Calculate the entropy change when a sample of 30 g of Ne is expanded from 1 L to 10L and simultaneously cooled from 350 K to 300 K. Cv = 3/2
Question
Calculate the entropy change when a sample of 30 g of Ne is expanded from 1 L to 10L and simultaneously cooled from 350 K to 300 K. Cv = 3/2
Solution
To calculate the entropy change, we can use the formula:
ΔS = Cv * ln(T2/T1) + R * ln(V2/V1)
Given that the sample of Ne is expanded from 1 L to 10 L and cooled from 350 K to 300 K, we can substitute the values into the formula.
ΔS = (3/2) * ln(300/350) + R * ln(10/1)
Now, we need to determine the value of R. The gas constant R is equal to 8.314 J/(mol·K). Since we have a sample of Ne with a mass of 30 g, we need to convert it to moles.
First, we calculate the molar mass of Ne. The molar mass of Ne is approximately 20.18 g/mol.
Next, we calculate the number of moles of Ne in the sample:
moles = mass / molar mass moles = 30 g / 20.18 g/mol
moles ≈ 1.485 mol
Now, we can substitute the value of R into the equation:
ΔS = (3/2) * ln(300/350) + 8.314 J/(mol·K) * ln(10/1)
Calculating the natural logarithm:
ΔS = (3/2) * ln(0.857) + 8.314 J/(mol·K) * ln(10)
Using a calculator, we find:
ΔS ≈ -0.225 J/K + 18.23 J/K
ΔS ≈ 18.005 J/K
Therefore, the entropy change when the sample of Ne is expanded from 1 L to 10 L and cooled from 350 K to 300 K is approximately 18.005 J/K.
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