Find the work for an isothermal expansion of 3.08 mol of an ideal gas at temperature of 18oC from pi = 4.55 atm to pf = 3.32 atm.
To find the work for an isothermal expansion, we can use the formula:
W = -nRT ln(Vf/Vi)
Where:
W - Work done by the system
n - Number of moles of gas
R - Ideal gas constant (0.0821 L•atm/(mol•K))
T - Temperature of the gas in Kelvin
Vi - Initial volume of the gas
Vf - Final volume of the gas
First, we need to convert the temperature from Celsius to Kelvin:
T = 18°C + 273.15 = 291.15 K
Next, we calculate the work using the given values:
n = 3.08 mol
R = 0.0821 L•atm/(mol•K)
Vi is not given, but we can solve for it using the ideal gas law at the initial conditions:
pi = 4.55 atm
Rearranging the ideal gas law gives us:
Vi = nRT/pi = (3.08 mol)(0.0821 L•atm/(mol•K))(291.15 K)/(4.55 atm)
Vi ≈ 61.7 L
Now we can use the formula to calculate the work:
W = -nRT ln(Vf/Vi) = -(3.08 mol)(0.0821 L•atm/(mol•K))(291.15 K) ln(3.32 atm/61.7 L)
W ≈ -17.07 atm L
Note: The negative sign indicates that work is done on the system during an expansion.