One mole of an ideal gas is sub jected to an isothermal increase in pr...
Calculation of Change in Gibbs Free Energy
To determine the change in Gibbs free energy (ΔG) of the system, we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
Step 1: Calculation of ΔH
Since the gas is ideal, we can assume that it follows the ideal gas law:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature
Given that the pressure increases from 100 kPa to 1000 kPa, and the number of moles is constant (1 mole), we can rearrange the equation to solve for the change in volume (ΔV):
ΔV = nRTΔP
Where:
ΔV is the change in volume
ΔP is the change in pressure
Substituting the values into the equation:
ΔV = (1 mol)(8.3 J/K.mol)(300 K)(1000 kPa - 100 kPa)
Simplifying the equation:
ΔV = 1 mol(8.3 J/K.mol)(300 K)(900 kPa)
Converting kilopascals (kPa) to pascals (Pa):
ΔV = 1 mol(8.3 J/K.mol)(300 K)(900,000 Pa)
Calculating the change in enthalpy (ΔH) using the equation:
ΔH = ΔU + PΔV
Where:
ΔU is the change in internal energy
P is the constant pressure
Assuming the process is isothermal, the change in internal energy (ΔU) is zero. Thus:
ΔH = PΔV
ΔH = (1000 kPa)(1 mol)(8.3 J/K.mol)(300 K)(900,000 Pa)
Step 2: Calculation of ΔS
Since the process is isothermal, the change in entropy (ΔS) can be calculated using the equation:
ΔS = nRln(P2/P1)
Where:
P2 is the final pressure
P1 is the initial pressure
Substituting the values into the equation:
ΔS = (1 mol)(8.3 J/K.mol)ln(1000 kPa/100 kPa)
Simplifying the equation:
ΔS = (1 mol)(8.3 J/K.mol)ln(10)
ΔS = (1 mol)(8.3 J/K.mol)(2.3026)
Step 3: Calculation of ΔG
Now that we have calculated ΔH and ΔS, we can substitute these values into the equation to find the change in Gibbs free energy:
ΔG = ΔH - TΔS
ΔG = (1000 kPa)(1 mol)(8.3 J/K.mol)(300 K)(900,000 Pa) - (300 K)(1 mol)(8.3 J/K.mol)(2.3026)
Calculating the