Suppose 1.8 Mol Of A Monatomic Ideal Gas Initially At 11 L And 300 K Is Heated At Constant Volume To

Suppose 1.8 mol of a monatomic ideal gas initially at 11 L and 300 K is heated at constant volume to a higher temperature. This scenario encapsulates fundamental principles of thermodynamics, particularly involving ideal gases, heat transfer, and the relationship between temperature, pressure, and volume. Understanding the behavior of gases under such conditions is essential for students and professionals working in physics, chemistry, and engineering fields. In this comprehensive guide, we will explore the various aspects of this process, including calculations of the final temperature, work done, heat absorbed, and the implications on pressure, along with relevant concepts and formulas.

Understanding the Initial Conditions

Initial Volume and Temperature

The initial volume (V₁) of the gas is given as 11 liters, which can be converted into cubic meters for SI unit consistency:

• V₁ = 11 L = 11 x 10^-3 m³

The initial temperature (T₁) is 300 K, a standard room temperature. The amount of gas is 1.8 mol, which informs us about the number of particles involved in the process.

Properties of a Monatomic Ideal Gas

Monatomic gases, such as helium, neon, or argon, have specific heat capacities:

• Molar heat capacity at constant volume, C_v = (3/2) R
• Molar heat capacity at constant pressure, C_p = (5/2) R

where R is the universal gas constant, R ≈ 8.314 J/(mol·K).

Fundamental Thermodynamic Relationships

Ideal Gas Law

The ideal gas law relates pressure (P), volume (V), temperature (T), and the amount of gas (n):

• PV = nRT

Using this law, we can determine initial and final pressures as the temperature changes, assuming constant volume.

Heat Capacity and Internal Energy

The change in internal energy (ΔU) for a monatomic ideal gas depends only on temperature:

• ΔU = n C_v ΔT = n (3/2) R ΔT

Since no work is done at constant volume (W = 0), the heat supplied (Q) equals the change in internal energy.

Calculating the Final Temperature

Suppose the gas is heated at constant volume to a final temperature T₂. The key is to find T₂ based on the heat supplied or other known quantities.

Method 1: Using Heat Input

If the heat added (Q) is known, the temperature change can be directly calculated:

• ΔT = Q / (n C_v)

However, if Q is not specified, we can analyze other aspects, such as the pressure change or the final temperature based on other conditions.

Method 2: Using Pressure Change

Since volume is constant, the pressure is proportional to temperature:

• P₁ / T₁ = P₂ / T₂
• P₂ = P₁ (T₂ / T₁)

If the initial pressure P₁ can be calculated via the ideal gas law:

• P₁ = nRT₁ / V₁

then the final pressure P₂ can be found once T₂ is known.

Calculations Step-by-Step

1. Calculate Initial Pressure (P₁)

Using the ideal gas law:

• P₁ = (n R T₁) / V₁

Plugging in values:

• n = 1.8 mol
• R = 8.314 J/(mol·K)
• T₁ = 300 K
• V₁ = 11 x 10^-3 m³

P₁ = (1.8 mol 8.314 J/(mol·K) 300 K) / (11 x 10^-3 m³) P₁ ≈ (1.8 8.314 300) / 0.011 P₁ ≈ (1.8 2494.2) / 0.011 P₁ ≈ 4489.56 / 0.011 P₁ ≈ 407,233 Pa ≈ 407 kPa

This is the initial pressure inside the container.

2. Determine Final Temperature (T₂)

If the process involves heating without doing work (constant volume), the final temperature T₂ can be derived if the final pressure P₂ is known or if the amount of heat added is specified.

For a typical problem, suppose the gas is heated until the pressure doubles:

• P₂ = 2 P₁ ≈ 814 kPa

Using the proportional relationship:

• T₂ = T₁ (P₂ / P₁) = 300 K 2 = 600 K

Alternatively, if the heat added is given, T₂ can be directly calculated using the internal energy change:

• ΔU = Q = n C_v (T₂ - T₁)

Note: The specific value of T₂ depends on the problem context—whether we're given heat input, final pressure, or temperature.

Work Done and Heat Transfer

Work Done (W)

At constant volume, the work done by the gas is zero:

• W = P ΔV = 0

since ΔV = 0.

Heat Absorbed (Q)

Heat absorbed is related to the change in internal energy:

• Q = ΔU = n C_v (T₂ - T₁)

Calculating C_v:

• C_v = (3/2) R = (3/2) 8.314 ≈ 12.471 J/(mol·K)

Thus:

• Q = 1.8 mol 12.471 J/(mol·K) (T₂ - 300 K)

Using the T₂ determined earlier (e.g., 600 K):

• Q ≈ 1.8 12.471 (600 - 300) = 1.8 12.471 300 ≈ 1.8 3,741.3 ≈ 6,734.3 J

The positive value indicates heat absorption.

Implications and Real-World Applications

Pressure and Temperature Relationship

Understanding how temperature changes influence pressure at constant volume is critical in designing pressurized systems, such as cylinders or reactors. The direct proportionality allows engineers to predict system behavior under heating.

Energy Considerations

The energy input required to reach a certain temperature or pressure can be calculated precisely, aiding in energy efficiency analyses.

Safety Aspects

Knowing the maximum pressure and temperature limits ensures safety in handling gases under various conditions.

Summary and Key Formulas

• Initial Pressure: P₁ = (n R T₁) / V₁
• Final Temperature (assuming pressure doubles): T₂ = T₁ (P₂ / P₁)
• Change in Internal Energy: ΔU = n C_v (T₂ - T₁)
• Heat Added: Q = ΔU
• Work Done at constant volume: W = 0

Understanding these relationships enables precise control and prediction of gas behavior under thermal processes.

Conclusion

Heating a monatomic ideal gas at constant volume involves changes primarily in temperature and pressure, with no work done due to volume constancy. By leveraging the ideal gas law and thermodynamic principles, one can determine the final state variables, energy exchanges, and system behavior. This fundamental knowledge serves as a cornerstone in various scientific and engineering applications, ranging from designing engines and turbines to understanding atmospheric phenomena and laboratory experiments.

Remember: Always consider the initial conditions and the specifics of the heating process to apply the appropriate calculations effectively.