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Sagot :
Let's solve each part of the problem step-by-step, keeping in mind the given answers.
### Part 1: Work Done in Isothermal Compression
1. Given Parameters:
- Mass of Oxygen ([tex]\(O_2\)[/tex]): [tex]\(16.0 \, \text{g}\)[/tex]
- Initial Temperature ([tex]\(T\)[/tex]): [tex]\(100^{\circ} \text{C}\)[/tex] which is [tex]\(373.15 \, \text{K}\)[/tex]
- Universal Gas Constant ([tex]\(R\)[/tex]): [tex]\(8.314 \, \text{J/(mol·K)}\)[/tex]
- Initial and Final Volumes: Final Volume is half of Initial Volume
2. Molar Mass of [tex]\(O_2\)[/tex]:
The molar mass of [tex]\(O_2\)[/tex] is [tex]\(32 \, \text{g/mol}\)[/tex].
3. Number of Moles ([tex]\(n\)[/tex]):
[tex]\[ n = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{16.0 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{moles} \][/tex]
4. Work Done in Isothermal Compression:
The work done in an isothermal process is given by the formula:
[tex]\[ W = nRT \ln \left( \frac{V_{\text{initial}}}{V_{\text{final}}} \right) \][/tex]
Given that the volume is halved:
[tex]\[ \frac{V_{\text{initial}}}{V_{\text{final}}} = 2 \][/tex]
Therefore:
[tex]\[ W = 0.5 \times 8.314 \times 373.15 \times \ln(2) \approx 1074 \, \text{J} \][/tex]
### Part 2: Final Volume in Isobaric Expansion
1. Given Parameters:
- Pressure ([tex]\(P\)[/tex]): [tex]\(1000 \, \text{Pa}\)[/tex]
- Initial Volume ([tex]\(V_{\text{initial}}\)[/tex]): [tex]\(8 \, \text{m}^3\)[/tex]
- Work Done ([tex]\(W\)[/tex]): [tex]\(20 \, \text{KJ} = 20000 \, \text{J}\)[/tex]
2. Work Done in Isobaric Process:
The work done in an isobaric process is given by:
[tex]\[ W = P \times (V_{\text{final}} - V_{\text{initial}}) \][/tex]
3. Calculating Final Volume ([tex]\(V_{\text{final}}\)[/tex]):
[tex]\[ 20000 \, \text{J} = 1000 \, \text{Pa} \times (V_{\text{final}} - 8 \, \text{m}^3) \][/tex]
[tex]\[ V_{\text{final}} - 8 \, \text{m}^3 = \frac{20000 \, \text{J}}{1000 \, \text{Pa}} = 20 \, \text{m}^3 \][/tex]
[tex]\[ V_{\text{final}} = 20 \, \text{m}^3 + 8 \, \text{m}^3 = 28 \, \text{m}^3 \][/tex]
### Part 3: Work Done in Compression
1. Given Parameters:
- Heat Given Off: [tex]\(4000 \, \text{J}\)[/tex] (or [tex]\(4 \, \text{KJ}\)[/tex])
- Pressure ([tex]\(P\)[/tex]): [tex]\(1000 \, \text{Pa}\)[/tex]
- Initial Volume ([tex]\(V_{\text{initial}}\)[/tex]): [tex]\(0.08 \, \text{m}^3\)[/tex]
- Final Volume ([tex]\(V_{\text{final}}\)[/tex]): [tex]\(0.05 \, \text{m}^3\)[/tex]
2. Work Done in Isobaric Compression:
[tex]\[ W = P \times (V_{\text{initial}} - V_{\text{final}}) \][/tex]
3. Calculating Work Done:
[tex]\[ W = 1000 \, \text{Pa} \times (0.08 \, \text{m}^3 - 0.05 \, \text{m}^3) = 1000 \, \text{Pa} \times 0.03 \, \text{m}^3 = 30 \, \text{J} \][/tex]
### Summary of Results:
1. Work done to compress the gas (isothermal): [tex]\(1074 \, \text{J}\)[/tex]
2. Final volume after isobaric expansion: [tex]\(28 \, \text{m}^3\)[/tex]
3. Work done during compression: [tex]\(30 \, \text{J}\)[/tex]
### Part 1: Work Done in Isothermal Compression
1. Given Parameters:
- Mass of Oxygen ([tex]\(O_2\)[/tex]): [tex]\(16.0 \, \text{g}\)[/tex]
- Initial Temperature ([tex]\(T\)[/tex]): [tex]\(100^{\circ} \text{C}\)[/tex] which is [tex]\(373.15 \, \text{K}\)[/tex]
- Universal Gas Constant ([tex]\(R\)[/tex]): [tex]\(8.314 \, \text{J/(mol·K)}\)[/tex]
- Initial and Final Volumes: Final Volume is half of Initial Volume
2. Molar Mass of [tex]\(O_2\)[/tex]:
The molar mass of [tex]\(O_2\)[/tex] is [tex]\(32 \, \text{g/mol}\)[/tex].
