1. Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (5.5 atm, 6.5 L) to a state with pressure 6.5 atm. For a monoatomic gas y=5/3. (a) Find the volume of the gas after compression. ✓final = 5.9 (b) Find the work done by the gas in the process. W= -3.9 Latm (c) Find the change in internal energy of the gas in the process. AE int=39 VL.atm. Check: What do you predict the signs of work and change in internal energy to be? Do the signs of work and change in internal energy match with your predictions?
1. Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (5.5 atm, 6.5 L) to a state with pressure 6.5 atm. For a monoatomic gas y=5/3. (a) Find the volume of the gas after compression. ✓final = 5.9 (b) Find the work done by the gas in the process. W= -3.9 Latm (c) Find the change in internal energy of the gas in the process. AE int=39 VL.atm. Check: What do you predict the signs of work and change in internal energy to be? Do the signs of work and change in internal energy match with your predictions?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Example Problem on Adiabatic Processes for an Educational Website**
**Problem Statement:**
Two moles of a monatomic ideal gas such as helium are compressed adiabatically and reversibly from a state (5.5 atm, 6.5 L) to a state with pressure 6.5 atm. For a monatomic gas, γ (gamma) = 5/3.
**Questions:**
1. **Find the volume of the gas after compression.**
**Answer:**
\[
V_{\text{final}} = 5.9 \, \text{L}
\]
2. **Find the work done by the gas in the process.**
**Answer:**
\[
W = -3.9 \, \text{L·atm}
\]
3. **Find the change in internal energy of the gas in the process.**
**Answer:**
\[
\Delta E_{\text{int}} = 3.9 \, \text{L·atm}
\]
**Check:**
What do you predict the signs of work and change in internal energy to be? Do the signs of work and change in internal energy match with your predictions?
**Explanation of Answers:**
- For (a), the final volume of the gas after adiabatic compression is calculated to be 5.9 L.
- For (b), the work done by the gas is -3.9 L·atm. This negative sign indicates that the work is done on the gas during compression.
- For (c), the change in internal energy is 3.9 L·atm. This positive value confirms that no heat is exchanged in adiabatic processes, meaning all work done results in a change in internal energy.
**Note:**
To determine these values, use the adiabatic condition for an ideal gas, which is \( PV^\gamma = \text{constant} \) and the first law of thermodynamics \( \Delta E_{\text{int}} = Q + W \), where \( Q = 0 \) for an adiabatic process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8f42db7f-5d63-42d4-89b9-bc5373ca5a94%2Fe43d20a3-d910-426c-a0df-90056c55e6f9%2Fswesn7l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example Problem on Adiabatic Processes for an Educational Website**
**Problem Statement:**
Two moles of a monatomic ideal gas such as helium are compressed adiabatically and reversibly from a state (5.5 atm, 6.5 L) to a state with pressure 6.5 atm. For a monatomic gas, γ (gamma) = 5/3.
**Questions:**
1. **Find the volume of the gas after compression.**
**Answer:**
\[
V_{\text{final}} = 5.9 \, \text{L}
\]
2. **Find the work done by the gas in the process.**
**Answer:**
\[
W = -3.9 \, \text{L·atm}
\]
3. **Find the change in internal energy of the gas in the process.**
**Answer:**
\[
\Delta E_{\text{int}} = 3.9 \, \text{L·atm}
\]
**Check:**
What do you predict the signs of work and change in internal energy to be? Do the signs of work and change in internal energy match with your predictions?
**Explanation of Answers:**
- For (a), the final volume of the gas after adiabatic compression is calculated to be 5.9 L.
- For (b), the work done by the gas is -3.9 L·atm. This negative sign indicates that the work is done on the gas during compression.
- For (c), the change in internal energy is 3.9 L·atm. This positive value confirms that no heat is exchanged in adiabatic processes, meaning all work done results in a change in internal energy.
**Note:**
To determine these values, use the adiabatic condition for an ideal gas, which is \( PV^\gamma = \text{constant} \) and the first law of thermodynamics \( \Delta E_{\text{int}} = Q + W \), where \( Q = 0 \) for an adiabatic process.
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