Question

A 1.75 mole sample of an ideal gas for which \(P=2.50\) bar and \(T=335 \mathrm{K}\) is expanded adiabatically against an external pressure of 0.225 bar until the final pressure is 0.225 bar. Calculate the final temperature, \(q, w, \Delta H\) and \(\Delta U\) for (a) \(C_{V, \text { m }}=3 R / 2,\) and (b) \(C_{V, m}=5 R / 2\)

Short answer

For case (a): Final temperature: \(T_{2a} = 245.30\,\text{K}\) Heat transfer (q): \(q = 0\) Work done (w): \(w_a = -204.5\,\text{J}\) Change in internal energy (\(\Delta U\)): \(\Delta U_a = -204.5\,\text{J}\) Change in enthalpy (\(\Delta H\)): \(\Delta H_a = -340.8\,\text{J}\) For case (b): Final temperature: \(T_{2b} = 260.48\,\text{K}\) Heat transfer (q): \(q = 0\) Work done (w): \(w_b = -204.5\,\text{J}\) Change in internal energy (\(\Delta U\)): \(\Delta U_b = -204.5\,\text{J}\) Change in enthalpy (\(\Delta H\)): \(\Delta H_b = -566.3\,\text{J}\)

Step-by-step solution

Understanding Adiabatic Expansion

In an adiabatic process, there is no exchange of heat with the surroundings (q=0). Therefore, we can write the first law of thermodynamics as ∆U = w. In case of ideal gas, we know that ∆U = n * Cv * ∆T, and for adiabatic process we have the formula: \(PV^{\gamma} = \text{constant}\), where γ = Cp/Cv.

Apply the Adiabatic Process Formula

We are given initial temperature (T1), initial pressure (P1), external pressure (Pext) and final pressure (P2). We need to find the final temperature (T2), ∆U, q, w, and ∆H. Using the adiabatic process formula, we can write: \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\) Since the volume ratio is given by: \(\frac{V_2}{V_1} = \frac{P_1}{P_2}\) We get: \(P_1 V_1 = P_2 V_2\)

Apply the Ideal Gas Law to Calculate V1 and V2

Applying ideal gas law we can calculate the initial and final volume of the gas for both cases: \(PV = nRT\) \(V_1 = \frac{n R T_1}{ P_1}\) and \(V_2 = \frac{n R T_2}{ P_2}\) We also know the moles, temperature and pressure of gas initially, so we can calculate the initial volume V1.

Solve for T2 for both cases

From the known volume ratio and adiabatic process equation, we can derive the relationship between initial and final temperatures: \(\frac{T2}{T1} = (\frac{P2}{P1})^{\frac{γ-1}{γ}}\) Case(a): γ = 1 + 2/3 = 5/3 Case(b): γ = 1 + 3/5 = 8/5 We can find T2 in each case using these values of γ.

Calculate q, w, ΔU, and ΔH

Since the process is adiabatic, q = 0. Next, calculate work done (w) using the formula: \(w = -\int_{V_1}^{V_2} P_{\text{ext}} dV = -P_{\text{ext}} (V_2 - V_1)\) Once we have w, we can find the change in internal energy as ΔU = w. Finally, we calculate the change in enthalpy using the following relation: \(\Delta H = \Delta U + P\Delta V\) where \(\Delta U = n C_{V,m} \Delta T\) and \(P\Delta V = n R \Delta T\) Using these relations, we can calculate the change of enthalpy \(ΔH_{\text {a }}\) and \(ΔH_{\text {b }}\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in the study of gases. It relates the pressure (P), volume (V), temperature (T), and amount of gas (n) through a simple equation:

• PV = nRT

In this equation, R is the universal gas constant, a value that remains consistent across different scenarios involving ideal gases.
In the given exercise, we utilize this law to determine the initial and final volumes of the gas.
By plugging in the initial conditions (pressure, temperature, and number of moles), we can calculate the initial volume (V₁). Similarly, using the final temperature and pressure, we can compute the final volume (V₂). This understanding helps us set the stage for further calculations in thermodynamic processes like adiabatic expansion.

First Law of Thermodynamics

The first law of thermodynamics is a key concept describing the conservation of energy. It states that the total energy in a closed system is constant. In simpler terms, energy cannot be created or destroyed; it can only change forms. This is expressed mathematically as:

• ΔU = q + w

where ΔU represents the change in the internal energy of the system, q is the heat exchanged, and w is the work done.
In an adiabatic process, like the one described in the exercise, there is no heat exchange with the surroundings. This makes q = 0, simplifying the equation to ΔU = w.
Thus, any work done by or on the gas leads to a change in its internal energy, underscoring the first law's role in energy transformation during gas expansion.

Internal Energy

Internal energy is the total energy contained within a system, due to both its molecular motion and its intermolecular forces. In the context of gases, it is primarily affected by temperature changes. For an ideal gas, the change in internal energy (ΔU) is calculated using the expression:

• ΔU = nC_{V, m}ΔT

where n is the number of moles, C_{V, m} is the molar heat capacity at constant volume, and ΔT is the temperature change.
Given that the process in the exercise is adiabatic, the work done by the gas reflects directly as a change in internal energy.
This relationship emphasizes how temperature variations impact the energy dynamics within a gas system as it expands or contracts adiabatically.

Enthalpy Change

Enthalpy change, often symbolized as ΔH, represents the total heat content change in a system under constant pressure. Though this exercise involves adiabatic expansion with no heat exchange, we can still compute ΔH using the relationship:

• ΔH = ΔU + PΔV

Here, ΔU is the change in internal energy, P is the pressure, and ΔV is the change in volume of the gas. The relation for ΔH highlights how changes in internal energy and volume work together to determine the enthalpy change.
For ideal gases, another useful relation emerges as part of this process:

• ΔH = nC_p, mΔT

where C_p, m is the molar heat capacity at constant pressure.
This is applicable because the change in enthalpy can be directly related to the change in temperature, even in adiabatic processes, underscoring the inherent connection between heat content, energy, and pressure-volume work.