Question

A 2.25 mole sample of an ideal gas with \(C_{V, m}=3 R / 2\) initially at \(310 . \mathrm{K}\) and \(1.25 \times 10^{5}\) Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is \(3.10 \times 10^{6}\) Pa. Calculate the final temperature of the gas. Calculate \(q, w, \Delta U,\) and \(\Delta H\) for this process.

Short answer

The final values for the adiabatic compression process are as follows: - Final temperature (T2) ≈ 462.24 K - Heat transferred (q) = 0 - Work done (w) ≈ -679.36 R - Change in internal energy (∆U) ≈ 679.36 R - Change in enthalpy (∆H) ≈ 1,132.27 R

Step-by-step solution

Calculate final temperature (T2)

For adiabatic compression of an ideal gas we have the relation, \(\frac{T_1^{\gamma-1}}{P_1^{\gamma-1}} = \frac{T_2^{\gamma-1}}{P_2^{\gamma-1}}\) where, γ = Cp, m / Cv, m (ratio of molar heat capacities) Since Cv, m = (3/2)R, we can calculate Cp, m as, Cp, m = Cv, m + R = (3/2)R + R = (5/2)R Now, we can find γ, γ = Cp, m / Cv, m = (5/2)R / (3/2)R = 5/3 Substituting the values of T1, P1, P2, and γ in the adiabatic relation, we can solve for T2: \(\frac{310^{(5/3-1)}}{1.25 \times 10^5^{(5/3-1)}} = \frac{T_2^{(5/3-1)}}{3.10 \times 10^6^{(5/3-1)}}\) Solve for T2: T2 ≈ 462.24 K

Calculate the heat transferred (q)

In an adiabatic process, no heat is exchanged between the system and its surroundings. Therefore, for this adiabatic compression, we have: q = 0

Calculate work done (w)

The work done, w, in an adiabatic process can be calculated using the equation: w = nCv, m (T1 - T2) Substituting the given values, we get: w = 2.25 (3/2 R) (310-462.24) w ≈ -679.36 R

Calculate change in internal energy (∆U)

The change in internal energy can be calculated using the equation: ∆U = nCv, m (T2 - T1) Substituting the given values, we get: ∆U = 2.25 (3/2 R) (462.24 - 310) ∆U ≈ 679.36 R

Calculate the change in enthalpy (∆H)

The change in enthalpy can be calculated using the equation: ∆H = nCp, m (T2 - T1) Substituting the given values, we get: ∆H = 2.25 (5/2 R) (462.24 - 310) ∆H ≈ 1,132.27 R The final values for the process are as follows: - Final temperature (T2) ≈ 462.24 K - Heat transferred (q) = 0 - Work done (w) ≈ -679.36 R - Change in internal energy (∆U) ≈ 679.36 R - Change in enthalpy (∆H) ≈ 1,132.27 R

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in physics and chemistry, describing how an ideal gas behaves under various conditions of pressure, volume, and temperature. It is expressed as: \( PV = nRT \), where \( P \) stands for pressure, \( V \) is volume, \( n \) represents the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
If you are analyzing processes involving gases, this law helps predict how a change in one property affects the others.
However, it is important to remember that the Ideal Gas Law makes some assumptions.

• The gas particles have no volume.
• There are no intermolecular forces.
• Collisions between gas particles are perfectly elastic.

While real gases do not perfectly align with these assumptions, especially under high pressure or low temperature, the Ideal Gas Law provides an excellent approximation in many scenarios. Understanding this law is crucial in physical chemistry and physics, especially for processes like adiabatic compression.

Molar Heat Capacity

Molar Heat Capacity is a property that indicates how much heat energy is needed to change the temperature of a mole of a substance by one Kelvin. For gases, particularly in the context of the Ideal Gas Law and thermodynamic processes, we often deal with two specific types: molar heat capacity at constant volume \( C_{V, m} \) and molar heat capacity at constant pressure \( C_{P, m} \).
Molar heat capacity plays a crucial role in determining changes in internal energy and enthalpy.

• For a monoatomic ideal gas: \( C_{V, m} = \frac{3}{2}R \), \( C_{P, m} = \frac{5}{2}R \).
• The ratio of \( C_{P, m} \) to \( C_{V, m} \) , often denoted as \( \gamma \), is critical for adiabatic processes.

For example, \( \gamma = \frac{C_{P, m}}{C_{V, m}} \), in this exercise is \( \frac{5}{3} \). Understanding molar heat capacities allows us to calculate the work and change in energy in processes like adiabatic compression, where no heat is exchanged with the surroundings.

Work in Thermodynamics

Work in thermodynamics relates to the transfer of energy that occurs when a force is applied over a distance. In the context of gases, and particularly in thermodynamics, work is often associated with volume changes against an external pressure.
During an adiabatic process, like the compression described, work done by or on a gas can be calculated as:

• For reversible processes: \( w = \int PdV \)
• For adiabatic processes involving ideal gases, it's simplified: \( w = nC_{V, m}(T_1 - T_2) \).

Within the scenario of adiabatic compression, the system does work on the gas (compressing it). Because no heat is exchanged in an adiabatic process, the work done directly affects the internal energy and temperature changes of the gas. It is negative when work is done on the gas—as in our exercise—because the gas is being compressed, resulting in an increase in internal energy.

Change in Enthalpy

Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is defined as \( H = U + PV \) where \( U \) is the internal energy, \( P \) stands for pressure, and \( V \) is the volume.
In thermodynamic processes involving gases, like the one in our exercise, enthalpy change \( \Delta H \) describes the heat exchange at constant pressure and can be calculated as:

• \( \Delta H = nC_{P, m}(T_2 - T_1) \)

Understanding enthalpy is vital when analyzing processes under constant pressure. In adiabatic processes, although no heat is transferred, the change in enthalpy still occurs due to work done and change in temperature.
In our example, the gas undergoes compression, resulting in an increase in enthalpy, illustrating the relationship between heat transfer concepts and thermodynamic work, even without direct heat exchange.