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5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K?
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5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80d...
**Given Data:**
- Number of moles (n) = 5 mol
- Initial volume (V1) = 8 dm^3
- Final volume (V2) = 80 dm^3
- Temperature (T) = 300 K
**Calculating Change in Entropy:**
The change in entropy (ΔS) of an ideal gas during an isothermal and reversible expansion can be calculated using the equation:
ΔS = nR ln(V2/V1)
where:
- n is the number of moles of the gas
- R is the molar gas constant (8.314 J/(mol·K))
- ln is the natural logarithm function
- V2/V1 is the ratio of final volume to initial volume
**Substituting the given values:**
ΔS = 5 mol * 8.314 J/(mol·K) * ln(80 dm^3 / 8 dm^3)
**Calculating the ln(80/8):**
ln(80/8) = ln(10) = 2.3026
**Substituting back into the equation:**
ΔS = 5 mol * 8.314 J/(mol·K) * 2.3026
**Simplifying:**
ΔS = 95.73 J/K
Therefore, the change in entropy is 95.73 J/K.
**Answer:**
The correct option is C) 95.73 J/K.
- Number of moles (n) = 5 mol
- Initial volume (V1) = 8 dm^3
- Final volume (V2) = 80 dm^3
- Temperature (T) = 300 K
**Calculating Change in Entropy:**
The change in entropy (ΔS) of an ideal gas during an isothermal and reversible expansion can be calculated using the equation:
ΔS = nR ln(V2/V1)
where:
- n is the number of moles of the gas
- R is the molar gas constant (8.314 J/(mol·K))
- ln is the natural logarithm function
- V2/V1 is the ratio of final volume to initial volume
**Substituting the given values:**
ΔS = 5 mol * 8.314 J/(mol·K) * ln(80 dm^3 / 8 dm^3)
**Calculating the ln(80/8):**
ln(80/8) = ln(10) = 2.3026
**Substituting back into the equation:**
ΔS = 5 mol * 8.314 J/(mol·K) * 2.3026
**Simplifying:**
ΔS = 95.73 J/K
Therefore, the change in entropy is 95.73 J/K.
**Answer:**
The correct option is C) 95.73 J/K.
Community Answer
5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80d...
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5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K?
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5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K?.
5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 mole of an ideal gas expand reversibly from a volume of 8dm^3 to 80dm^3 at a temperature of 300K. The change in entropy is A) 41.57J/K B) -95.73J/K C) 95.73J/K D) -41.57J/K?.
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