Question

Calculate \(q, w, \Delta E,\) and \(\Delta H\) for the process in which 88.0 g of nitrous oxide (laughing gas, \(\mathrm{N}_{2} \mathrm{O} )\) is cooled from \(165^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) at a constant pressure of 5.00 \(\mathrm{atm} .\) The molar heat capacity for \(\mathrm{N}_{2} \mathrm{O}(g)\) is 38.7 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot\) mol.

Short answer

The heat transfer (q) for the given process is -8522.0 J, the work done (w) cannot be calculated due to insufficient information, the change in internal energy (ΔE) is -8522.0 J + w, where w is the work done, and the change in enthalpy (ΔH) is -8522.0 J.

Step-by-step solution

1. Convert mass of N2O to moles

To calculate moles of N2O, we need to use the molar mass of N2O. The molar mass of Nitrous Oxide (N2O) is: 28.0134 g/mol (N) + 2 * 15.9994 g/mol (O) = 44.0128 g/mol Given mass = 88.0 g Moles of N2O = Mass (g) / Molar Mass Moles of N2O = 88.0 g / 44.0128 g/mol = 2.0 mol

2. Calculate the change in temperature

ΔT = Tfinal - Tinitial ΔT = 55°C - 165°C = -110°C

3. Calculate the heat transfer (q)

q = n * C_p * ΔT Where n = moles of N2O, and C_p = molar heat capacity q = (2.0 mol) * (38.7 J/mol °C) * (-110 °C) q = -8522.0 J

4. Calculate the work done (w)

At constant pressure, the work done is given as: w = -P * ΔV Since the information on the change in volume is not provided, we can proceed to calculate ΔE without w.

5. Calculate the change in internal energy (ΔE)

ΔE = q + w Without the work (w) value, we can't find the exact value for ΔE. However, we can find the relationship between ΔE and the work (w). ΔE = -8522.0 J + w

6. Calculate the change in enthalpy (ΔH)

At constant pressure, the change in enthalpy (ΔH) can be calculated as: ΔH = q Since the heat transfer (q) is -8522.0 J, ΔH = -8522.0 J

Results

: q = -8522.0 J w = Not enough information ΔE = -8522.0 J + w ΔH = -8522.0 J

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics where energy is exchanged between systems due to a temperature difference. In this exercise, when nitrous oxide is cooled from 165°C to 55°C, it loses heat to the surroundings.
This process highlights the exothermic nature of cooling, as energy is released while the gas transitions to a lower temperature state. Heat transfer is often quantified using the formula:

• \(q = n \cdot C_p \cdot \Delta T\)

Here:

• \(q\) is the heat transfer.
• \(n\) represents the number of moles.
• \(C_p\) is the molar heat capacity.
• \(\Delta T\) refers to the change in temperature.

The negative sign in the calculated heat transfer \(q = -8522.0\ J\) indicates that the system is losing energy, which is expected during a cooling process.
The expression "constituent parts" refers to the factoring of these variables to find the total energy exchanged in the form of heat.

Molar Heat Capacity

Molar heat capacity \(C_p\) is a property that defines the amount of heat needed to raise the temperature of one mole of a substance by one degree Celsius at constant pressure. It is a critical factor in determining the heat transfer in thermodynamic processes, such as the one described in this exercise.
For nitrous oxide (2O), the molar heat capacity given is 38.7 J/mol °C. This unit expresses the energy requirement per mole for each degree of temperature change.
Understanding the molar heat capacity allows us to predict how substances will interact with heat energy. The higher the value, the more energy needed to change the temperature, indicating greater energy storage within the substance for a given temperature change. In this case:

• This specific value informs us how much heat transfer will occur when the temperature of nitrous oxide is altered.
• This parameter is crucial for accurately calculating the heat exchange \(q\) in the equation \(q = n \cdot C_p \cdot \Delta T\).
  As such, molar heat capacity serves as a bridge between molecular properties and macroscopic thermodynamic behavior.

Internal Energy

Internal energy \(\Delta E\) of a system is the sum of kinetic and potential energies of its molecules. It captures all the microscopic energy forms that contribute to the system's overall energy state. When a system undergoes changes like heat transfer or work being done, its internal energy changes.
In this exercise, we're interested in how internal energy changes when nitrous oxide is cooled at constant pressure:

• \(\Delta E = q + w\)

Where:

• \(q\) is the heat transfer.
• \(w\) is the work done by or on the system.

Although work \(w\) is not provided, this expression highlights the interdependency between heat, work, and internal energy.
During such changes, internal energy helps us understand how the system's state evolves. Without precise work data, the internal energy change is indefinite, represented as \(\Delta E = -8522.0\ J + w\), allowing us to recognize that any added or removed work affects the energy change directly.

Enthalpy Change

Enthalpy \(\Delta H\) is a thermodynamic property representing the total heat content of a system. At constant pressure, it reflects not only internal energy changes but also the value associated with work done by the system, usually in pressure-volume work. In essence, it quantifies the energy exchange under constant pressure situations.
For this exercise, the change in enthalpy is computed directly with heat exchange due to constant pressure conditions:

• \(\Delta H = q\)

Here:

• The heat transfer to the environment is \(\Delta H = -8522.0\ J\), indicating an exothermic process as the substance releases energy during cooling.

This relationship underlines the simplicity of calculating enthalpy when pressure is held constant, demonstrating how enthalpy can intuitively correlate with observed energy flow in or out of a system depending on thermal interaction with the surroundings.