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One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]
Correct answer is '-8.314'. Can you explain this answer?
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One mole of an ideal gas is subjected to a change from 10 atm to 1 atm...
Explanation:
To calculate the change in entropy of the surroundings (Ssurrounding), we can use the equation:
Ssurrounding = -ΔH/T
where ΔH is the enthalpy change and T is the temperature.
Step 1: Calculate the enthalpy change (ΔH)
We can calculate the enthalpy change using the ideal gas equation:
ΔH = ΔnRT
where Δn is the change in the number of moles, R is the gas constant, and T is the temperature.
In this case, since we have one mole of gas, the change in the number of moles is 1.
ΔH = (1 mol)(8.314 JK-1mol-1)(300 K)
= 2494.2 J
Step 2: Calculate the change in entropy of the surroundings (Ssurrounding)
Now we can substitute the values into the equation for Ssurrounding:
Ssurrounding = -ΔH/T
= -(2494.2 J)/(300 K)
≈ -8.314 JK-1
Final answer:
The change in entropy of the surroundings (Ssurrounding) is approximately -8.314 JK-1.
To calculate the change in entropy of the surroundings (Ssurrounding), we can use the equation:
Ssurrounding = -ΔH/T
where ΔH is the enthalpy change and T is the temperature.
Step 1: Calculate the enthalpy change (ΔH)
We can calculate the enthalpy change using the ideal gas equation:
ΔH = ΔnRT
where Δn is the change in the number of moles, R is the gas constant, and T is the temperature.
In this case, since we have one mole of gas, the change in the number of moles is 1.
ΔH = (1 mol)(8.314 JK-1mol-1)(300 K)
= 2494.2 J
Step 2: Calculate the change in entropy of the surroundings (Ssurrounding)
Now we can substitute the values into the equation for Ssurrounding:
Ssurrounding = -ΔH/T
= -(2494.2 J)/(300 K)
≈ -8.314 JK-1
Final answer:
The change in entropy of the surroundings (Ssurrounding) is approximately -8.314 JK-1.
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One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. Can you explain this answer?
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One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. 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 One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. Can you explain this answer?.
One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. 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 One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal gas is subjected to a change from 10 atm to 1 atm at 300 K. Calculate (in JK-1 rounded up to three decimal places) ∆Ssurrounding = ? [R = 8.314 JK-1mol-1]Correct answer is '-8.314'. Can you explain this answer?.
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