -10.0 C to
liquid water at +10.0 C.
3
——————————————————————————————————————————Model 2 – Heating and cooling curve
By defining the freezing (Tf) and boiling (Tb) points, the generic curve shown below can be used
to investigate the heating (adding thermal energy) or cooling (removing thermal energy) for any
substance.
Note: the curve represents a general profile, sections are not to scale for any one substance.
The curve sections and their related energy equations for each are noted in the table.
Section
Adding thermal energy
Description
Equation
Section
Removing thermal energy
Description
Equation
1
Solid Phase
𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻
5
Gas Phase
𝒒𝟓 = 𝒎𝑪𝒔,𝒈𝒂𝒔 ∆𝑻
2
Solid→Liquid
(melting)
𝒒𝟐 = 𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏
*4
Gas→Liquid
(condensation)
𝒒𝟒 = −𝒏∆𝑯𝒗𝒂𝒑𝒐𝒓𝒊𝒛𝒂𝒕𝒊𝒐𝒏
3
Liquid Phase
𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻
3
Liquid Phase
𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻
4
Liquid→Gas
(vaporization)
𝒒𝟒 = 𝒏∆𝑯𝒗𝒂𝒑𝒐𝒓𝒊𝒛𝒂𝒕𝒊𝒐𝒏
*2
Liquid→Solid
(freezing)
𝒒𝟐 = −𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏
5
Gas Phase
𝒒𝟓 = 𝒎𝑪𝒔,𝒈𝒂𝒔 ∆𝑻
1
Solid Phase
𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻
*Note: The processes of vaporization and condensation as well as melting and freezing are
opposite processes. As H values are only reported for the endothermic processes, the
addition of a negative sign is required in using them for the corresponding exothermic
processes.
——————————————————————————————————————————4
Key Questions – Use the information provided in Model 2 to answer these questions.
KQ 6) The curve in Model 2 notes 4 points, labeled A-D. Describe the substance at each of these
points, point A has been done as an example.
Example: At point A, the substance is in its solid state and is at the freezing/melting transition
temperature.
KQ 7) Explain why moving from point C to point D on the curve does not require the inclusion
of a negative sign in the formula but moving from point D to point C does require a negative
sign in the formula.
KQ 8) Explain why there is no need to include a negative sign in the formula when moving from
point B to point C OR from point C to point B.
KQ 9) If a sample starts as a gas, with a temperature above its boiling point, how many
individual calculations of thermal energy (q) are required to determine the total amount of
thermal energy removed in converting the entire sample to a solid at its freezing temperature?
5
Exercises – These questions will help you develop your understanding of the relationships
presented in Model 2. For each Exercise, report numerical answers with the correct number of
significant digits. Use the generic curve in Model 2 and the tabulated data below when solving
these exercises.
Value (units)
𝑇𝑓 (℃)
𝑇𝑏 (℃)
𝑘𝐽
∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 (
)
𝑚𝑜𝑙
𝑘𝐽
∆𝐻𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 (
)
𝑚𝑜𝑙
𝐽
𝐶𝑠,𝑠𝑜𝑙𝑖𝑑 (
)
𝑔∙℃
𝐽
𝐶𝑠,𝑙𝑖𝑞𝑢𝑖𝑑 (
)
𝑔∙℃
𝐽
𝐶𝑠,𝑔𝑎𝑠 (
)
𝑔∙℃
Water, H2O
0.00
100.0
Benzene, C6H6
5.40
80.1
6.02
9.90
40.7
30.7
2.09
1.51
4.18
1.73
2.01
1.33
EX 5) Determine the total amount of thermal energy that is evolved when a 10.0 g sample of
water vapor at 105.0 C is cooled to liquid water at 60.0 C.
EX 6) Determine the total amount of thermal energy required to convert a 25.0 g sample of
solid benzene at -2.40 C to benzene vapor at 108.0 C.
6
Vapor Pressure – In a sealed container at a given temperature, an
equilibrium is established between the vaporization and condensation
processes. The amount of gas above the liquid exerts a pressure called the
‘vapor pressure’ for the liquid.
