Alcohol baths used to be used to decrease the body temperature of a person with

Question

Alcohol baths used to be used to decrease the body temperature of a person with a fever.This was done by rubbing the person down with alcohol.

1) explain the fact given the following thermochemical equation C2H5OH (l) = C2H5OH (g) dH =42.2kJ/mol

2) How much heat exchange is occuring if a person is rubbed down with 250mL of alcohol? Density of alcohol is 0.786 g/mL at 20C.

3) Viewing the alcohol liquid as a system, what was the initial volume of gaseous alcohol?

4) What was the final volume of gaseous alcohol with atmospheric pressure of 600mmHg?

5) Calculate work done as the alcohol vapor expanded into the environment with atmospheric pressure of 600mmHg?

6) From this information, calculate the change in internal energy for the "system" we defined above.

Is it positive or nagative? What is the system in this example?

Explanation / Answer

C2H5OH (l) = C2H5OH (g) deltaH = 42.2 kJ/mol

1.) This is endothermic reaction. The heat required (42.2 kJ/mol) to convert alcohal from liquid state to gaseous state is provided by heat of the body. The alchal (l), when rubbed with the body, absorb heat and get converted to gaseous state.

2.) Molecular Weight of the alcohal, M = 46 g/mol
At 20 oC, density = 0.786 g/mL
Volume = 250 mL
Thus, mass of alcohal = density * volume = 0.786 * 250 = 196.5 g
number of moles = mass/M = 196.5/46 = 4.27
deltaH = 42.2 kJ/mol
Heat exchange = deltaH * no. of moles = 42.2 * 4.27 = 180.27 kJ

3.) By viewing the liquid alcohal as a system, intial volume of gaseous alcohal = 0.

4.) For 250 mL of the liquid alcohal, n = 4.27.
P = 600 mm Hg = 1.013 * 600/760 atm = 0.800 atm
T = 20 oC = 293 K
Assume the gaseous alcohal to be an ideal gas.
PV = nRT => V = nRT/P = 4.27 * 0.0821 * 293 / 0.800 = 128.40 L

5.) Work done by alcohal during expansion,
W = P2V2 - P1V1 = (0.8 * 1.013 * 105 Pa)*(128.4 * 10-3 m3) - 0 = 104.055 * 102 J = 10.4 kJ

6.) Q = 180.27 kJ, W = 10.4 kJ
From 1st Law of Thermodynamcis,
delta U = Q - W
delta U = 180.27 - 10.4 = 169.87 kJ

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