Question

For 1.25 mol of an ideal gas, \(P_{\text {external }}=P=\) \(350 . \times 10^{3} \mathrm{Pa} .\) The temperature is changed from \(135^{\circ} \mathrm{C}\) to \(21.2^{\circ} \mathrm{C},\) and \(C_{V, m}=3 R / 2 .\) Calculate \(q, w, \Delta U,\) and \(\Delta H\)

Short answer

In conclusion, the heat transferred (q) during this process is -17,643 J, the work done (w) is 1540 J, the change in internal energy (∆U) is -16,103 J, and the change in enthalpy (∆H) is -17,643 J.

Step-by-step solution

Convert the given temperatures to Kelvin scale

Convert the initial and final temperatures from Celsius to Kelvin by adding 273.15 to each value: \(T_1 = 135^{\circ} \mathrm C + 273.15 = 408.15\mathrm K\) \(T_2 = 21.2^{\circ} \mathrm C + 273.15 = 294.35\mathrm K\)

Use the ideal gas law to find the initial and final volume

Since we have information on pressure (P), temperature (T), and moles (n), we can use the ideal gas law, PV = nRT, to find the initial and final volumes (V1 and V2): \(V_1 = \frac{nRT_1}{P} = \frac{1.25 \times 8.314 \times 408.15}{350 \times 10^3} = 0.0148 \mathrm{m^3}\) Similarly, for final volume V2: \(V_2 = \frac{nRT_2}{P} = \frac{1.25 \times 8.314 \times 294.35}{350 \times 10^3} = 0.0104 \mathrm{m^3}\)

Calculate the work done during the process

During an isobaric process (constant pressure), the work done can be calculated using the formula: \(w = -P\Delta V\) Here, \(\Delta V = V_2 - V_1\) So, the work done is: \(w = -P\Delta V = -350 \times 10^3 \times (0.0104 - 0.0148)\mathrm{J} = 1540 \mathrm{J}\)

Calculate the change in internal energy

Since we are given the molar heat capacity at constant volume (Cv,m), we can calculate the change in internal energy using the formula: \(\Delta U = nC_{V,m} \Delta T\) Here, \(\Delta T = T_2 - T_1\) So, the change in internal energy is: \(\Delta U = 1.25 \times \frac{3}{2}R \times (294.35 - 408.15) = -16,103 \mathrm{J}\)

Calculate the heat transferred during the process

We can use the first law of thermodynamics to find the heat transferred (q). The first law states: \(\Delta U = q + w\) Solving for q, we get: \(q = \Delta U - w = -16,103 - 1,540 = -17,643 \mathrm{J}\)

Calculate the change in enthalpy

In an isobaric process (constant pressure), we can relate enthalpy change (∆H) to heat transfer (q) by: \(\Delta H = q\) So, \(\Delta H = -17,643 \mathrm{J}\) In conclusion, the heat transferred (q) during this process is -17,643 J, the work done (w) is 1540 J, the change in internal energy (∆U) is -16,103 J, and the change in enthalpy (∆H) is -17,643 J.

Key concepts

Isobaric Process

An isobaric process is a thermodynamic process in which the pressure remains constant. This type of process is common in certain gas-related operations or reactions, where the system undergoes a transformation or exchange of heat while maintaining a steady pressure.

During an isobaric process involving an ideal gas, such as the one detailed in the exercise, work done by the gas can be easily calculated. Remembering that work (\(w\)) done during expansion or compression at constant pressure is the product of the external pressure and the change in volume, expressed mathematically as w = -P\Delta V. It's important to note that the sign convention here is that work done by the system is negative, as the system is losing energy.

Understanding this concept is crucial, as it underpins the calculation of work in the given exercise where the volume changes at a constant pressure, signifying an isobaric process.

Molar Heat Capacity

Molar heat capacity of a substance is the amount of heat required to raise the temperature of one mole of the substance by one Kelvin. For an ideal gas, the heat capacity can be specified at constant volume (\(C_{V}\)) or at constant pressure (\(C_{P}\)).

In the exercise, the molar heat capacity at constant volume (\(C_{V,m}\)) is given as \(C_{V,m}=\frac{3R}{2}\)), indicating how much energy is needed to change the temperature of a mole of the gas without changing its volume. This parameter is an intrinsic property of a substance and is vital to calculating the change in internal energy (\(\Delta U\)) of a system when temperature changes occur, as seen in the exercise solution steps.

First Law of Thermodynamics

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. It states that the energy of an isolated system is constant and that energy can be transformed from one form to another, but cannot be created or destroyed.

Mathematically, it is represented as \(\Delta U = q + w\)), where \(\Delta U\)) is the change in internal energy, \(q\) is the heat added to the system, and \(w\) is the work done by the system. In the context of the provided exercise, this fundamental principle allows us to determine the heat transferred during the gas's temperature change by rearranging the equation to solve for \(q\). Thus, comprehending this law is essential for analyzing energy changes within physical and chemical processes.

Enthalpy Change

Enthalpy change, denoted as \(\Delta H\)), is the amount of heat released or absorbed in a system at constant pressure. It's a measure of the total energy change of a system, including both internal energy and the energy associated with pressure-volume work.

In many thermodynamic processes, especially those occurring at constant pressure, the change in enthalpy is equal to the heat transferred to or from the system. This relationship simplifies calculations, as shown in the exercise where \(\Delta H = q\)), under the conditions of an isobaric process. Being able to calculate enthalpy change is important as it helps predict how much energy is required or produced during a reaction or phase transition at constant pressure.