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Chapter 19: Problem 5

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0\(^\circ\)C, 392 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a \(pV\)-diagram for the process.

Short Answer

Expert verified
(a) The initial pressure was approximately 1.87 atm. (b) Sketch the curve as hyperbolic from high volume, low pressure to low volume, high pressure on the pV-diagram.

Step by step solution

01

Understanding Isothermal Compression

In an isothermal process, the temperature remains constant. For an ideal gas undergoing isothermal compression, the internal energy doesn't change, meaning the work done on the gas is equal to the heat transferred out. We'll use the ideal gas equation to find the initial pressure.
02

Using the Ideal Gas Equation

For an ideal gas, the equation is given by \( PV = nRT \). Since the temperature is constant, \( P_1V_1 = P_2V_2 \) applies. The work done on the gas \( W = nRT \ln\left(\frac{V_2}{V_1}\right) \) helps find the ratio \( \frac{V_2}{V_1} \). Use \( W = 392 \text{ J} \) and the gas constant \( R = 8.314 \text{ J/mol K} \) to calculate \( T \) in Kelvin, \( n = 0.305 \text{ mol} \), and \( T = 22.0 + 273.15 \text{ K} = 295.15 \text{ K} \).
03

Calculating Temperature in Kelvin

Convert the given temperature in Celsius to Kelvin: \( T = 22.0 + 273.15 = 295.15 \text{ K} \). This allows the use of consistent units throughout the calculations.
04

Finding the Pressure-Volume Ratio

Calculate the volume ratio from work done: \[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \implies 392 = 0.305 \times 8.314 \times 295.15 \times \ln\left(\frac{V_2}{V_1}\right) \]. Solve for \( \ln\left(\frac{V_2}{V_1}\right) \), which provides \( \ln\left(\frac{V_2}{V_1}\right) \approx 0.058 \).
05

Solving for the Volume Ratio

From the previous step, \( \ln\left(\frac{V_2}{V_1}\right) = 0.058 \) implies \( \frac{V_2}{V_1} = e^{0.058} \approx 1.06 \). Since \( P_1V_1 = P_2V_2 \), then \( P_1 = P_2 \frac{V_2}{V_1} \).
06

Calculating Initial Pressure

Using the ratio \( \frac{V_2}{V_1} = 1.06 \) and \( P_2 = 1.76 \text{ atm} \), find \( P_1 = 1.76 \times 1.06 \approx 1.87 \text{ atm} \).
07

Drawing the pV-Diagram

In a \( pV \)-diagram, plot the pressure on the y-axis and volume on the x-axis. The curve will show a hyperbolic shape (since the product \( PV \) is constant), starting from a higher volume and lower pressure (\( P_1 \)) to a lower volume and higher pressure (\( P_2 \)).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of ideal gases. It is represented by the formula: \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant, approximately 8.314 J/mol路K
  • \( T \) is the temperature in Kelvin
This law assumes that the gas particles do not interact and that the volume of the gas particles is negligible compared to the volume of the container. It is especially useful for predicting the behavior of gases under various conditions of pressure, volume, and temperature. When applying the Ideal Gas Law, it is crucial to ensure that all units are consistent, particularly the temperature which must be in Kelvin. This law applies very well to many common gases at low pressure and high temperature.
Isothermal Process
An isothermal process is a thermodynamic process that occurs at a constant temperature. This means that the heat transferred into or out of the system adjusts to keep the temperature constant. For ideal gases, there are some important implications:
  • The internal energy of the gas remains unchanged because internal energy depends on temperature.
  • The work done by or on the gas equals the heat exchanged during the process; in symbols, \( Q = W \).
  • The equation \( P_1V_1 = P_2V_2 \) holds true, demonstrating that the product of pressure and volume remains constant during the process.
During an isothermal compression, like the one presented in the exercise, the gas is compressed by the surroundings doing work on it, while any heat produced is simultaneously transferred out of the gas to maintain constant temperature. Such processes are graphically represented by a hyperbolic curve on a Pressure-Volume diagram, indicating that as the volume decreases, the pressure increases.
Work Done on Gas
Work done on a gas during an isothermal process is an important concept in thermodynamics. During such a process, the total work done can be calculated by:\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]where:
  • \( W \) is the work done
  • \( n \) is the number of moles of gas
  • \( R \) is the gas constant
  • \( T \) is the temperature held constant in Kelvin
  • \( V_1 \) and \( V_2 \) are the initial and final volumes respectively
It shows that work done on the gas depends on the ratio of final volume to initial volume. When calculating the work done, the sign convention can be important: work done on the gas is considered positive, while work done by the gas is considered negative. Understanding these principles helps in analyzing various thermodynamic processes and predicting how a system will respond under isothermal conditions.
Pressure-Volume Diagram
A Pressure-Volume (PV) diagram is a graphical representation of the relationship between the pressure and volume of a gas. In these diagrams:
  • The y-axis typically represents pressure.
  • The x-axis represents volume.
  • Each point on the curve represents a different state of the gas during the process.
For an isothermal process involving an ideal gas, the PV diagram displays a hyperbolic curve. This curve reflects the fact that during isothermal compression (or expansion), the product \( PV \) stays constant: as the volume decreases, pressure increases proportionately.To sketch a PV diagram for an isothermal process like the one in the exercise:
  • Start at the initial pressure \( P_1 \) and volume \( V_1 \).
  • The curve will slope downwards towards the final pressure \( P_2 \) and volume \( V_2 \).
  • The curve follows a hyperbolic shape due to the inverse relationship between pressure and volume when temperature remains constant.
By analyzing these diagrams, we gain insight into how gases behave under different thermodynamic conditions, which is crucial for designing and evaluating engineering systems.

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Most popular questions from this chapter

Heat \(Q\) flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

You have a cylinder that contains 500 L of the gas mixture pressurized to 2000 psi (gauge pressure). A regulator sets the gas flow to deliver 8.2 \(L\)/min at atmospheric pressure. Assume that this flow is slow enough that the expansion is isothermal and the gases remain mixed. How much time will it take to empty the cylinder? (a) 1 h; (b) 33 h; (c) 57 h; (d) 140 h.

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Starting with 2.50 mol of N\(_2\) gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0\(^\circ\)C, a chemist first heats the gas at constant volume, adding 1.36 \(\times\) 10\(^4\) J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

Five moles of monatomic ideal gas have initial pressure 2.50 \(\times\) 10\(^3\) Pa and initial volume 2.10 m\(^3\). While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?
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