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1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:
- a)3.775
- b)37.256
- c)372.56
- d)3725.6
Correct answer is option 'C'. Can you explain this answer?
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Understanding the Problem
In this scenario, we are dealing with 1 mole of an ideal gas heated reversibly at constant pressure. The specific heat at constant volume (CV) is given as 3/2 R.
Key Known Values
- Number of moles (n): 1 mole
- Initial temperature (T1): 25°C = 298.15 K
- Final temperature (T2): 100°C = 373.15 K
- Constant pressure (P): 1 atm
- Universal gas constant (R): 8.314 J/(mol·K)
Calculation of Enthalpy Change (ΔH)
The enthalpy change (ΔH) for an ideal gas at constant pressure can be calculated using the formula:
ΔH = n * Cp * ΔT
Where:
- ΔT = T2 - T1
- Cp is the heat capacity at constant pressure.
Since CV is given, we need to find Cp. For a monatomic ideal gas:
Cp = CV + R = (3/2 R) + R = (5/2 R)
Substituting R:
Cp = (5/2) * 8.314 = 20.785 J/(mol·K)
Now, calculate ΔT:
ΔT = 373.15 K - 298.15 K = 75 K
Final Calculation
Now we can calculate ΔH:
ΔH = 1 mole * 20.785 J/(mol·K) * 75 K
ΔH = 1558.875 J
To convert to calories (1 cal = 4.184 J):
ΔH = 1558.875 J * (1 cal / 4.184 J) = 372.56 cal
Conclusion
Thus, the enthalpy change (ΔH) is 372.56 cal, confirming that the correct answer is option 'C'.
In this scenario, we are dealing with 1 mole of an ideal gas heated reversibly at constant pressure. The specific heat at constant volume (CV) is given as 3/2 R.
Key Known Values
- Number of moles (n): 1 mole
- Initial temperature (T1): 25°C = 298.15 K
- Final temperature (T2): 100°C = 373.15 K
- Constant pressure (P): 1 atm
- Universal gas constant (R): 8.314 J/(mol·K)
Calculation of Enthalpy Change (ΔH)
The enthalpy change (ΔH) for an ideal gas at constant pressure can be calculated using the formula:
ΔH = n * Cp * ΔT
Where:
- ΔT = T2 - T1
- Cp is the heat capacity at constant pressure.
Since CV is given, we need to find Cp. For a monatomic ideal gas:
Cp = CV + R = (3/2 R) + R = (5/2 R)
Substituting R:
Cp = (5/2) * 8.314 = 20.785 J/(mol·K)
Now, calculate ΔT:
ΔT = 373.15 K - 298.15 K = 75 K
Final Calculation
Now we can calculate ΔH:
ΔH = 1 mole * 20.785 J/(mol·K) * 75 K
ΔH = 1558.875 J
To convert to calories (1 cal = 4.184 J):
ΔH = 1558.875 J * (1 cal / 4.184 J) = 372.56 cal
Conclusion
Thus, the enthalpy change (ΔH) is 372.56 cal, confirming that the correct answer is option 'C'.
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1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer?.
1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer?.
Solutions for 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The ΔH(cal) is:a)3.775b)37.256c)372.56d)3725.6Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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