The boiling point for a liquid is defined as the temperature at which the
liquid’s vapor pressure is equal to the external pressure (e.g., atmospheric
pressure). Hence, the boiling point of a liquid changes as the external
pressure changes. For example, pure water has a boiling point of 100.0 C
at an atmospheric pressure of 760 torr (‘sea level’). However, its boiling
point is elevation dependent and drops to 94.0 C in Denver, Colorado
(altitude = 5280 ft, approximate atmospheric pressure = 631 torr).
——————————————————————————————————————————Model 3 – Vapor pressure curves
The vapor pressure for four different pure liquids are plotted as a function of their
temperature. The curves vary due to the intermolecular forces present in each liquid.
——————————————————————————————————————————Key Questions – Use the information provided in Model 3 to answer these questions.
KQ 10) Which molecule has the weakest intermolecular forces present in its liquid sample?
KQ 11) Which molecule has the lowest normal boiling point?
7
KQ 12) If each of the pure liquids in Model 3 were heated to 20.0 C, which would have the
highest vapor pressure?
Exercises – These questions will help you develop your understanding of the relationships
presented in Model 3.
EX 7) Acetone (C3H6O) has a normal boiling point of 56.1 C. Compare the strength of its
intermolecular forces and its vapor pressure to those of ethyl alcohol.
EX 8) Use the information about three unknown compounds to complete each statement.
Compound A
Compound B
Compound C
Molar Mass (g/mol)
100.16
100.20
102.17
Normal Boiling Point (C)
128
98
159
i) Compound ____ experiences hydrogen bonding.
ii) Compound ____ is polar but unable to experience hydrogen bonding.
iii) Compound ____ is neither polar nor able to experience hydrogen bonding.
iv) Compound ____ will have the highest vapor pressure of the three compounds.
EX 9) Using Model 3, determine the normal boiling point of diethyl ether and ethyl alcohol
Normal Tboil
Tboil at 200 torr
diethyl ether
_______________°C
_______________°C
ethyl alcohol
_______________°C
_______________°C
——————————————————————————————————————————Model 3– More Information Vapor pressure vs. temperature curves
The relationship between vapor pressure and temperature is an exponential growth curve and
can be modeled by the Clausius-Clapeyron Equation. Manipulation of the Clausius-Clapeyron
Equation shows that the logarithm of the vapor pressure vs. inverse absolute temperature is a
linear function whose slope is proportional to the ΔHvap. Further manipulation can yield a form
of the Clausius-Clapeyron Equation in a two-point form (needing just two measurements of
8
vapor pressure and temperature), see Equation 1. The two-point form generally gives less
accurate results but is quite nice in a classroom setting. The two-point form can also be used to
estimate the boiling point at various ambient atmospheric pressures as long as you know ΔHvap
and a vapor pressure at a particular temperature (remember: the vapor pressure at the normal
boiling point is 760 torr).
 P  − H vap  1 1 
 − 
ln  2  =
R  T2 T1 
 P1 
Equation 1: Clausius-Clapeyron Equation in a two-point form. P1, T1 and P2, T2 represent
some initial and final set of temperature and vapor pressure for a particular substance.
Pressure can be in any unit as long as P1 and P2 have the same units. Temperature must be
in units of Kelvin. ΔHvap is the enthalpy of vaporization in units of J/mol. R is the ideal gas
constant 8.314 J/mol-K. ln indicates that you will be taking the natural log of the value in
parentheses.
Ex 10) Use your values from EX 9 and equation 1 to estimate the ΔHvap for diethyl ether and
ethyl alcohol.
diethyl ether
ethyl alcohol
9
Challenge Problem – This problem requires you to combine your prior knowledge of
calorimetry with your newly acquired knowledge of phase change energy.
1. A 53.5 g sample of ice at exactly 0 C is placed in 115. g of water at 75. C. If no thermal
energy is lost to the surroundings, what is the final temperature of the mixture?
2. The ΔHvap for mercury is 59.1 kJ/mol. Its normal boiling point is 357°C. What is the
vapor pressure at 25°C? Express your answer to 2 sig figs.
3. At 25°C, how many grams of mercury vapor would fill up a room that is 3.00 m X 4.00
m x 5.00 m?
10