Question

Calculate \(w\) and \(\Delta E\) when \(1\) mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H\) for the vaporization of the liquid is \(30.7 \mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C}\). Assume the volume of \(1\) mole of liquid is negligible as compared to the volume of \(1\) mole of gas at \(80 .^{\circ} \mathrm{C}\) and \(1.00\) atm.

Short answer

In summary, when 1 mole of a liquid is vaporized at its boiling point of 80°C and 1.00 atm pressure, the work done during vaporization (\(w\)) is -2929 J, and the change in internal energy (\(\Delta E\)) is 27771 J/mol.

Step-by-step solution

Calculate the work done during vaporization

To find the work done (\(w\)) during the vaporization process, we'll use the formula: \[ w = -P_{ext} \Delta V \] where \(P_{ext}\) is the external pressure and \(\Delta V\) is the change in volume. As we're assuming that the volume of 1 mole of the liquid is negligible compared to the volume of 1 mole of gas, and the volume of 1 mole of gas is determined using 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, and T is the temperature in Kelvin. We can rearrange this formula to find the volume (V) of 1 mole of gas: \[ V = \frac{nRT}{P} \] Given the gas constant R = 0.0821 L atm/mol K, we can now plug in the values into the equation to find the volume of 1 mole of gas. \[ V = \frac{(1\,\text{mol})(0.0821\,\text{L atm/mol K})(80 + 273.15\,\text{K})}{1.00\,\text{atm}} \] Now, let's calculate the volume.

Calculate the volume of 1 mole of gas

\[ V = \frac{(1)(0.0821)(353.15)}{1.00} \] \[ V = 28.93\,\text{L} \] Since the volume of 1 mole of liquid is negligible, the change in volume during vaporization is equal to the volume of 1 mole of gas: \[ \Delta V = 28.93\,\text{L} \]

Calculate the work done during vaporization

Now, let's plug the values into the work formula: \[ w = -(1.00\,\text{atm})(28.93\,\text{L}) \] \[ w = -28.93\,\text{L atm} \] Convert the work done to Joules using the conversion factor (1 L atm = 101.3 J): \[ w = -28.93\,\text{L atm} \times \frac{101.3\,\text{J}}{1\,\text{L atm}} \] \[ w = -2929\,\text{J} \] The work done during vaporization is -2929 J.

Calculate the change in internal energy using the first law of thermodynamics

The first law of thermodynamics states that: \[ \Delta E = q + w \] where \(\Delta E\) is the change in internal energy, \(q\) is the heat absorbed, and \(w\) is the work done in the system. We know that the work done \(w\) is -2929 J, and we are given the enthalpy change for vaporization \(\Delta H\) as 30700 J/mol (as 1 kJ = 1000 J). Since we are dealing with a constant pressure process, the heat absorbed (\(q\)) is equal to the enthalpy change: \[ q = \Delta H = 30700\,\text{J/mol} \] Now, let's calculate the change in internal energy:

Calculate the change in internal energy

\[ \Delta E = 30700\,\text{J/mol} - 2929\,\text{J/mol} \] \[ \Delta E = 27771\,\text{J/mol} \] The change in internal energy during the vaporization process is 27771 J/mol. To summarize, the work done during vaporization is -2929 J, and the change in internal energy is 27771 J/mol.

Key concepts

Enthalpy of Vaporization

Enthalpy of vaporization, often denoted as \( \Delta H_{vap} \), is a crucial concept in the study of phase changes, particularly when a substance transitions from its liquid phase to its vapor phase at the boiling point. It represents the amount of energy required to vaporize one mole of a substance at constant pressure without any change in temperature.

In the example provided, the enthalpy of vaporization is given as \(30.7\) kJ/mol at \(80\) degrees Celsius. This value indicates the heat required to convert one mole of the liquid into gas form at its boiling point under the provided conditions. It's important to note that while the enthalpy of vaporization is a measure of the heat absorbed during the process, it differs from the total change in internal energy, which also accounts for work done on or by the system.

To understand this concept fully, students should recognize the use of the enthalpy of vaporization in calculating the heat absorbed during phase transitions, particularly at constant pressure. This ties directly into the first law of thermodynamics, with the heat absorbed being a component of the energy changes within the system.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that the total energy of an isolated system is constant. Energy can be transformed from one form to another, but it cannot be created or destroyed.

In the context of our exercise, the first law can be expressed as \( \Delta E = q + w \), where \( \Delta E \) represents the change in internal energy of the system, \( q \) is the heat added to the system, and \( w \) is the work done by the system against the external environment. When a substance undergoes vaporization, it absorbs heat and potentially does work, expanding against atmospheric pressure.

The detailed solution to the exercise demonstrates the application of the first law by using the given enthalpy of vaporization (which equates to \( q \) at constant pressure) and the calculated work done during the vaporization to determine the change in internal energy of the molten liquid. This approach underlines the intertwined nature of heat and work in thermodynamic processes and the importance of considering both when calculating energy changes.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics which connects the pressure \( P \), volume \( V \), temperature \( T \), and the amount of gas in moles \( n \), along with the ideal gas constant \( R \). The law is usually stated as:

\[ PV = nRT \]
This relationship is critical when studying gases and their behavior under different conditions of pressure, temperature, and volume. In the context of our problem, it allows us to calculate the volume of a gas when it vaporizes from its liquid state. Given that we’re dealing with the vaporization process of one mole of liquid, the change in volume \( \Delta V \) can be accurately predicted using the Ideal Gas Law, assuming that the behavior of the vapor is ideally following the gas laws.

To adhere to the Ideal Gas Law, the solution uses the known values of temperature, the ideal gas constant, and the number of moles of the vapor. It assumes that the volume of the liquid is negligible, simplifying the calculation to determine the volume of gas produced during vaporization, which in turn is used to calculate the work done during the phase change.