# The tutor implicates the Clausius-Clapeyron equation to explain why combustion is more difficult at lower temperatures.

To burn, a fuel must evaporate.1

In colder temperatures, fuel has less tendency to evaporate. For two specific temperatures (T1 and T2), the Clausius-Clapeyron equation can give a comparison of the vapor pressures (p2 vs p1) of a liquid with molar enthalpy of evaporation ΔHvap:

ln(p2/p1) = ΔHvap(T2 – T1)/(R*T2*T1), where

R = 8.314 J/(K*mol), the idea gas constant.

T1, T2 are absolute temperatures.

Note, from my post from Feb 20, that ΔHvap for gasoline is approximately 38.1kJ/mol.

Example: Compare the vapor pressure of gasoline at 12°C (285K) vs -5°C (268K).

Solution: Let T2 be 268K (-5°C). Then

ln(p2/p1) = 38100(268 – 285)/(268*285) = -1.02

Taking the antilog gives

p2/p1 = 0.36 = 36%

Apparently, gasoline is about 36% as likely to evaporate at -5°C as at 12°C. Not much surprise, really, that a car can be more difficult to start in colder temperatures.

Source:

Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

White, J. Edmund. Physical Chemistry, Harcourt Brace Jovanovich College Outline Series. San Diego: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor introduces an interesting law from chemistry.

Amonton’s Law states that the pressure of a gas is directly related to its absolute temperature:

P=kT, where k is some constant.

Amonton’s Law needs the conditions of constant moles of gas present and constant volume, which means it is only feasible for calculating pressure response to temperature change, rather than the other way. (For the increase of temperature due to compression, see my post here.)

With refrigeration, Amonton’s Law is useful for considering what happens after compression. During compression, the refrigerant’s temperature rises. Afterwards, however, it radiates heat to the environment; its temperature plummets. During that time, its pressure decreases as well, by Amonton’s Law. The loss of heat and decrease in pressure prepare the refrigerant for condensation to liquid state.

I’ll be talking further about refrigeration:)

Source:

Mortimer, Charles E. Chemistry, 6th ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor looks more specifically into the effect of compression on gas temperature.

In my January 20 post I began about thermodynamics and the effect of compressing a gas. Today, I’ll give more specific coverage.

The temperature rise a gas experiences (without change in entropy) due to pressure is given by the formula

T2 = T1(P2/P1)[1-1/γ]

where

T1,T2 are initial and final temperatures

P1,P2 are initial and final temperatures

γ = Cp/Cv, where

Cp = gas specific heat at constant pressure

Cv = gas specific heat at constant volume

Typcially, γ might be around 1.4. Therefore, imagining a diesel engine with 17:1 compression, at starting temperature 298K (25°C) the resulting temp, T2, might be

T2=298(17/1)[1-1/1.4]

T2=670K

Source:

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor offers definitions of diffuse and effuse, two related terms from chemistry.

To diffuse is to spread out, while to effuse is to escape. A gas will diffuse through the atmosphere, while it will effuse through a hole in a container.

Source:

White, J. Edmund. Physical Chemistry, College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor examines the idea that internal combustion engines are more efficient in cold weather.

An upper limit for efficiency of an internal combustion engine is

eff = (Tcombust – Tsurrounding)/Tcombust

where

Tcombust is the temp of the combustion cylinder

Tcombust, Tsurrounding both in degrees Kelvin (Celsius + 273).

Let’s imagine a diesel engine, whose average internal cylinder temperature might be around 1600°C. Then at outdoor temp of 25°C (293K) we have

eff = (1873 – 298)/1873 = 84.1%

Likewise, at outdoor temp -25°C we have

eff = (1873 – 248)/1873 = 86.8%

The 2.7% increase in efficiency at -25°C vs 25°C may be noticeable to an operator.

Source:

www.ncert.nic.in

White, J. Edmund. Physical Chemistry: College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor defines the term state function.

A state function is one that is independent of path. Another way of explaining it: for a state function, if you end up where you started, nothing has changed.

An example of a state function is heat content of a cup of water. Let’s imagine it starts at 25°C. If you heat it from 25°C to 99°C, then let it cool back to 85°C, its ending heat content will be the same as if you just heat it from 25°C to 85°C. The heat content of an object of fixed mass depends only on its temperature, not how it arrived at that temperature.

Gravitational potential energy is also a state function. It depends on mass and height according to

Egp=mgh

Therefore, a given object’s gravitational potential energy depends on its current height, but not how it arrived there.

Position is a state function, but distance is not. A runner who travels exactly one lap of a track has not changed position, but has travelled 400m (typically). Running another lap, the runner’s position will be restored, but their distance will increase to 800m.

Source:

White, J. Edmund. Physical Chemistry: College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.