Calculate delta G, delta A, and delta Stotal for each of the following processes and state any
approximations made:
(a) Reversible vaporization of 39 g of C6H6 at its normal boiling point of 80.1°C and 1
atm;
(b) Adiabatic reversible expansion of 0.100 mol of a perfect gas with initial
temperature of 300 K, initial volume of 2.00 L, to final volume of 6.00 L.
(c) Adiabatic expansion of 0.100 mol of a perfect gas into with initial temperature of
300 K, initial volume of 2.00 L, to final volume of 6.00 L under vacuum
(a) To calculate delta G, delta A, and delta Stotal for the reversible vaporization of 39 g of C6H6, we need to use the following formulas:
delta G = delta H - T * delta S
delta A = delta U - T * delta S
delta Stotal = delta S(system) + delta S(surrounding)
First, we need to determine delta H, delta U, and delta S for the vaporization process. We can look up the values for these parameters in a reference table or use the given data.
Assuming ideal behavior for the gas, we can approximate the heat change (delta H) as the enthalpy of vaporization of C6H6 at its boiling point. We will also assume that the entropy change (delta S) is constant over the temperature range of interest.
Given:
- Mass of C6H6 = 39 g
- Boiling point of C6H6 = 80.1°C = 353.25 K
- Pressure = 1 atm
The enthalpy of vaporization (delta H) of C6H6 at its normal boiling point is approximately 30.8 kJ/mol.
delta H = 30.8 kJ/mol
To calculate delta U, we need to know the initial and final states of the substance. However, since the problem does not provide this information, we cannot calculate delta U accurately.
Assuming constant heat capacity (C), we can approximate delta U using the formula:
delta U = C * delta T
However, since we don't know the specific heat capacity of C6H6 or the change in temperature (delta T) accurately, we cannot calculate delta U.
Next, we can calculate delta S using the formula:
delta S = delta H / T
delta S = (30.8 kJ/mol) / (353.25 K)
Now we can calculate delta G using the formula:
delta G = delta H - T * delta S
delta G = (30.8 kJ/mol) - (353.25 K) * [(30.8 kJ/mol) / (353.25 K)]
(b) To calculate the delta G, delta A, and delta Stotal for the adiabatic reversible expansion of 0.100 mol of a perfect gas, we use the following formulas:
delta G = 0 (for an adiabatic process)
delta A = delta U - T * delta S
delta Stotal = 0 (for an adiabatic process)
Since the process is adiabatic, there is no heat exchange (delta Q = 0), and therefore, delta G is equal to zero.
To calculate delta A, we need to know the initial and final states of the gas. However, the problem only provides initial and final volumes, so we cannot calculate delta A accurately.
We cannot calculate delta Stotal either since it depends on the values of delta S(system) and delta S(surrounding), which we do not have.
(c) For the adiabatic expansion of 0.100 mol of a perfect gas into a vacuum, the process occurs without any heat exchange. Therefore:
delta G = 0
delta A = delta U - T * delta S
delta Stotal = 0
Since the expansion is adiabatic, there is no heat exchange (delta Q = 0), and therefore, delta G is equal to zero.
To calculate delta A, we need to know the initial and final states of the gas. However, the problem only provides the initial and final volumes, and the process is occurring under vacuum, so we cannot calculate delta A accurately.
Similarly, we cannot calculate delta Stotal since it depends on the values of delta S(system) and delta S(surrounding), which we do not have.
To calculate delta G, delta A, and delta Stotal for each process, you need to use the following formulas:
1. Delta G (Gibbs free energy change):
Delta G = Delta H - T * Delta S
2. Delta A (Helmholtz free energy change):
Delta A = Delta U - T * Delta S
3. Delta Stotal (total entropy change):
Delta Stotal = Delta Ssystem + Delta Ssurroundings
Now, let's calculate the values for each process:
(a) Reversible vaporization of 39 g of C6H6 at its normal boiling point of 80.1°C and 1 atm:
To calculate delta G for this process, we need to know the enthalpy change (Delta H), the entropy change (Delta S), and the temperature (T). The enthalpy change can be obtained from reference tables or calculated using the heat of vaporization. The entropy change can also be obtained from reference tables or calculated using the molar entropy values.
Approximation: Assuming that the vapor behaves as an ideal gas, which is a reasonable approximation for this process.
(b) Adiabatic reversible expansion of 0.100 mol of a perfect gas with an initial temperature of 300 K, an initial volume of 2.00 L, to a final volume of 6.00 L:
To calculate delta G and delta A, we need to know the internal energy change (Delta U), the entropy change (Delta S), and the temperature (T). The internal energy change can be obtained from the ideal gas law or by considering the work done during the expansion. The entropy change can be calculated using the ideal gas equation and molar entropy values.
Approximation: Assuming the gas behaves as a perfect gas and follows the ideal gas law.
(c) Adiabatic expansion of 0.100 mol of a perfect gas with an initial temperature of 300 K, an initial volume of 2.00 L, to a final volume of 6.00 L under vacuum:
Since the expansion is adiabatic and occurs under vacuum, there is no heat transfer between the system and the surroundings. Therefore, the values of delta G and delta A are zero for this process.
Approximation: Assuming the gas behaves as a perfect gas and follows the ideal gas law.
Please note that the calculations for each process may involve additional steps and equations depending on the specific properties and behavior of the substances involved.