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One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm?
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One mole of an ideal gas at 300K is expanded isothermally from an init...
Answer:
Given:
Initial volume (V1) = 1 liter
Final volume (V2) = 10 liters
Temperature (T) = 300K
Gas constant (R) = 2 cal Mol^-1 K^-1
To find the change in internal energy (ΔE) for the process, we can use the equation:
ΔE = nRT ln(V2/V1)
where:
n = number of moles of the gas (1 mole in this case)
R = gas constant
T = temperature
V2 = final volume
V1 = initial volume
Calculating the change in internal energy:
ΔE = 1 * 2 * 300 ln(10/1)
ΔE = 600 ln(10)
Calculating the value of ln(10):
ln(10) = 2.3026
Substituting the value of ln(10) in the equation:
ΔE = 600 * 2.3026
ΔE = 1381.56 cal
Rounding off to the nearest cal:
ΔE ≈ 1382 cal
Therefore, the change in internal energy (ΔE) for the process is approximately 1382 cal.
- Given values:
- Initial volume (V1) = 1 liter
- Final volume (V2) = 10 liters
- Temperature (T) = 300K
- Gas constant (R) = 2 cal Mol^-1 K^-1
- Equation to calculate change in internal energy:
- ΔE = nRT ln(V2/V1)
- Substituting the given values:
- ΔE = 1 * 2 * 300 ln(10/1)
- Calculating ln(10):
- ln(10) = 2.3026
- Substituting the value of ln(10) in the equation:
- ΔE = 600 * 2.3026
- ΔE = 1381.56 cal
- Rounding off to the nearest cal:
- ΔE ≈ 1382 cal
- Therefore, the change in internal energy (ΔE) for the process is approximately 1382 cal.
Given:
Initial volume (V1) = 1 liter
Final volume (V2) = 10 liters
Temperature (T) = 300K
Gas constant (R) = 2 cal Mol^-1 K^-1
To find the change in internal energy (ΔE) for the process, we can use the equation:
ΔE = nRT ln(V2/V1)
where:
n = number of moles of the gas (1 mole in this case)
R = gas constant
T = temperature
V2 = final volume
V1 = initial volume
Calculating the change in internal energy:
ΔE = 1 * 2 * 300 ln(10/1)
ΔE = 600 ln(10)
Calculating the value of ln(10):
ln(10) = 2.3026
Substituting the value of ln(10) in the equation:
ΔE = 600 * 2.3026
ΔE = 1381.56 cal
Rounding off to the nearest cal:
ΔE ≈ 1382 cal
Therefore, the change in internal energy (ΔE) for the process is approximately 1382 cal.
- Given values:
- Initial volume (V1) = 1 liter
- Final volume (V2) = 10 liters
- Temperature (T) = 300K
- Gas constant (R) = 2 cal Mol^-1 K^-1
- Equation to calculate change in internal energy:
- ΔE = nRT ln(V2/V1)
- Substituting the given values:
- ΔE = 1 * 2 * 300 ln(10/1)
- Calculating ln(10):
- ln(10) = 2.3026
- Substituting the value of ln(10) in the equation:
- ΔE = 600 * 2.3026
- ΔE = 1381.56 cal
- Rounding off to the nearest cal:
- ΔE ≈ 1382 cal
- Therefore, the change in internal energy (ΔE) for the process is approximately 1382 cal.
Community Answer
One mole of an ideal gas at 300K is expanded isothermally from an init...
A) 1381.1 Cal
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One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm?
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One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm?.
One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 litres. The Del(E) for the process is (R = 2cal Mol^-1 K^-1) A) 1381.1 cal B) zero C) 163.7 cal D) 9 Latm?.
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