ideal gas .450 mole initial pressure 16 atm and 290 K expands isothermally to a final pressure 1 atm. find work , if expansion is against a vacuum, a constant external pressure of 1 atm and reviersibly
W(isothermal) = nRT*ln(V2/V1)
or
W(isothermal) = nRT*ln(P1/P2)
See also:
http://www.diracdelta.co.uk/science/source/i/s/isothermal%20expansion/source.html
To find the work done during the isothermal expansion of an ideal gas from an initial pressure of 16 atm to a final pressure of 1 atm, we can use the formula:
Work = -nRT * ln(Pf/Pi)
Where:
- n is the number of moles of the gas (0.450 mole in this case)
- R is the ideal gas constant (0.0821 L.atm/K.mol)
- T is the temperature in Kelvin (290 K in this case)
- Pf is the final pressure (1 atm in this case)
- Pi is the initial pressure (16 atm in this case)
Let's plug in the values and calculate the work:
Work = -0.450 mole * 0.0821 L.atm/K.mol * 290 K * ln(1 atm / 16 atm)
First, we simplify the ratio inside the natural logarithm function:
Work = -0.450 mole * 0.0821 L.atm/K.mol * 290 K * ln(1/16)
Now we can calculate the natural logarithm of the ratio:
Work = -0.450 mole * 0.0821 L.atm/K.mol * 290 K * ln(0.0625)
Using a calculator, we find that ln(0.0625) is approximately -2.7726:
Work = -0.450 mole * 0.0821 L.atm/K.mol * 290 K * (-2.7726)
Calculating the numerical values:
Work ≈ -0.450 mole * 0.0821 L.atm/K.mol * 290 K * (-2.7726)
≈ 10.56 L.atm
Therefore, the work done during the isothermal expansion against a vacuum and a constant external pressure of 1 atm, and reversibly, is approximately 10.56 L.atm.