3. Number of Moles ([tex]\(n\)[/tex]):
[tex]\[ n = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{16.0 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{moles} \][/tex]
4. Work Done in Isothermal Compression:
The work done in an isothermal process is given by the formula:
[tex]\[ W = nRT \ln \left( \frac{V_{\text{initial}}}{V_{\text{final}}} \right) \][/tex]
Given that the volume is halved:
[tex]\[ \frac{V_{\text{initial}}}{V_{\text{final}}} = 2 \][/tex]
Therefore:
[tex]\[ W = 0.5 \times 8.314 \times 373.15 \times \ln(2) \approx 1074 \, \text{J} \][/tex]
### Part 2: Final Volume in Isobaric Expansion
1. Given Parameters:
- Pressure ([tex]\(P\)[/tex]): [tex]\(1000 \, \text{Pa}\)[/tex]
- Initial Volume ([tex]\(V_{\text{initial}}\)[/tex]): [tex]\(8 \, \text{m}^3\)[/tex]
- Work Done ([tex]\(W\)[/tex]): [tex]\(20 \, \text{KJ} = 20000 \, \text{J}\)[/tex]
2. Work Done in Isobaric Process:
The work done in an isobaric process is given by:
[tex]\[ W = P \times (V_{\text{final}} - V_{\text{initial}}) \][/tex]
3. Calculating Final Volume ([tex]\(V_{\text{final}}\)[/tex]):
[tex]\[ 20000 \, \text{J} = 1000 \, \text{Pa} \times (V_{\text{final}} - 8 \, \text{m}^3) \][/tex]
[tex]\[ V_{\text{final}} - 8 \, \text{m}^3 = \frac{20000 \, \text{J}}{1000 \, \text{Pa}} = 20 \, \text{m}^3 \][/tex]
[tex]\[ V_{\text{final}} = 20 \, \text{m}^3 + 8 \, \text{m}^3 = 28 \, \text{m}^3 \][/tex]
### Part 3: Work Done in Compression
1. Given Parameters:
- Heat Given Off: [tex]\(4000 \, \text{J}\)[/tex] (or [tex]\(4 \, \text{KJ}\)[/tex])
- Pressure ([tex]\(P\)[/tex]): [tex]\(1000 \, \text{Pa}\)[/tex]
- Initial Volume ([tex]\(V_{\text{initial}}\)[/tex]): [tex]\(0.08 \, \text{m}^3\)[/tex]
- Final Volume ([tex]\(V_{\text{final}}\)[/tex]): [tex]\(0.05 \, \text{m}^3\)[/tex]
2. Work Done in Isobaric Compression:
[tex]\[ W = P \times (V_{\text{initial}} - V_{\text{final}}) \][/tex]
3. Calculating Work Done:
[tex]\[ W = 1000 \, \text{Pa} \times (0.08 \, \text{m}^3 - 0.05 \, \text{m}^3) = 1000 \, \text{Pa} \times 0.03 \, \text{m}^3 = 30 \, \text{J} \][/tex]
### Summary of Results:
1. Work done to compress the gas (isothermal): [tex]\(1074 \, \text{J}\)[/tex]
2. Final volume after isobaric expansion: [tex]\(28 \, \text{m}^3\)[/tex]
3. Work done during compression: [tex]\(30 \, \text{J}\)[/tex]
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