What is the change in entropy when 1 mole of an ideal gas expands isothermally f

₹0.0

Question: What is the change in entropy when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K?

Options:

Correct Answer: 1.0 J/K

Solution:

ΔS = nR ln(Vf/Vi) = 1 * 8.314 * ln(20/10) = 1.0 J/K.

What is the change in entropy when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K?

What is the change in entropy when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K?

Step 1: Identify the variables needed for the calculation. We have 1 mole of gas (n = 1), the gas constant (R = 8.314 J/(mol·K)), the initial volume (Vi = 10 L), and the final volume (Vf = 20 L).
Step 2: Use the formula for change in entropy (ΔS) for an ideal gas during isothermal expansion: ΔS = nR ln(Vf/Vi).
Step 3: Substitute the values into the formula: ΔS = 1 * 8.314 * ln(20/10).
Step 4: Calculate the natural logarithm: ln(20/10) = ln(2).
Step 5: Calculate ln(2) which is approximately 0.693.
Step 6: Multiply the values: ΔS = 1 * 8.314 * 0.693.
Step 7: Perform the multiplication: ΔS ≈ 5.764 J/K.
Step 8: Round the result to one decimal place: ΔS ≈ 5.8 J/K.

Entropy Change in Ideal Gases – The calculation of entropy change (ΔS) for an ideal gas during isothermal expansion using the formula ΔS = nR ln(Vf/Vi), where n is the number of moles, R is the ideal gas constant, and Vf and Vi are the final and initial volumes.
Isothermal Processes – Understanding that isothermal processes occur at constant temperature, which affects the behavior of gases and the calculation of entropy.