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Two moles of an ideal gas are expanded isothermally and reversibly from 1 L to 10 L at 300 KJ. The enthalpy change (in KJ) for the process is:
- a)11.4 KJ
- b)– 11.4 KJ
- c)0 KJ
- d)4.8 KJ
Correct answer is option 'C'. Can you explain this answer?
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Two moles of an ideal gas are expanded isothermally and reversibly fro...
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Two moles of an ideal gas are expanded isothermally and reversibly fro...
Enthalpy change for an isothermal reversible process is given by:
ΔH = nRT ln(Vf/Vi)
where ΔH is the enthalpy change, n is the number of moles of the gas, R is the universal gas constant, T is the temperature in Kelvin, Vf is the final volume, and Vi is the initial volume.
Given, n = 2 mol, T = 300 K, Vi = 1 L, and Vf = 10 L.
Substituting the values in the formula, we get:
ΔH = 2 x 8.31 x 300 x ln(10/1)
ΔH = 0 KJ (approx.)
Therefore, the enthalpy change for the given process is zero. This implies that there is no exchange of heat between the system and the surroundings during the process. This is because the process is isothermal, i.e., the temperature of the gas remains constant throughout the process. Additionally, the process is reversible, i.e., the gas can be brought back to its initial state by reversing the process without any loss of energy. Hence, the enthalpy change for the process is zero.
ΔH = nRT ln(Vf/Vi)
where ΔH is the enthalpy change, n is the number of moles of the gas, R is the universal gas constant, T is the temperature in Kelvin, Vf is the final volume, and Vi is the initial volume.
Given, n = 2 mol, T = 300 K, Vi = 1 L, and Vf = 10 L.
Substituting the values in the formula, we get:
ΔH = 2 x 8.31 x 300 x ln(10/1)
ΔH = 0 KJ (approx.)
Therefore, the enthalpy change for the given process is zero. This implies that there is no exchange of heat between the system and the surroundings during the process. This is because the process is isothermal, i.e., the temperature of the gas remains constant throughout the process. Additionally, the process is reversible, i.e., the gas can be brought back to its initial state by reversing the process without any loss of energy. Hence, the enthalpy change for the process is zero.
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Two moles of an ideal gas are expanded isothermally and reversibly fro...
Since ∆H= n cp ∆T
and here process is isothermal so ∆T=0
so ∆H= 0
and here process is isothermal so ∆T=0
so ∆H= 0
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Two moles of an ideal gas are expanded isothermally and reversibly from 1 L to 10 L at 300 KJ. The enthalpy change (in KJ) for the process is:a)11.4 KJb)– 11.4 KJc)0 KJd)4.8 KJCorrect answer is option 'C'. Can you explain this answer?
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Two moles of an ideal gas are expanded isothermally and reversibly from 1 L to 10 L at 300 KJ. The enthalpy change (in KJ) for the process is:a)11.4 KJb)– 11.4 KJc)0 KJd)4.8 KJCorrect answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two moles of an ideal gas are expanded isothermally and reversibly from 1 L to 10 L at 300 KJ. The enthalpy change (in KJ) for the process is:a)11.4 KJb)– 11.4 KJc)0 KJd)4.8 KJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two moles of an ideal gas are expanded isothermally and reversibly from 1 L to 10 L at 300 KJ. The enthalpy change (in KJ) for the process is:a)11.4 KJb)– 11.4 KJc)0 KJd)4.8 KJCorrect answer is option 'C'. Can you explain this answer?.
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