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One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.
(Round off to one decimal place)
[Given: Gas constant (R) = 8.3 J K–1 mol–1]
(Round off to one decimal place)
[Given: Gas constant (R) = 8.3 J K–1 mol–1]
Correct answer is between '5.6,5.8'. Can you explain this answer?
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One mole of an ideal gas is subjected to an isothermal increase in pre...
Since the process is isothermal, the temperature (T) is constant and we can use the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change and ΔS is the entropy change.
For an ideal gas, the enthalpy change is given by:
ΔH = nCpΔT
where n is the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.
Since the temperature is constant, ΔT = 0 and ΔH = 0.
The entropy change is given by:
ΔS = nRln(P2/P1)
where R is the gas constant and P1 and P2 are the initial and final pressures, respectively.
Substituting the values, we get:
ΔS = (1 mol)(8.314 J mol^-1 K^-1) ln(1000 kPa/100 kPa) = 16.06 J K^-1
Converting to kJ:
ΔS = 0.01606 kJ K^-1
Finally, substituting these values in the equation for ΔG, we get:
ΔG = ΔH - TΔS = 0 - (300 K)(0.01606 kJ K^-1) = -4.82 kJ mol^-1
Therefore, the change in Gibbs free energy of the system is -4.82 kJ mol^-1.
ΔG = ΔH - TΔS
where ΔH is the enthalpy change and ΔS is the entropy change.
For an ideal gas, the enthalpy change is given by:
ΔH = nCpΔT
where n is the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.
Since the temperature is constant, ΔT = 0 and ΔH = 0.
The entropy change is given by:
ΔS = nRln(P2/P1)
where R is the gas constant and P1 and P2 are the initial and final pressures, respectively.
Substituting the values, we get:
ΔS = (1 mol)(8.314 J mol^-1 K^-1) ln(1000 kPa/100 kPa) = 16.06 J K^-1
Converting to kJ:
ΔS = 0.01606 kJ K^-1
Finally, substituting these values in the equation for ΔG, we get:
ΔG = ΔH - TΔS = 0 - (300 K)(0.01606 kJ K^-1) = -4.82 kJ mol^-1
Therefore, the change in Gibbs free energy of the system is -4.82 kJ mol^-1.
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One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer?
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One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer?.
One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is __________ kJ mol–1.(Round off to one decimal place)[Given: Gas constant (R) = 8.3 J K–1 mol–1]Correct answer is between '5.6,5.8'. Can you explain this answer?.
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