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Chapter 6: Problem 18

The work done to compress a gas is \(74 \mathrm{~J}\). As a result, \(26 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in energy of the gas.

Short Answer

Expert verified
The change in energy of the gas is \(48 J\).

Step by step solution

01

Understand the formula

To solve this problem, use the first law of thermodynamics, which states that the change in internal energy of a system can be calculated by subtracting the heat lost by the system from the work done on the system. The formula is as follows: \(\Delta U = Work done + Heat lost\)
02

Insert the given values

From the exercise, we know that the work done to compress the gas is 74 J and the heat given off to the surroundings by the gas is 26 J. It should be noted that this value is negative since it is losing heat. Therefore, insert the given values into the formula: \(\Delta U = 74J - 26J\)
03

Calculate the change in energy of the gas

By subtraction of 26 J from 74 J, one gets a change in internal energy of \(\Delta U = 48J\). This is therefore the change in the total collective kinetic and potential energy of the molecules of the gas.

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Most popular questions from this chapter

The \(\Delta H_{\mathrm{f}}^{\circ}\) values of the two allotropes of oxygen, \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\), are 0 and \(142.2 \mathrm{~kJ} / \mathrm{mol}\), respectively, at \(25^{\circ} \mathrm{C}\). Which is the more stable form at this temperature?

A 44.0-g sample of an unknown metal at \(99.0^{\circ} \mathrm{C}\) was placed in a constant-pressure calorimeter containing \(80.0 \mathrm{~g}\) of water at \(24.0^{\circ} \mathrm{C}\). The final temperature of the system was found to be \(28.4^{\circ} \mathrm{C}\). Calculate the specific heat of the metal. (The heat capacity of the calorimeter is \(\left.12.4 \mathrm{~J} /{ }^{\circ} \mathrm{C} .\right)\)

Define calorimetry and describe two commonly used calorimeters. In a calorimetric measurement, why is it important that we know the heat capacity of the calorimeter? How is this value determined?

Consider these changes: (a) \(\mathrm{Hg}(l) \longrightarrow \mathrm{Hg}(g)\) (b) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Describe the interconversions of forms of energy occurring in these processes: (a) You throw a softball up into the air and catch it. (b) You switch on a flashlight. (c) You ride the ski lift to the top of the hill and then ski down. (d) You strike a match and let it burn down.
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