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Chapter 5: Problem 7

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2\(]\)

Short Answer

Expert verified
For the process of compressing air in a bicycle pump: (a) the sign of work (w) is negative since work is being done on the system, (b) the sign of heat (q) is negative since heat is being removed from the system, and (c) the sign of ∆E is generally positive due to the increased kinetic energy of the molecules as air is compressed, despite heat transfer losses.

Step by step solution

01

(a) Determine the sign of w when compressing air.

Since we are compressing the air in the bicycle pump, work is being done on the system (air inside the pump). According to the definition, when work is done on the system, the work (w) is negative. Therefore, the sign of w is negative.
02

(b) Determine the sign of q for this process.

As we compress the air in the bicycle pump and the pump gets warm, we observe that heat is being transferred from the system (air) to the surroundings (pump body). Since heat is being removed from the system, the sign of heat (q) is negative.
03

(c) Determine the sign of ΔE and explain your reasoning.

We can use the thermodynamic relation \(ΔE = q + w\) to determine the sign of ΔE. In this case, both q and w are negative, which means \(ΔE\) is negative. However, we should also consider the physical process of compression. When air is compressed, its internal energy, mainly due to the increased kinetic energy of the molecules, should increase. This results in a positive change in internal energy (∆E). In reality, we could suppose that due to heat transfer some part of the internal energy increase is lost, but our sign for ∆E would still remain positive in general. So, to summarize, for the process of compressing air in the pump, \(w < 0\), \(q < 0\), and \(\Delta E > 0\).

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