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Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will be
- a)- 1870.6 J
- b)- 1870.6J
- c)-18.47 J
- d)+18.47 J
Correct answer is option 'A'. Can you explain this answer?
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Five moles of an ideal gas at 27° C are allowed to expand isother...
Work done during isothermal irreversible process is given by the formula w = -pext (nrt/p2-nrt/p1)
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Five moles of an ideal gas at 27° C are allowed to expand isother...
Understanding the Problem
To calculate the work done by the gas during isothermal expansion, we can use the formula for work done in a gas expansion against a constant external pressure.
Given Data
- Moles of gas (n): 5 moles
- Initial pressure (P1): 10.0 atm
- Final pressure (P2): 4.0 atm
- External pressure (Pext): 1.0 atm
- Temperature (T): 27°C (or 300 K)
Work Done Calculation
The work done (W) by the gas during expansion against a constant external pressure is given by the formula:
W = -Pext * ΔV
Where ΔV is the change in volume. Since the process is isothermal, we can relate the pressures and volumes using the ideal gas law:
- Initial volume (V1) can be calculated as:
V1 = (nRT) / P1
- Final volume (V2) can be calculated as:
V2 = (nRT) / P2
Thus, the change in volume (ΔV) is:
ΔV = V2 - V1
Substituting the values and solving for the volumes:
- V1 = (5 moles * 0.0821 L·atm/K·mol * 300 K) / 10 atm = 12.315 L
- V2 = (5 moles * 0.0821 L·atm/K·mol * 300 K) / 4 atm = 30.7875 L
Now, the change in volume (ΔV) is:
ΔV = 30.7875 L - 12.315 L = 18.4725 L
Now, converting liters to cubic meters (1 L = 0.001 m³), we have:
ΔV = 0.0184725 m³
Now, substituting into the work done formula:
W = -Pext * ΔV = -1 atm * 0.0184725 m³ = -1 * 101.325 J/atm * 0.0184725 m³ = -1870.6 J
Thus, the work done by the gas is -1870.6 J, indicating work done by the system (expansion).
Conclusion
The correct answer is option 'A': 1870.6 J (work done by the gas).
To calculate the work done by the gas during isothermal expansion, we can use the formula for work done in a gas expansion against a constant external pressure.
Given Data
- Moles of gas (n): 5 moles
- Initial pressure (P1): 10.0 atm
- Final pressure (P2): 4.0 atm
- External pressure (Pext): 1.0 atm
- Temperature (T): 27°C (or 300 K)
Work Done Calculation
The work done (W) by the gas during expansion against a constant external pressure is given by the formula:
W = -Pext * ΔV
Where ΔV is the change in volume. Since the process is isothermal, we can relate the pressures and volumes using the ideal gas law:
- Initial volume (V1) can be calculated as:
V1 = (nRT) / P1
- Final volume (V2) can be calculated as:
V2 = (nRT) / P2
Thus, the change in volume (ΔV) is:
ΔV = V2 - V1
Substituting the values and solving for the volumes:
- V1 = (5 moles * 0.0821 L·atm/K·mol * 300 K) / 10 atm = 12.315 L
- V2 = (5 moles * 0.0821 L·atm/K·mol * 300 K) / 4 atm = 30.7875 L
Now, the change in volume (ΔV) is:
ΔV = 30.7875 L - 12.315 L = 18.4725 L
Now, converting liters to cubic meters (1 L = 0.001 m³), we have:
ΔV = 0.0184725 m³
Now, substituting into the work done formula:
W = -Pext * ΔV = -1 atm * 0.0184725 m³ = -1 * 101.325 J/atm * 0.0184725 m³ = -1870.6 J
Thus, the work done by the gas is -1870.6 J, indicating work done by the system (expansion).
Conclusion
The correct answer is option 'A': 1870.6 J (work done by the gas).
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Five moles of an ideal gas at 27° C are allowed to expand isother...
-1870.6J
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Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will bea)- 1870.6 Jb)- 1870.6Jc)-18.47 Jd)+18.47 JCorrect answer is option 'A'. Can you explain this answer?
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Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will bea)- 1870.6 Jb)- 1870.6Jc)-18.47 Jd)+18.47 JCorrect answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will bea)- 1870.6 Jb)- 1870.6Jc)-18.47 Jd)+18.47 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will bea)- 1870.6 Jb)- 1870.6Jc)-18.47 Jd)+18.47 JCorrect answer is option 'A'. Can you explain this answer?.
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