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, we can use the formula:
w = -nRT ln(V₂/V₁)
where:
w = work done by the gas
n = number of moles of the gas
R = ideal gas constant (8.314 J/mol·K)
T = temperature in Kelvin
V₁ = initial volume
V₂ = final volume
In this case, we are given:
n = 1 mol
R = 8.314 J/mol·K
T = 298.15 K
V₁ = 1.00 L
V₂ = unknown
To find V₂, we can use Boyle's law, which states that at constant temperature, the pressure and volume of a gas are inversely proportional. Mathematically, it can be expressed as:
P₁V₁ = P₂V₂
where:
P₁ = initial pressure
P₂ = final pressure
We are given:
P₁ = 2.00 atm
P₂ = 1.00 atm
V₁ = 1.00 L
V₂ = unknown
Rearranging the equation, we have:
V₂ = (P₁V₁) / P₂
Substituting the given values, we get:
V₂ = (2.00 atm * 1.00 L) / 1.00 atm
= 2.00 L
Now we can substitute the values of n, R, T, V₁, and V₂ into the expansion work formula:
w = - (1 mol) * (8.314 J/mol·K) * (298.15 K) * ln(2.00 L / 1.00 L)
Simplifying, we get:
w = - (1 mol) * (8.314 J/mol·K) * (298.15 K) * ln(2.00)
Using a calculator, we can evaluate the natural logarithm of 2.00, which is approximately 0.6931.
w = - (1 mol) * (8.314 J/mol·K) * (298.15 K) * 0.6931
= -101.325 J
Therefore, the expansion work is -101.325 J.