Question

When a gas expands rapidly through a valve, you often feel the valve get colder. This is an adiabatic expansion \((q=0)\). Calculate the decrease in temperature of \(1.0\) mole of ideal gas as it is expanded from \(0.20\) to \(1.00\) liter under the conditions given below. Assume a constant volume molar heat capacity, \(C_{V}\) of \(\frac{3}{2} R\). Note that the energy, \(E\), of an ideal gas depends only on the temperature: It is independent of the volume of the system. A. The expansion is irreversible with an external pressure of 1 atmosphere and an initial temperature of \(300 \mathrm{~K}\), B. The expansion is reversible with an initial temperature of \(300 \mathrm{~K}\). C. Calculate \(\Delta E\) for the changes in state described in parts \(\mathrm{A}\) and \(\mathrm{B}\). D. Assume the expansion is carried out isothermally at \(300 \mathrm{~K}\), rather than adiabatically. Calculate the work done if the expansion is carried out irreversibly with an external pressure of \(1.0\) atmosphere. E. Calculate the work done if the isothermal expansion is carried out reversibly. F. Calculate \(q\) and \(\Delta E\) for the changes in state described in parts \(D\) and \(E\)

Short answer

Temperature decreases by 6.50 K for part A; work done is -81.06 J (irreversible) and -382.83 J (reversible) in isothermal expansion. \( q = 81.06 \, \text{J}, \Delta E = 0 \) for irreversible and \( q = 382.83 \, \text{J} \) for reversible isothermal expansion.

Step-by-step solution

Understand Adiabatic Process

In an adiabatic expansion, the system does not exchange heat with its surroundings, so \( q = 0 \). The change in internal energy (\( \Delta E \)) of the gas is equal to the work done on or by the system (\( \Delta E = -W \)). For an ideal gas, \( E \) only depends on temperature, so \( \Delta E = C_{V} \Delta T \).

Calculate \( \Delta E \) for Adiabatic Irreversible Expansion (Part A)

For irreversible expansion, work done \( W = -P_{ext} \Delta V = -1.0 \, \text{atm} \times (1.00-0.20) \, \text{L} = -0.80 \, \text{L atm} = -0.80 \, \times 101.325 \, \text{J/L atm} = -81.06 \, \text{J} \). Since \( \Delta E = -W \), \( \Delta E = 81.06 \, \text{J} \).

Calculate \( \Delta E \) for Adiabatic Reversible Expansion (Part B)

Reversibly, for ideal gases in adiabatic conditions, \( \Delta E = C_{V} \Delta T \). The molar heat capacity \( C_{V} = \frac{3}{2} R \approx 12.47 \, \text{J/mol K} \). To find \( \Delta T \), solve \( \Delta E = C_{V} \Delta T \); we already determined \( \Delta E = 81.06 \, \text{J} \), giving \( \Delta T = \frac{81.06}{12.47} \approx 6.50 \, \text{K} \).

Isothermal Expansion Work (Part D)

For irreversible isothermal expansion with external pressure, work done \( W = -P_{ext} \Delta V = -1.0 \, \text{atm} \times 0.80 \, \text{L} = -81.06 \, \text{J} \), remaining the same as for adiabatic due to constant external pressure.

Reversible Isothermal Expansion Work (Part E)

For reversible isothermal expansion, work \( W = -nRT \ln \frac{V_f}{V_i} \). Using \( n = 1.0 \, \text{mol}, R = 8.314 \, \text{J/mol K}, T = 300 \, \text{K}, V_i = 0.20 \), \( V_f = 1.00 \), we find \( W = -(1)(8.314)(300) \ln \frac{1.00}{0.20} = -382.83 \text{J} \).

Calculate \( q \) and \( \Delta E \) (Part F)

Since it's isothermal, \( \Delta E = 0 \). For part D, since \( \Delta E = q + W \), \( q = -W \) and thus \( q = 81.06 \, \text{J} \). For part E, \( q = 382.83 \, \text{J} \) because \( W = -q \) (since \( \Delta E = 0 \)).

Key concepts

Ideal Gas Law

An ideal gas is a theoretical gas that consists of many randomly moving point particles. These particles interact through perfectly elastic collisions and don't have intermolecular forces. The ideal gas law is an equation of state for an ideal gas and is expressed as:
\[ PV = nRT \]where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (approximately 8.314 J/mol K), and
• \( T \) is the temperature in Kelvin.

This equation helps us understand how gases behave under different conditions of pressure, temperature, and volume. In the context of adiabatic processes, although the gas follows the ideal gas law, the change in these parameters is often calculated differently due to no heat exchange with the surroundings.

Molar Heat Capacity

Molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin. For gases, it can be measured at constant volume (\( C_V \)) or constant pressure (\( C_P \)). These capacities describe how a substance stores energy as heat.
For this exercise, we're concerned with the molar heat capacity at constant volume, noted as \( C_V \). It is given by the formula:
\[ C_V = \frac{3}{2} R \]where \( R \) is the gas constant. In adiabatic conditions for an ideal gas, the change in internal energy is directly related to the molar heat capacity at constant volume. The formula for change in internal energy \( \Delta E \) becomes:
\[ \Delta E = C_V \Delta T \]where \( \Delta T \) is the change in temperature.

Isothermal Process

An isothermal process is a thermodynamic process where the temperature remains constant. This means any changes in pressure and volume occur without a change in temperature. For an ideal gas undergoing an isothermal process, the internal energy remains constant since it depends solely on temperature.
During isothermal expansion, the work done by the system can be significant since the system remains at a consistent temperature by exchanging heat with the surroundings. Unlike in adiabatic processes, heat is added or removed to maintain constant temperature, making the change in internal energy \( \Delta E \) equal to zero.
The work done in a reversible isothermal process is determined by the formula:
\[ W = -nRT \ln \frac{V_f}{V_i} \]where \( V_i \) and \( V_f \) are the initial and final volumes, respectively. This scenario captures how an engine or refrigerator involves isothermal processes by carefully controlling heat exchange.

Reversible Process

A reversible process is an ideal concept in thermodynamics where all changes occur infinitely slowly, allowing the system to be in thermodynamic equilibrium at every stage. This process is never fully realized in practice, but it provides a baseline for evaluating the efficiency of real processes.
In a reversible process, the work done is maximized because the system’s transformations are done so gradually that it can adapt to any changes in pressure and temperature in the surroundings. For gases, reversible processes often involve careful adjustments to maintain these equilibrium conditions.

• For adiabatic reversible processes, the change in internal energy is straightforward to calculate using molar heat capacity.
• In isothermal reversible processes, the work done is over a continuous range of states, translating into a logarithmic expression for work calculation.

The concept of reversibility also ties into understanding how machines and engines can approach, but never fully achieve, maximum possible efficiency.