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Suppose 1800 J of heat are added to 3.6 mol of argon gas at a constant pressure of \(120 \mathrm{kPa}\). Find the change in \((\mathrm{a})\) internal energy and (b) temperature for this gas. (c) Calculate the change in volume of the gas. (Assume that the argon can be treated as an ideal monatomic gas.)

Short Answer

Expert verified
(a) \(\Delta U = 1080 \text{ J}\); (b) \(\Delta T \approx 90 \text{ K}\); (c) \(\Delta V \approx 0.225 \text{ m}^3\).

Step by step solution

01

Understand the Problem

We need to find the change in internal energy, temperature, and volume of argon gas given heat added, constant pressure, and amount of moles. The problem assumes an ideal monatomic gas.
02

Use Formula for Internal Energy Change

For a monatomic ideal gas, the change in internal energy (\( \Delta U \)) is given by \( \Delta U = \frac{3}{2} n R \Delta T \). Use the first law of thermodynamics, \(Q = \Delta U + W\), where \(Q\) is the heat added, and \(W\) is the work done by the system at constant pressure, given by \(W = P \Delta V\). Since \(Q = 1800 \text{ J}\), rearrange to find \( \Delta U = Q - W \).
03

Calculate Change in Temperature

From the first law, expression for work done \(W = P \Delta V\) and \(Q = \Delta U + P \Delta V\), solve for \( \Delta T \). Knowing \( \Delta U = \frac{3}{2} nR \Delta T \), we calculate \( \Delta T = \frac{Q}{\frac{5}{2}nR} \), using \(R = 8.314 \text{ J/mol K} \), \(n = 3.6 \text{ mol} \), and provided heat \(Q = 1800 \text{ J}\).
04

Determine Change in Volume

Using the ideal gas law \( PV = nRT \), calculate the change in volume. Rearrange it experience the change: \( \Delta V = \frac{nR \Delta T}{P} \). Substitute \(R = 8.314 \text{ J/mol K}\) and the calculated \(\Delta T\), \(P = 120 \text{ kPa}\) to find \(\Delta V\).
05

Perform Calculations

Calculate \(\Delta U\), \(\Delta T\), and \(\Delta V\) using values from previous steps:1. From \(Q = 1800 \text{ J}\), first calculate \(\Delta U = 3 \times 8.314 \text{ J/mol K} \times \Delta T \).2. Plug into \(\Delta T = \frac{1800}{\frac{5}{2} \times 3.6 \times 8.314}\) to get \(\Delta T \).3. Finally, calculate \(\Delta V = \frac{3.6 \times 8.314 \times \Delta T}{120000}\) using \(\Delta T\) from the previous calculation.

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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 measure of the energy stored within a system due to the motion and interactions of particles. In the context of an ideal monatomic gas like argon, the internal energy change when heat is added can be calculated using a specific formula: \( \Delta U = \frac{3}{2} n R \Delta T \). This formula represents the sum of the kinetic energies of all the molecules. Here, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(\Delta T\) is the change in temperature.
For such a monatomic gas, only translational kinetic energy (movement of particles through space) contributes to the internal energy. The atoms do not partake in rotating or vibrating in a way that affects this term because they are considered point masses. This is why you only see the \(\frac{3}{2}\) factor, which accounts for the three degrees of freedom each atom possesses.
To find the change in internal energy, you need to calculate the work done by the gas and use the relation from the first law of thermodynamics: \(Q = \Delta U + W\).
This tells us how much of the added heat goes into increasing the internal energy versus doing work on the surroundings.
Temperature Change
The change in temperature \(\Delta T\) of a gas when heat is applied is a crucial factor that directly affects internal energy and, by extension, volume. For our argon gas, we can determine \(\Delta T\) using the rearranged first law of thermodynamics in combination with the formula for internal energy mentioned earlier.
Since we are assuming a constant pressure setting in this problem, the formula \( \Delta T = \frac{Q}{\frac{5}{2}nR} \) is used. This accounts for both the kinetic energy change and the work done as the gas expands, given the heat provided.
Plugging in:
  • \(Q = 1800 \text{ J}\)
  • \(n = 3.6 \text{ mol}\)
  • \(R = 8.314 \text{ J/mol K}\)
you can solve for the resulting change in temperature.
Volume Change
Volume change in gases is closely linked to changes in temperature under constant pressure, as dictated by Charles's Law, which is a special case of the ideal gas law. The volume of a gas will expand or compress to maintain consistent pressure when temperature shifts.
To find this change in volume \(\Delta V\), the ideal gas law is rearranged to handle dynamics: \( \Delta V = \frac{nR \Delta T}{P} \). This relationship tells us how much the gas will expand when it's heated under a constant pressure of 120 kPa, considering our given values for \(R\), \(n\), and \(\Delta T\).
With the expansion, heat causes the particles in the gas to move faster, increasing pressure unless the volume increases proportionally. The volume changes to accommodate the new energy levels per the ideal gas computations.
Monatomic Gas
A monatomic gas consists of single atoms that are not bonded to each other, like argon. This simplicity makes their thermodynamic model intuitive because it involves only straightforward motion without molecular rotations or vibrations.
For calculations involving heat and work, monatomic gases allow us to assume ideal behavior more easily. The system's energy changes can be conveniently analyzed using three degrees of freedom, as they modeled through straightforward kinetic terms.
Studies involving monatomic gases, particularly under ideal conditions, extrapolate to more complex systems but start with basic ideas like those related to internal energy and volume response to heating.
  • Monatomic gases are robust models for applying ideal gas law concepts.
  • Energetics remain basic due to their non-complicated structure.
Understanding monatomic gases paves the way for diving deeper into broader gas dynamics and thermodynamics by establishing clear foundational concepts.

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Most popular questions from this chapter

A cylinder with a movable piston holds 2.75 mol of argon at a constant temperature of \(295 \mathrm{K}\). As the gas is compressed isothermally, its pressure increases from 101 kPa to 121 kPa. Find \((a)\) the final volume of the gas, \((b)\) the work done by the gas, and \((c)\) the heat added to the gas.

(a) If 535 J of heat are added to 45 moles of a monatomic gas at constant volume, how much does the temperature of the gas increase? (b) Repeat part (a), this time for a constant-pressure process.

When \(1210 \mathrm{J}\) of heat are added to one mole of an ideal monatomic gas, its temperature increases from \(272 \mathrm{K}\) to \(276 \mathrm{K}\). Find the work done by the gas during this process.

Energy from Gasoline Burning a gallon of gasoline releases \(1.19 \times 10^{8} \mathrm{J}\) of internal energy. If a certain car requires \(5.20 \times 10^{5} \mathrm{J}\) of work to drive one mile, (a) how much heat is given off to the atmosphere each mile, assuming the car gets 25.0 miles to the gallon? (b) If the miles per gallon of the car is increased, does the amount of heat released to the atmosphere increase, decrease, or stay the same? Explain.

Entropy and the Sun The surface of the Sun has a tempera ture of \(5500^{\circ} \mathrm{C}\) and the temperature of deep space is \(3.0 \mathrm{K}\). (a) Find the entropy increase produced by the Sun in one day, given that it radiates heat at the rate of \(3.80 \times 10^{26} \mathrm{W}\). (b) How much work could have been done if this heat had been used to run an ideal heat engine?

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