Question

During an adiabatic expansion the temperature of 0.450 mol of argon (Ar) drops from 66.0\(^\circ\)C to 10.0\(^\circ\)C. The argon may be treated as an ideal gas. (a) Draw a \(pV\)-diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

Short answer

The gas does 485.4 J of work, and the change in internal energy is -485.4 J.

Step-by-step solution

Convert Temperature to Kelvin

First, convert the initial and final temperatures from Celsius to Kelvin. Use the formula \( T(K) = T(\degree C) + 273.15 \). For the initial temperature, \( T_i = 66 + 273.15 = 339.15\,K \). For the final temperature, \( T_f = 10 + 273.15 = 283.15\,K \).

Draw a pV-Diagram

In an adiabatic process, the pressure \( p \) and volume \( V \) of an ideal gas change without heat exchange with the surroundings. On a \( pV \)-diagram, this process is represented by a curve that slopes downward from left to right, as volume increases and pressure decreases.

Apply the Ideal Gas Law

Use the relation \( PV^\gamma = \text{constant} \) for adiabatic processes, where \( \gamma = C_p / C_v \). For a monoatomic gas like argon, \( \gamma = 5/3 \). This relationship helps in understanding the shape and behavior of the curve in the \( pV \)-diagram.

Calculate Work Done by the Gas

For an adiabatic process, work done \( W \) can be calculated using the formula for reversible adiabatic processes: \( W = \frac{nR}{\gamma - 1} (T_i - T_f) \). Substituting \( n = 0.450 \) mol, \( R = 8.314 \) J/mol K, \( \gamma = 5/3 \), \( T_i = 339.15 \) K, and \( T_f = 283.15 \) K into the equation: \( W = \frac{0.450 \times 8.314}{(5/3) - 1} (339.15 - 283.15) \). Solve this to find \( W = 485.4 \) J.

Calculate the Change in Internal Energy

The change in internal energy \( \Delta U \) for a monatomic ideal gas can be found using the formula \( \Delta U = \frac{3}{2}nR(T_f - T_i) \). Substituting the known values: \( \Delta U = \frac{3}{2} \times 0.450 \times 8.314 \times (283.15 - 339.15) \). Solve this to find \( \Delta U = -485.4 \) J, which matches the work done but opposite in sign, as expected due to energy conservation in adiabatic processes.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics. It is used to relate the pressure, volume, and temperature of a gas, assuming the gas behaves ideally. This means we assume the gas particles have no significant interactions and occupy no volume themselves.

The Ideal Gas Law is formulated as: \( PV = nRT \) where:

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

In adiabatic processes, like the one described in the exercise, there is no heat exchange with the surroundings. Therefore, the product \( PV^\gamma \) remains constant, where \( \gamma = \frac{C_p}{C_v} \) is specific to the type of gas. For monoatomic gases like argon, \( \gamma \) is typically \( \frac{5}{3} \). This constant relationship influences the pressure-volume behavior during the adiabatic expansion or compression.
Understanding the Ideal Gas Law and its application in adiabatic processes allows one to predict how gases respond to changes in temperature, volume, and pressure.

Internal Energy

Internal energy is the total energy contained within a system due to microscopic motions and interactions of its molecules. For ideal gases, internal energy is primarily dependent on the kinetic energy of the molecules since they are assumed to have no potential interactions.

In an adiabatic process, the change in internal energy \( \Delta U \) is directly related to the change in temperature. This is calculated using the formula: \( \Delta U = \frac{3}{2}nR(T_f - T_i) \) for a monoatomic ideal gas. Here, the number of moles \( n \), the ideal gas constant \( R \), and the initial and final temperatures \( T_i \) and \( T_f \) are used as inputs.

As seen in the exercise, the internal energy change \( \Delta U \) for the argon gas was calculated to be \(-485.4 \) J. This implies that the system lost energy, reflecting the decrease in temperature through the adiabatic expansion. No heat was absorbed or emitted, affirming that work done by the system led to its lowering internal energy.

Work Done by Gas

Work done by gas in an adiabatic process depends on the initial and final states of the system. More specifically, it is the energy transferred due to the expansion or compression of the gas without heat exchange with the surroundings.

For a reversible adiabatic process, we compute the work done \( W \) using \( W = \frac{nR}{\gamma - 1} (T_i - T_f) \). This incorporates the number of moles \( n \), the ideal gas constant \( R \), and other specific parameters including the type-specific heat capacity ratio \( \gamma \). For the argon gas in the exercise, this resulted in \( W = 485.4 \) J.

Notice how the work done matches in magnitude with the change in internal energy but is opposite in sign. This highlights the principle of conservation of energy, confirming that, in an adiabatic process, the work done by the gas is equal in magnitude but opposite to the energy change, since no heat is transferred.