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Chapter 6: Problem 101

Consider the following cyclic process carried out in two steps on a gas: Step 1: \(45 \mathrm{~J}\) of heat is added to the gas, and \(10 . \mathrm{J}\) of expansion work is performed. Step \(2: 60 . \mathrm{J}\) of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in Step \(2 .\)

Short Answer

Expert verified
The work done for gas compression in Step 2 is -25 J, indicating that work is done on the gas during the compression process.

Step by step solution

01

Consider the first law of thermodynamics

The first law of thermodynamics states that the change in internal energy (∆U) of a system is equal to the heat (Q) added to the system minus the work done by the system (W). Mathematically, this can be written as: \[∆U = Q - W\]
02

Apply the first law of thermodynamics to Step 1

In Step 1, we are given that 45 J of heat is added to the gas, and 10 J of expansion work is performed. Since the gas expands and does work, we consider it to be a positive value. Applying the first law of thermodynamics: \[∆U_1 = Q_1 - W_1 = 45 \mathrm{~J} - 10\mathrm{~J} = 35 \mathrm{~J}\]
03

Apply the first law of thermodynamics to Step 2

In step 2, 60 J of heat is removed from the gas, and work is done on the gas as it is compressed. We represent the heat removed as a negative value since it is taken out of the system: \(Q_2 = -60\mathrm{~J}\). The work done for gas compression is represented by \(W_2\). Applying the first law of thermodynamics to Step 2: \[∆U_2 = Q_2 - W_2 = -60 \mathrm{~J} - W_2\]
04

Find total change in internal energy in a cyclic process

In a cyclic process, the total change in internal energy is zero since the system returns to its initial state. Therefore, \[∆U_\mathrm{total} = ∆U_1 + ∆U_2 = 0\]
05

Calculate the work done in step 2

Using the equations for the change in internal energy in step 1 and step 2 as well as the fact that the total change in internal energy is zero, we can find the work done in step 2. \begin{align*} ∆U_1 + ∆U_2 &= 0 \\ (35\mathrm{~J}) + (-60\mathrm{~J} - W_2) &= 0 \\ 35\mathrm{~J} - 60\mathrm{~J} &= W_2 \\ -25\mathrm{~J} &= W_2 \end{align*} The work done for gas compression in Step 2 is -25 J. Since the work is negative, it means that work is done on the gas during the compression process.

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Most popular questions from this chapter

Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A \(0.1964\) -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter with a heat capacity of \(1.56 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C} .\) Calculate the energy of combustion of quinone per gram and per mole.

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

The specific heat capacity of silver is \(0.24 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{~g}\) Ag from \(273 \mathrm{~K}\) to \(298 \mathrm{~K}\). b. Calculate the energy required to raise the temperature of \(1.0\) mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{~kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample of silver.

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5 \mathrm{~L}\). Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .{ }^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H_{\text {vap }}\) for the liquid is \(30.7 \mathrm{~kJ} / \mathrm{mol}\) at \(80 .{ }^{\circ} \mathrm{C}\).
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