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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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