Part 2: Change in Internal Energy, Heat, and Pressure-Volume Work - Edubirdie

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

2019
Views:

158

Pages:

1
Author:

BobN

In this example, we find the change in internal energy given heat and pressure-volume work data. Our problem reads, "A combustion reaction occurs within a cylinder equipped with a piston. The reaction releases 0.438 kilojoules of thermal energy and pushes the piston out against a constant external pressure of 2.50 atmospheres from 0.600 liters to 2.400 liters. What is the change in internal energy for the reaction in joules? Hint: 101.3 joules equals one liter atmosphere." We can start with our equation for the change in internal energy: ΔE equals heat, which is Q, plus work, which is W. In the problem, we're told 0.438 kilojoules of thermal energy is released by the reaction. When energy is released by a reaction, the reaction is exothermic and Q is negative. And since we want to calculate the change in internal energy in joules, we convert our kilojoules to joules. Our equation for work is W equals negative P delta V, where P is the external pressure and delta V is the change in volume. Plugging in our numbers from the problem, we can calculate the change in volume as the final volume minus the initial volume. Our work is -4.5 liter atmospheres. A negative work means the system is performing work on the surroundings, which in this case is the expansion of the gas against an external pressure. We can't plug this value directly into our equation because we need work in joules, so we will use dimensional analysis to convert -4.50 liter atmospheres to joules using the conversion factor given in the problem: 1 liter atmosphere is equal to 101.3 joules. Completing the calculation, our work is -456 joules. Plugging in our values for heat and work into the equation at the top, our change in internal energy for the reaction is -894 joules.

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