1. Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (2 atm, 5.5 L) to a state with pressure 6 atm. For a monoatomic gas y = 5/3. (a) Find the volume of the gas after compression. V final= L (b) Find the work done by the gas in the process. W= L.atm (c) Find the change in internal energy of the gas in the process. L.atm AE int = 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 (2 atm, 5.5 L) to a state with pressure 6 atm. For a monoatomic gas y = 5/3. (a) Find the volume of the gas after compression. V final= L (b) Find the work done by the gas in the process. W= L.atm (c) Find the change in internal energy of the gas in the process. L.atm AE int = 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?
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![### Thermodynamics - Adiabatic Compression
1. **Problem Statement:**
Two moles of a monatomic ideal gas, such as helium, are compressed adiabatically and reversibly from a state (2 atm, 5.5 L) to a state with pressure 6 atm. For a monatomic gas, the adiabatic index (\( \gamma \)) is 5/3.
a. **Find the volume of the gas after compression.**
\[
V_{\text{final}} = \Box \text{ L}
\]
b. **Find the work done by the gas in the process.**
\[
W = \Box \text{ L·atm}
\]
c. **Find the change in internal energy of the gas in the process.**
\[
\Delta E_{\text{int}} = \Box \text{ L·atm}
\]
**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:
When a gas undergoes adiabatic compression, no heat is exchanged with the surroundings. The work done by the gas is related to its volume and pressure changes, and the internal energy change is associated with temperature changes in the process. For a monatomic ideal gas, the internal energy can be calculated using the ideal gas law and specific heat capacities. The process follows the adiabatic relation \( P V^\gamma = \text{constant} \).
- **Volume Calculation:**
Using the initial state and the adiabatic condition:
\[
P_1 V_1^\gamma = P_2 V_2^\gamma
\]
\[
2 \times (5.5)^{5/3} = 6 \times V_{\text{final}}^{5/3}
\]
- **Work Done:**
The work done in an adiabatic process can be computed using:
\[
W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}
\]
- **Change in Internal Energy:**
The change in internal energy for a monatomic gas is:
\[
\Delta E_{\text{int}} = \frac{3}{2} n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bc1e865-88f5-4594-837c-35f6919f2fc7%2F485f837d-74aa-422c-b8cf-11caab40af34%2F8w60tzp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Thermodynamics - Adiabatic Compression
1. **Problem Statement:**
Two moles of a monatomic ideal gas, such as helium, are compressed adiabatically and reversibly from a state (2 atm, 5.5 L) to a state with pressure 6 atm. For a monatomic gas, the adiabatic index (\( \gamma \)) is 5/3.
a. **Find the volume of the gas after compression.**
\[
V_{\text{final}} = \Box \text{ L}
\]
b. **Find the work done by the gas in the process.**
\[
W = \Box \text{ L·atm}
\]
c. **Find the change in internal energy of the gas in the process.**
\[
\Delta E_{\text{int}} = \Box \text{ L·atm}
\]
**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:
When a gas undergoes adiabatic compression, no heat is exchanged with the surroundings. The work done by the gas is related to its volume and pressure changes, and the internal energy change is associated with temperature changes in the process. For a monatomic ideal gas, the internal energy can be calculated using the ideal gas law and specific heat capacities. The process follows the adiabatic relation \( P V^\gamma = \text{constant} \).
- **Volume Calculation:**
Using the initial state and the adiabatic condition:
\[
P_1 V_1^\gamma = P_2 V_2^\gamma
\]
\[
2 \times (5.5)^{5/3} = 6 \times V_{\text{final}}^{5/3}
\]
- **Work Done:**
The work done in an adiabatic process can be computed using:
\[
W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}
\]
- **Change in Internal Energy:**
The change in internal energy for a monatomic gas is:
\[
\Delta E_{\text{int}} = \frac{3}{2} n
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