calculate the change in entropy when 1.00g of oxygen at 27 degree celsius is expanded from 1 dm 3 to 4 dm3
To calculate the change in entropy, we need to use the formula:
ΔS = nR ln(V2/V1)
Where:
ΔS: Change in entropy
n: Quantity of gas in moles
R: Ideal gas constant (8.314 J/(mol•K))
V1: Initial volume
V2: Final volume
First, we need to determine the quantity of gas in moles. We can use the ideal gas equation:
PV = nRT
Where:
P: Pressure
V: Volume
n: Quantity of gas in moles
R: Ideal gas constant (8.314 J/(mol•K))
T: Temperature in Kelvin
Rearranging the equation to solve for n:
n = PV / (RT)
Given:
P is not provided in the question, but we can assume it to be constant, since the process is an expansion.
T = 27 degrees Celsius = 300 Kelvin (convert to Kelvin by adding 273.15)
Now, we can substitute the values into the equation to solve for n.
n = PV / (RT)
Next, substitute the known values to calculate the value of n.
n = (constant pressure) × (initial volume) / (constant gas constant) × (temperature)
Since the pressure is constant and not given, it cancels out in the equation, and we can simply use the ratio of the volumes.
n = (initial volume) / (constant gas constant × temperature)
Next, calculate the value of n using the provided values.
n = (1 dm^3) / [(8.314 J/(mol•K)) × (300 K)]
Calculate the value of n:
n = 0.00134 mol
Now, with the value of n, we can calculate the change in entropy using the formula:
ΔS = nR ln(V2/V1)
Substitute the known values:
ΔS = (0.00134 mol) × (8.314 J/(mol•K)) × ln(4 dm^3 / 1 dm^3)
Calculate the natural logarithm:
ΔS = (0.00134 mol) × (8.314 J/(mol•K)) × ln(4)
Now, calculate the final result:
ΔS ≈ 0.052 J/K
Therefore, the change in entropy when 1.00g of oxygen at 27 degrees Celsius is expanded from 1 dm^3 to 4 dm^3 is approximately 0.052 J/K.