What is the work done when 1 mole of an ideal gas expands isothermally from 10 L

₹0.0

Question: What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at a temperature of 300 K? (2020)

Options:

Correct Answer: 2.5 kJ

Exam Year: 2020

Solution:

Work done (W) = nRT ln(Vf/Vi) = 1 mol * 8.314 J/(mol K) * 300 K * ln(20/10) = 2.5 kJ.

What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at a temperature of 300 K? (2020)

What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at a temperature of 300 K? (2020)

Step 1: Identify the variables needed for the formula. We have: n (number of moles) = 1, R (ideal gas constant) = 8.314 J/(mol K), T (temperature) = 300 K, Vi (initial volume) = 10 L, and Vf (final volume) = 20 L.
Step 2: Write down the formula for work done during isothermal expansion of an ideal gas: W = nRT ln(Vf/Vi).
Step 3: Substitute the values into the formula: W = 1 mol * 8.314 J/(mol K) * 300 K * ln(20/10).
Step 4: Calculate the natural logarithm: ln(20/10) = ln(2).
Step 5: Calculate the value of ln(2) which is approximately 0.693.
Step 6: Now substitute ln(2) back into the equation: W = 1 * 8.314 * 300 * 0.693.
Step 7: Perform the multiplication: 8.314 * 300 = 2494.2 J, then multiply by 0.693 to get approximately 1727.5 J.
Step 8: Convert the work done from Joules to kilojoules: 1727.5 J = 1.7275 kJ.
Step 9: Round the answer to two decimal places: The work done is approximately 1.73 kJ.

No concepts available.