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Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.
(final answer should be rounded off to two decimal places)
[Given: R = 8.314 J mol–1 K–1]
(final answer should be rounded off to two decimal places)
[Given: R = 8.314 J mol–1 K–1]
Correct answer is '14.89'. Can you explain this answer?
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Most Upvoted Answer
Consider an isothermal reversible compression of one mole of an ideal ...
The entropy change of the surroundings can be calculated using the formula:
ΔSsurroundings = -ΔH/T
where ΔH is the enthalpy change of the system and T is the temperature of the surroundings. Since the compression is isothermal, the temperature of the surroundings is also 300 K.
To calculate ΔH, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
At the initial state, the pressure is 5 atm and the volume is V1. At the final state, the pressure is 30 atm and the volume is V2.
Since the compression is reversible, we can assume that the process is quasi-static and that the gas is always in equilibrium with the surroundings. Therefore, we can use the formula for reversible work:
Wrev = -nRT ln(V2/V1)
The enthalpy change is equal to the work done at constant pressure:
ΔH = Wrev = -nRT ln(V2/V1)
Substituting the values, we get:
ΔH = -(1 mol)(0.08206 L atm/mol K)(300 K) ln(V2/V1)
ΔH = -24.62 ln(V2/V1) J
The ratio of volumes is given by the ratio of pressures, since the number of moles and temperature are constant:
V2/V1 = P1/P2 = 5/30 = 1/6
Substituting this value, we get:
ΔH = -24.62 ln(1/6) J
ΔH = 24.62 ln(6) J
Now we can calculate the entropy change of the surroundings:
ΔSsurroundings = -ΔH/T
ΔSsurroundings = -(24.62 ln(6) J)/(300 K)
ΔSsurroundings = -0.041 JK^-1
Therefore, the entropy change of the surroundings is -0.041 JK^-1.
ΔSsurroundings = -ΔH/T
where ΔH is the enthalpy change of the system and T is the temperature of the surroundings. Since the compression is isothermal, the temperature of the surroundings is also 300 K.
To calculate ΔH, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
At the initial state, the pressure is 5 atm and the volume is V1. At the final state, the pressure is 30 atm and the volume is V2.
Since the compression is reversible, we can assume that the process is quasi-static and that the gas is always in equilibrium with the surroundings. Therefore, we can use the formula for reversible work:
Wrev = -nRT ln(V2/V1)
The enthalpy change is equal to the work done at constant pressure:
ΔH = Wrev = -nRT ln(V2/V1)
Substituting the values, we get:
ΔH = -(1 mol)(0.08206 L atm/mol K)(300 K) ln(V2/V1)
ΔH = -24.62 ln(V2/V1) J
The ratio of volumes is given by the ratio of pressures, since the number of moles and temperature are constant:
V2/V1 = P1/P2 = 5/30 = 1/6
Substituting this value, we get:
ΔH = -24.62 ln(1/6) J
ΔH = 24.62 ln(6) J
Now we can calculate the entropy change of the surroundings:
ΔSsurroundings = -ΔH/T
ΔSsurroundings = -(24.62 ln(6) J)/(300 K)
ΔSsurroundings = -0.041 JK^-1
Therefore, the entropy change of the surroundings is -0.041 JK^-1.
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Consider an isothermal reversible compression of one mole of an ideal ...
There is a direct formula for finding the entropy of the system (del S sys = nR ln (pi/pf)....this value will be negative.... As entropy of the system is decreased.... So entropy of the surrounding is the positive of the above value (calculations are not shown here)
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Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer?
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Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. 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 Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer?.
Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. 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 Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer?.
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Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer?, a detailed solution for Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer? has been provided alongside types of Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in JK–1) is __________.(final answer should be rounded off to two decimal places)[Given: R = 8.314 J mol–1 K–1]Correct answer is '14.89'. Can you explain this answer? theory, EduRev gives you an
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