One mole of O2(g) undergoes the following change in state.
O2(g, 299 K, 4.38 bar) ⟶ O2(g, 252 K, 6.79 bar)
What is ΔS for the gas? Assume ideal gas behaviour.
The constant pressure molar heat capacity for O2(g) is Cp,m = 29.10 J K−1 mol−1.
use eqn cpln(t2/t1)-nRln(p2/p1)
To calculate the change in entropy (ΔS) for the gas, we can use the equation:
ΔS = nCp, m log(T2/T1) + nR log(V2/V1)
Where:
ΔS = change in entropy
n = number of moles
Cp, m = molar heat capacity at constant pressure
T1, T2 = initial and final temperatures in kelvin
V1, V2 = initial and final volumes
In this case, we are given the molar heat capacity (Cp, m) for O2(g), as well as the initial and final temperatures (299 K and 252 K). However, we are not given the volume, but we can assume that the volume remains constant.
Since there is no change in volume (V1 = V2), the second term of the equation becomes zero.
Therefore, the equation simplifies to:
ΔS = nCp,m log(T2/T1)
Now we can plug in the values and calculate ΔS:
Given:
n = 1 mole
Cp, m = 29.10 J K−1 mol−1
T1 = 299 K
T2 = 252 K
ΔS = (1 mol)(29.10 J K−1 mol−1) log(252 K / 299 K)
ΔS = 29.10 J K−1 × (-0.0412)
Calculating this, we find:
ΔS = -1.2 J K−1 mol−1
Therefore, the change in entropy for the gas is approximately -1.2 J K−1 mol−1.