Question

Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from 3.20 \(\times\) 10\(^{-2}\) m\(^3\) to 4.50 \(\times\) 10\(^{-2}\) m\(^3\). Calculate (a) the initial and final temperatures of the gas; (b) the amount of work the gas does in expanding; (c) the amount of heat added to the gas; (d) the change in internal energy of the gas.

Short answer

Initial temperature: 324.15 K, final: 455.74 K; work done: 32914.18 J; heat added: 54685.94 J; internal energy change: 21771.76 J.

Step-by-step solution

Understand Given Values

We are given that the number of moles \( n = 3 \), the initial volume \( V_i = 3.20 \times 10^{-2} \text{ m}^3 \), the final volume \( V_f = 4.50 \times 10^{-2} \text{ m}^3 \), and the constant pressure \( P = 2.50 \text{ atm} \). We need to convert pressure to SI units: \( 2.50 \text{ atm} \times 101325 \text{ Pa/atm} = 253312.5 \text{ Pa} \).

Use Ideal Gas Law for Initial Temperature

Use the ideal gas law \( PV = nRT \) to find the initial temperature \( T_i \). Rearrange to \( T_i = \frac{PV_i}{nR} \). Substituting known values, \( T_i = \frac{253312.5 \times 3.20 \times 10^{-2}}{3 \times 8.314} = 324.15 \text{ K} \).

Use Ideal Gas Law for Final Temperature

Apply the same formula for the final temperature \( T_f = \frac{PV_f}{nR} \). Substituting the final volume, \( T_f = \frac{253312.5 \times 4.50 \times 10^{-2}}{3 \times 8.314} = 455.74 \text{ K} \).

Calculate Work Done by Gas

The work done \( W \) by the gas at constant pressure is given by \( W = P(V_f - V_i) \). Substituting the values in, \( W = 253312.5(4.50 \times 10^{-2} - 3.20 \times 10^{-2}) = 32914.18 \text{ J} \).

Calculate Heat Added to Gas

For a monatomic gas and constant pressure, the heat added \( Q \) is given by \( Q = nC_p\Delta T \), where \( \Delta T = T_f - T_i \) and \( C_p = \frac{5}{2}R \). Thus, \( Q = 3 \times \frac{5}{2} \times 8.314 \times (455.74 - 324.15) = 54685.94 \text{ J} \).

Find Change in Internal Energy

The change in internal energy \( \Delta U \) for a monatomic ideal gas is given by \( \Delta U = nC_v\Delta T \), where \( C_v = \frac{3}{2}R \). So, \( \Delta U = 3 \times \frac{3}{2} \times 8.314 \times (455.74 - 324.15) = 21771.76 \text{ J} \).

Final Summary

The initial temperature is 324.15 K, the final temperature is 455.74 K. The work done by the gas is 32914.18 J, the heat added to the gas is 54685.94 J, and the change in internal energy of the gas is 21771.76 J.

Key concepts

Monatomic Gas

Monatomic gases consist of single atoms, typical examples include noble gases like helium, neon, and argon. These gases are considered ideal when they follow the ideal gas law well, which states that pressure, volume, and temperature are related such that \(PV = nRT\), where \(R\) is the ideal gas constant.

These gases are significant in thermodynamics studies due to their simplicity. The absence of bonds between atoms means there are no molecular vibrations or rotations contributing to internal energy. Only the kinetic energy of the gas particles needs to be considered.

• Monatomic gases have specific heat capacities that depend only on translation movements.
• At constant pressure, the molar heat capacity \(C_p\) for a monatomic gas is \(\frac{5}{2}R\).
• At constant volume, \(C_v\) is \(\frac{3}{2}R\).

This simplicity makes them an excellent model for exploring gas behaviors and thermodynamic processes.

Work Done by Gas

When a gas expands, it does work on its surroundings. This concept is crucial in understanding energy transfers in thermodynamics. For an ideal gas expanding at constant pressure, the work done \(W\) can be calculated using:
\[W = P(V_f - V_i)\]
where:

• \(P\) is the constant pressure
• \(V_f\) and \(V_i\) are the final and initial volumes, respectively.

In the exercise, this formula allows us to calculate the work done as the gas expands its volume from 3.20 \(\times\) 10\(^{-2}\) m\(^3\) to 4.50 \(\times\) 10\(^{-2}\) m\(^3\) under a constant pressure.

This work is a form of energy transfer from the gas to its surroundings. Understanding this helps in analyzing the efficiency of engines and other thermodynamic systems.

Internal Energy Change

The internal energy of a gas changes when it receives heat or does work. For monatomic ideal gases, the change in internal energy \(\Delta U\) is directly related to the temperature change and can be calculated by:
\[\Delta U = nC_v\Delta T\]
where:

• \(\Delta T = T_f - T_i\) is the change in temperature
• \(C_v = \frac{3}{2}R\) is the specific heat at constant volume

For the ideal monatomic gas in the exercise, substituting the known values into this equation gives us the internal energy change.
This concept is vital because the sum of energy added as heat and energy done as work equals the change in the internal energy, embodying the first law of thermodynamics, \(\Delta U = Q - W\).

Constant Pressure Expansion

During constant pressure expansion, a gas undergoes a change in volume while maintaining the same pressure. This situation is common in real-world applications like pistons in engines. For such a process, the behavior of the gas can be described by the ideal gas law and aids in answering several types of thermodynamic questions.

In constant pressure processes, the amount of heat \(Q\) added to the system can be calculated from:
\[Q = nC_p\Delta T\]
where:

• \(C_p = \frac{5}{2}R\) is the specific heat at constant pressure
• \(n\) is the number of moles
• \(\Delta T\) is the temperature change

This provides a deeper insight into how energy is distributed between work done by the gas and changes in its internal energy. Understanding constant pressure expansion is pivotal for designing efficient thermodynamic cycles.