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Unformatted text preview: CHAPTER 15 THERMODYNAMICS CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1. SSM REASONING AND SOLUTION The plunger of a bicycle tire pump is pushed down rapidly with the end of the pump sealed so that no air escapes. Since the compression occurs rapidly, there is no time for heat to flow into or out of the system. Therefore, to a very good approximation, the process may be treated as an adiabatic compression that is described by Equation 15.4: W = (3 /2) nR ( T i − T f ) The person who pushes the plunger down does work on the system, therefore W is negative. It follows that the term ( T i − T f ) must also be negative. Thus, the final temperature T f must be greater than the initial temperature T i . This increase in temperature is evidenced by the fact that the pump becomes warm to the touch. Alternate Explanation: Since the compression occurs rapidly, there is no time for heat to flow into or out of the system. Therefore, to a very good approximation, the process may be treated as an adiabatic compression. According to the first law of thermodynamics, the change in the internal energy is ∆ , since Q = 0 for adiabatic processes. Since work is done on the system, W is negative; therefore the change in the internal energy, ∆ U , is positive. The work done by the person pushing the plunger is manifested as an increase in the internal energy of the air in the pump. The internal energy of an ideal gas is proportional to the Kelvin temperature. Since the internal energy of the gas increases, the temperature of the air in the pump must also increase. This increase in temperature is evidenced by the fact that the pump becomes warm to the touch. U = Q − W = − W ____________________________________________________________________________________________ 2 . REASONING AND SOLUTION The work done in an isobaric process is given by Equation 15.2: W = P ( V f − V i ). According to the first law of thermodynamics, the change in the internal energy is ∆ U = Q − W = Q − P ( V f − V i ). One hundred joules of heat is added to a gas, and the gas expands at constant pressure (isobarically). Since the gas expands, the final volume will be greater than the initial volume. Therefore, the term P ( V f − V i ) will be positive. Since Q = +100 J, and the term P ( V f − V i ) is positive, the change in the internal energy must be less than 100 J. It is not possible that the internal energy increases by 200 J. ____________________________________________________________________________________________ 3 . REASONING AND SOLUTION The internal energy of an ideal gas is proportional to its Kelvin temperature (see Equation 14.7). In an isothermal process the temperature remains constant; therefore, the internal energy of an ideal gas remains constant throughout an isothermal process. Thus, if a gas is compressed isothermally and its internal energy increases, the gas is not an ideal gas. increases, the gas is not an ideal gas....
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 Spring '08
 holland
 Thermodynamics, Entropy, Heat

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