The volume of 0.100 mol of helium gas at 34°C is increased isothermally from 1.00 L to 4.50 L. Assuming the gas to be ideal, calculate the entropy change for the process.
It would appear from the site below that delta S = ln(v2/v1) since T is constant.
http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node39.html
To calculate the entropy change (ΔS) for the isothermal expansion of helium gas, we need to use the equation:
ΔS = nR ln(Vf/Vi)
Where:
- ΔS is the entropy change
- n is the number of moles of gas
- R is the ideal gas constant (8.314 J/(mol·K))
- Vf is the final volume of the gas
- Vi is the initial volume of the gas
Given:
- n = 0.100 mol
- R = 8.314 J/(mol·K)
- Vi = 1.00 L
- Vf = 4.50 L
Now, let's substitute these values into the equation to calculate ΔS:
ΔS = (0.100 mol) * (8.314 J/(mol·K)) * ln(4.50 L / 1.00 L)
Calculating the natural logarithm:
ΔS = (0.100 mol) * (8.314 J/(mol·K)) * ln(4.50)
Using a calculator, we find that ln(4.50) ≈ 1.5041.
ΔS ≈ (0.100 mol) * (8.314 J/(mol·K)) * 1.5041
Now, we can calculate the entropy change:
ΔS ≈ 0.125 J/K
Therefore, the entropy change for the isothermal expansion of helium gas is approximately 0.125 J/K.