Change in Internal Energy, Heat, and Pressure-Volume Work | University of Arkansas - Edubirdie

University:
University of Arkansas
Course:
CHEM 1123 | University Chemistry II
Academic year:

2019
Views:

488

Pages:

1
Author:

BobN

This example problem will involve the change in internal energy, heat, and pressure-volume work. Our problem reads a combustion reaction in a piston decreases the internal energy of the reactants by seven hundred and fifty joules and increases the volume in the piston from 0.2 liters to 0.6 liters against a constant external pressure of one atmosphere. To calculate the heat for this combustion reaction, we will use the equation ΔE = Q + W, where ΔE is the change in internal energy, Q is heat, and W is work. We can solve for heat by subtracting work from the change in internal energy. In the problem, we are told that the piston decreases the internal energy of the reactants by seven hundred and fifty joules, so our change in internal energy is -750 joules. We can use the equation work = -pressure x change in volume. In this problem, we have a constant external pressure of one atmosphere, and our final volume minus our initial volume is 0.6 liters minus 0.2 liters. Multiplying these values, we get -0.4 liter-atmospheres for work. Next, we need to convert liter-atmospheres into joules in order to use it in our equation for heat. We can use dimensional analysis and the conversion factor given in our hint, which is 1 liter-atmosphere equals 101.3 joules. Converting -0.4 liter-atmospheres, we get -40.52 joules for work. Now, we can plug our change in internal energy (-750 joules) and our work (-40.52 joules) into our equation for heat: ΔE = Q + W. Solving for heat, we get Q = -750 joules - (-40.52 joules) = -709.48 joules. Since heat is negative, it means that our combustion reaction is exothermic, giving off this amount of heat (-709.48 joules).

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