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Chapter 5: Problem 27

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} ;(\mathbf{b})\) a system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system; (c) the system absorbs \(7.25 \mathrm{~kJ}\) of heat from the surroundings while its volume remains constant (assume that only \(P-V\) work can be done).

Short Answer

Expert verified
For the given cases, we have calculated the following values for \(\Delta E\): - Case (a): \(\Delta E = -77 \mathrm{~J}\), indicating an exothermic process. - Case (b): \(\Delta E = -22100 \mathrm{~J}\), indicating an exothermic process. - Case (c): \(\Delta E = 7250 \mathrm{~J}\), indicating an endothermic process.

Step by step solution

01

Case (a)

Calculate \(\Delta E\) and determine if the process is endothermic or exothermic. We have q = 0.763 kJ and w = -840 J. First, we need to convert q into joules: 0.763 kJ * 1000 J/kJ = 763 J Now we can use the formula: \[\Delta E = q + w = 763 \mathrm{~J} + (-840 \mathrm{~J}) = -77 \mathrm{~J}\] Since \(\Delta E\) is negative, the process is exothermic, which means it releases heat.
02

Case (b)

First, let's establish the signs of q and w. Since the system releases heat, q is negative. Since the surroundings do work on the system, w is positive. Now, convert the given kJ to J: q = -66.1 kJ * 1000 J/kJ = -66100 J w = 44.0 kJ * 1000 J/kJ = 44000 J Next, we'll calculate \(\Delta E\): \[\Delta E = q + w = (-66100 \mathrm{~J}) + (44000 \mathrm{~J}) = -22100 \mathrm{~J}\] With a negative \(\Delta E\), the process is exothermic, meaning it releases heat.
03

Case (c)

Since the volume remains constant and only \(P-V\) work can be done, there is no work being done in this case (w = 0). The system absorbs heat from the surrounding so q is positive: q = 7.25 kJ * 1000 J/kJ = 7250 J Now compute \(\Delta E\): \[\Delta E = q + w = 7250 \mathrm{~J} + 0 = 7250 \mathrm{~J}\] Since \(\Delta E\) is positive, the process is endothermic, which means it absorbs heat.

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