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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

Expert verified
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

01

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 鈭哢 = w. In case of ideal gas, we know that 鈭哢 = n * Cv * 鈭員, and for adiabatic process we have the formula: \(PV^{\gamma} = \text{constant}\), where 纬 = Cp/Cv.
02

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), 鈭哢, q, w, and 鈭咹. 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: \(P_1 V_1 = P_2 V_2\)
03

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.
04

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: \(\frac{T2}{T1} = (\frac{P2}{P1})^{\frac{纬-1}{纬}}\) Case(a): 纬 = 1 + 2/3 = 5/3 Case(b): 纬 = 1 + 3/5 = 8/5 We can find T2 in each case using these values of 纬.
05

Calculate q, w, 螖U, and 螖H

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 螖U = 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 \(螖H_{\text {a }}\) and \(螖H_{\text {b }}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (V鈧). Similarly, using the final temperature and pressure, we can compute the final volume (V鈧). 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:
  • 螖U = q + w
where 螖U 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 螖U = 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 (螖U) is calculated using the expression:
  • 螖U = nC_{V, m}螖T
where n is the number of moles, C_{V, m} is the molar heat capacity at constant volume, and 螖T 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 螖H, 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 螖H using the relationship:
  • 螖H = 螖U + P螖V
Here, 螖U is the change in internal energy, P is the pressure, and 螖V is the change in volume of the gas. The relation for 螖H 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:
  • 螖H = nC_p, m螖T
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.

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Most popular questions from this chapter

A 2.50 mole sample of an ideal gas, for which \(C_{V, m}=3 R / 2,\) is subjected to two successive changes in state: (1) From \(25.0^{\circ} \mathrm{C}\) and \(125 \times 10^{3} \mathrm{Pa}\), the gas is expanded isothermally against a constant pressure of \(15.2 \times 10^{3} \mathrm{Pa}\) to twice the initial volume. (2) At the end of the previous process, the gas is cooled at constant volume from \(25.0^{\circ} \mathrm{C}\) to \(-29.0^{\circ} \mathrm{C} .\) Calculate \(q, w, \Delta U,\) and \(\Delta H\) for each of the stages. Also calculate \(q, w, \Delta U,\) and \(\Delta H\) for the complete process.

An ideal gas undergoes an expansion from the initial state described by \(P_{i}, V_{i}, T\) to a final state described by \(P_{f}, V_{f}, T\) in (a) a process at the constant external pressure \(P_{f}\), and (b) in a reversible process. Derive expressions for the largest mass that can be lifted through a height \(h\) in the surroundings in these processes.

The formalism of the Young's modulus is sometimes used to calculate the reversible work involved in extending or compressing an elastic material. Assume a force \(F\) is applied to an elastic rod of cross-sectional area \(A_{0}\) and length \(L_{0}\). As a result of this force the rod changes in length by \(\Delta L\). The Young's modulus \(E\) is defined as $$E=\frac{\text { tensile stress }}{\text { tensile strain }}=\frac{F / A_{0}}{\Delta L / L_{0}}=\frac{F L_{0}}{A_{0} \Delta L}$$ a. Relate \(k\) in Hooke's Law to the Young's modulus expression just given. b. Using your result in part (a) show that the magnitude of the reversible work involved in changing the length \(L_{0}\) of an elastic cylinder of cross- sectional area \(A_{0}\) by \(\Delta L\) is \(w=\frac{1}{2}\left(\frac{\Delta L}{L_{0}}\right)^{2} E A_{0} L_{0}\)

A 1.50 mole sample of an ideal gas at \(28.5^{\circ} \mathrm{C}\) expands isothermally from an initial volume of \(22.5 \mathrm{dm}^{3}\) to a final volume of \(75.5 \mathrm{dm}^{3} .\) Calculate \(w\) for this process (a) for expansion against a constant external pressure of \(1.00 \times 10^{5} \mathrm{Pa}\) and (b) for a reversible expansion.

Calculate \(\Delta H\) and \(\Delta U\) for the transformation of \(2.50 \mathrm{mol}\) of an ideal gas from \(19.0^{\circ} \mathrm{C}\) and 1.00 atm to \(550 .^{\circ} \mathrm{C}\) and \(19.5 \mathrm{atm}\) if \(C_{P, m}=20.9+0.042 \frac{T}{\mathrm{K}}\) in units of \(\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\)

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