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(III) A 1.00 -mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of \(420 \mathrm{~K}\) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are \(720 \mathrm{~K}\) and 1.60 atm. Determine \((a)\) the change in internal energy, (b) the work done by the gas, and (c) the heat added to the gas. (Assume five active degrees of freedom.)

Short Answer

Expert verified
(a) \( \Delta U \) can be found using \( \Delta U = \frac{5}{2}nR(T_2 - T_1) \). (b) Work \( W \) is calculated as the area under the curve from initial to final states. (c) Heat added \( Q \) uses \( Q = \Delta U + W \).

Step by step solution

01

Understand the Problem

We have a diatomic ideal gas that changes its pressure linearly with temperature, starting from \( P_1 = 1.00 \text{ atm} \) and \( T_1 = 420 \text{ K} \). Its final state is at \( P_2 = 1.60 \text{ atm} \) and \( T_2 = 720 \text{ K} \). We need to find the change in internal energy, work done by the gas, and heat added.
02

Calculate Initial and Final Volumes

Using the ideal gas law \( PV = nRT \), calculate the initial and final volumes. Start with initial volume: \( V_1 = \frac{nRT_1}{P_1} \) and final volume: \( V_2 = \frac{nRT_2}{P_2} \). Given \( n = 1.00 \text{ mol} \), \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). Substitute the values to find \( V_1 \) and \( V_2 \).
03

Find Change in Internal Energy

For a diatomic gas with five degrees of freedom, the internal energy change is \( \Delta U = \frac{5}{2} nR \Delta T \). Compute \( \Delta T = T_2 - T_1 = 720 \text{ K} - 420 \text{ K} \) and substitute to find \( \Delta U \).
04

Calculate Work Done by the Gas

Work done, \( W \), during a linear process can be calculated using the area under the \( P \) vs. \( V \) curve. Since \( P = mT + b \) (a line), find the slope \( m \) from the data. Then find \( W = \int P \, dV = \int (mT + b) \, dV \), and solve this integral using known limits and transformations through temperature changes.
05

Determine Heat Added to the Gas

Using the first law of thermodynamics, \( \Delta U = Q - W \), solve for heat added \( Q \). Use results from steps 3 and 4 to find \( Q \).
06

Verify Calculations

Go through each step, confirm calculations, and ensure that results satisfy thermodynamic laws, such that \( \Delta U = Q - W \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy
Internal energy is a core concept of thermodynamics that represents the total energy contained within a system. In an ideal gas, this energy is related to the random motion of the gas molecules. In our problem, the gas is diatomic, meaning it consists of two atoms per molecule, like oxygen or nitrogen. This structure provides it with five active degrees of freedom at the given temperatures. These include three translational movements and two rotational movements.

To determine the change in internal energy (\( \Delta U \)), we can use the equation:
  • \( \Delta U = \frac{5}{2} nR \Delta T \)
Here, \( n \) denotes the number of moles, \( R \) is the ideal gas constant, and \( \Delta T \) is the change in temperature from the initial to the final state. This formula stems directly from the kinetic theory of gases, considering the heat capacity at constant volume for diatomic gases. As the temperature chance dictates the change in internal energy, no work (pressure-volume work) directly affects \( \Delta U \). Thus, the internal energy change is solely linked to the temperature difference of the system.
Work Done
When considering work done by or on a gas, it's important to understand that this is related to the expansion or compression of the gas. The work (\( W \)) done by the gas is formulated as:
  • \( W = \int P \, dV \)
This expression calculates the area under the pressure-volume (\( P-V \)) graph for the specific process. Plotting pressure versus volume typically results in a curve or line, depending on how pressure relates to temperature and volume.

In the given exercise, pressure increases linearly with temperature. This relationship can usually be represented as a linear equation, \( P = mT + b \), where \( m \) is the slope obtained from the change in pressure and temperature, and \( b \) is the intercept. By integrating this linear equation over the changes in volume, we can calculate the total work done by the gas during its transition from the initial to the final state. This mathematical evaluation takes into consideration the path taken by the process within the \( P-V \) plane.
Heat Transfer
Heat transfer in a thermodynamic process describes how energy is exchanged between a system and its surroundings due to temperature differences. In the context of ideal gases, heat transfer (\( Q \)) is linked with both the change in internal energy and the work done by the system.

The first law of thermodynamics establishes the relation:
  • \( \Delta U = Q - W \)
This equation states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system. In practical terms, if we know \( \Delta U \) from the change in the state of the system and \( W \) from work calculations, we can easily solve for the heat added using this fundamental relationship. Understanding this concept is crucial in identifying how systems exchange energy and how processes such as heating or cooling impact a gas at molecular levels.
Degrees of Freedom
Degrees of freedom in the context of ideal gases refer to the number of independent ways in which the gas molecules can move. For a diatomic gas, this involves translational and rotational motion. These degrees are crucial in determining how energy is partitioned within a gas.

Let's break it down:
  • Translational movement: 3 degrees (movement along x, y, z axes)
  • Rotational movement for diatomic molecules: 2 degrees (rotation around two perpendicular axes)
Adding these up, the gas has 5 degrees of freedom. This is important because the internal energy of a gas depends on these degrees; specifically, each degree of freedom contributes equal amounts of energy, amounting to \( \frac{1}{2}k_B T \) per molecule per degree of freedom (where \( k_B \) is Boltzmann's constant). The enumeration of degrees of freedom is pivotal in correctly applying formulas related to energy calculations, such as determining the internal energy change in a gas.

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Most popular questions from this chapter

(III) A house thermostat is normally set to \(22^{\circ} \mathrm{C}\), but at night it is turned down to \(12^{\circ} \mathrm{C}\) for \(9.0 \mathrm{~h} .\) Estimate how much more heat would be needed (state as a percentage of daily usage) if the thermostat were not turned down at night. Assume that the outside temperature averages \(0^{\circ} \mathrm{C}\) for the \(9.0 \mathrm{~h}\) at night and \(8^{\circ} \mathrm{C}\) for the remainder of the day, and that the heat loss from the house is proportional to the difference in temperature inside and out. To obtain an estimate from the data, you will have to make other simplifying assumptions; state what these are.

[The Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter also has a group of General Problems that are not arranged by Section and not ranked. \(\begin{array}{l}{\text { (I) To what temperature will } 8700 \mathrm{J} \text { of heat raise } 3.0 \mathrm{kg} \text { of }} \\ {\text { water that is initially at } 10.0^{\circ} \mathrm{C} \text { ? }}\end{array}\)

(II) Samples of copper, aluminum, and water experience the same temperature rise when they absorb the same amount of heat. What is the ratio of their masses?

A marathon runner has an average metabolism rate of about 950 \(\mathrm{kcal} / \mathrm{h}\) during a race. If the runner has a mass of 55 \(\mathrm{kg}\) , estimate how much water she would lose to evaporation from the skin for a race that lasts 2.2 \(\mathrm{h}\) .

(II) When a \(290-\) g piece of iron at \(180^{\circ} \mathrm{C}\) is placed in a 95 -g aluminum calorimeter cup containing 250 \(\mathrm{g}\) of glycerin at \(10^{\circ} \mathrm{C},\) the final temperature is observed to be \(38^{\circ} \mathrm{C}\) Estimate the specific heat of glycerin.

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