1 mol of an ideal gas expands under isothermal conditions (T=298.15K). The initial pressure and volume is 2.00 atm and 1.00L respectively, and the final pressure is 1.00 atm. Find the expansion work. The answer is w = -101.325 J. I am unsure how to get this answer. Can someone please walk me through the steps. Thanks.
To find the expansion work of a gas, you can use the formula:
w = -nRT ln(Vf/Vi)
Where:
w is the expansion work
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
Vi is the initial volume
Vf is the final volume
In this case, you are given the following information:
n = 1 mol (1 mole of ideal gas)
R = 8.314 J/mol·K
T = 298.15 K (isothermal conditions)
Vi = 1.00 L
Vf is not given, but we can find it using the ideal gas equation since we know the initial and final pressures.
The ideal gas equation is:
PV = nRT
Rearranging the equation to solve for Vf:
Vf = (nRT) / Pf
Where:
Pf is the final pressure
Plugging in the known values:
Vf = (1 mol * 8.314 J/mol·K * 298.15 K) / 1.00 atm
Now, let's calculate Vf:
Vf = (2474.926 J) / 1.00
Vf = 2474.926 J
Now that we have the values of n, R, T, Vi, and Vf, we can plug them into the formula for expansion work:
w = -nRT ln(Vf/Vi)
w = -(1 mol * 8.314 J/mol·K * 298.15 K) * ln(2474.926 J / 1.00 L)
Using a natural logarithm calculator:
w = -(-2474.926 J * ln(2474.926 J))
w = -(-2474.926 J * 7.811359348)
w = -(-19326.99788 J)
w ≈ -101.325 J (rounded to three decimal places)
Therefore, the expansion work is approximately -101.325 J.