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# Ask AI: Culculate the entropy change when 3 moles of a perfect gas is allowed to expand at 300k from a volume of 2cm3 to 13 cm3

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Question: Culculate the entropy change when 3 moles of a perfect gas is allowed to expand at 300k from a volume of 2cm3 to 13 cm3

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To calculate the entropy change ΔS, you first need to know the general formula for entropy change for a perfect or ideal gas: ΔS = n*R*ln(V2/V1), where: n = number of moles R = Universal gas constant (8.314 J/(K*mol) V1 = Initial volume V2 = Final volume Note that volumes should be in the same units. Now, applying the data from your question: n = 3 moles R = 8.314 J/(K*mol) (universal gas constant) V1 = 2 cm³ V2 = 13 cm³ Even though it is not usually necessary to convert the volume to larger units such as m³ when calculating entropy changes (since the ratio V2/V1 would remain the same), we'll do it here to avoid possible confusion: 1 m³ = 1,000,000 cm³, so: V1 = 2 cm³ = 2 * 10^-6 m³ V2 = 13 cm³ = 13 * 10^-6 m³ Now you can substitute these values into the entropy change formula: ΔS = n * R * ln(V2/V1) = 3 * 8.314 * ln((13 * 10^-6) / (2 * 10^-6)) = 24.942 * ln(6.5) Now using the natural logarithm of 6.5, which is approximately 1.8718: ΔS = 24.942 * 1.8718 ΔS ≈ 46.7 J/K. So, the entropy change, when 3 moles of a perfect gas is allowed to expand at 300K from a volume of 2 cm³ to 13 cm³, is approximately 46.7 joules per kelvin. I hope this helps! If you have any other questions, don't hesitate to ask.

Feb. 5, 2024, 12:03 a.m.

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