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Chapter 5: Problem 28

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is cooled by removing \(0.655 \mathrm{~kJ}\) of heat. It shrinks on cooling, and the atmosphere does \(382 \mathrm{~J}\) of work on the balloon. (b) A 100.0 -g bar of gold is heated from \(25^{\circ} \mathrm{C}\) to \(50{ }^{\circ} \mathrm{C}\) during which it absorbs \(322 \mathrm{~J}\) of heat. Assume the volume of the gold bar remains constant. (c) The surroundings do \(1.44 \mathrm{~kJ}\) of work compressing gas in a perfectly insulated cylinder.

Short Answer

Expert verified
(a) The change in internal energy is \(-273 \mathrm{~J}\), and the process is exothermic. (b) The change in internal energy is \(322 \mathrm{~J}\), and the process is endothermic. (c) The change in internal energy is \(1440 \mathrm{~J}\), and the process is endothermic.

Step by step solution

01

(a) Calculate the change in internal energy

First, we need to convert the given values into the same units. The heat removed (Q) is given in kJ and needs to be converted to J: Q = -0.655 kJ * (1000 J/1 kJ) = -655 J (Since the heat is removed, it has a negative sign) The work done on the balloon (W) is given as 382 J. Now, we can use the first law of thermodynamics to find the change in internal energy: \(ΔU=Q+W\) \(ΔU=-655 J + 382 J\) \(ΔU=-273 J\) Since the internal energy decreased, the process is exothermic.
02

(b) Calculate the change in internal energy

In this case, the volume remains constant, so the work done on the system(W) is 0. The heat absorbed by the gold bar (Q) is given as 322 J. Now, use the first law of thermodynamics to find the change in internal energy: \(ΔU=Q+W\) \(ΔU= 322 J + 0\) \(ΔU= 322 J\) Since the internal energy increased, the process is endothermic.
03

(c) Calculate the change in internal energy

In this case, the cylinder is perfectly insulated, so no heat is exchanged with the surroundings (Q=0). The work done on the system by the surroundings (W) is given as 1.44 kJ, so we need to convert this to J: W = 1.44 kJ * (1000 J/1 kJ) = 1440 J Now, use the first law of thermodynamics to find the change in internal energy: \(ΔU=Q+W\) \(ΔU=0 + 1440 J\) \(ΔU=1440 J\) Since the internal energy increased, the process is endothermic.

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Most popular questions from this chapter

(a) What is work? (b) How do we determine the amount of work done, given the force associated with the work?

(a) When a \(0.235-\mathrm{g}\) sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19 ), the temperature rises \(1.642{ }^{\circ} \mathrm{C}\). When a 0.265 -g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to 0.001 \(\mathrm{g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Using values from Appendix \(\mathrm{C},\) calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

In what two ways can an object possess energy? How do these two ways differ from one another?

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