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 (\( P_2 \)).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
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·K
- \( T \) is the temperature in Kelvin
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.
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
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.
- 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.