Chapter 19: Problem 48
(II) The specific heat at constant volume of a particular gas is 0.182 \(\mathrm{kcal} / \mathrm{kg} \cdot \mathrm{K}\) at room temperature, and its molecular mass is 34 . (a) What is its specific heat at constant pressure? (b) What do you think is the molecular structure of this gas?
Short Answer
Expert verified
(a) \(C_P = 0.24024 \, \text{kcal/kg} \cdot \text{K}\). (b) The gas likely has a linear molecular structure.
Step by step solution
01
Understanding the Problem
We are given the specific heat at constant volume, \(C_V = 0.182 \, \text{kcal/kg} \cdot \text{K}\), and the molecular mass is \(M = 34 \, \text{g/mol}\). We need to find the specific heat at constant pressure \(C_P\) and discuss the molecular structure of the gas.
02
Use of the Molecular Mass to find R (Specific Heat of Gas)
Convert the molecular mass from grams per mole to kilograms per mole: \(M = 0.034 \, \text{kg/mol}\). The universal gas constant is \(R = 8.314 \, \text{J/mol} \cdot \text{K}\) or \(0.00198 \, \text{kcal/mol} \cdot \text{K}\) when converted into kcal. We use this converted \(R\) to calculate \(r = \frac{R}{M} = \frac{0.00198 \text{kcal/mol} \cdot \text{K}}{0.034 \text{kg/mol}} = 0.05824 \, \text{kcal/kg} \cdot \text{K}\).
03
Calculate Specific Heat at Constant Pressure (Cp)
The relationship between the specific heats \(C_P\) and \(C_V\) is given by \(C_P = C_V + r\). Substitute \(C_V = 0.182 \, \text{kcal/kg} \cdot \text{K}\) and \(r = 0.05824 \, \text{kcal/kg} \cdot \text{K}\) to find \(C_P = 0.182 + 0.05824 = 0.24024 \, \text{kcal/kg} \cdot \text{K}\).
04
Determine the Molecular Structure
The ratio \(\gamma = \frac{C_P}{C_V}\) indicates the degrees of freedom of the molecules. Let’s compute \(\gamma = \frac{0.24024}{0.182} \approx 1.32\). This ratio suggests that the gas has 5 degrees of freedom, which is typical for a linear molecular structure.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and temperature. It involves studying how energy moves and transforms within different systems. At its core, thermodynamics involves several key principles:
- Energy cannot be created or destroyed.
- Energy transformation and transfer are central phenomena.
- System properties such as temperature and pressure are critical.
Molecular Structure
Molecular structure describes how atoms are arranged within a molecule. The arrangement impacts the chemical and physical properties of a gas, such as its specific heat capacity. In the context of the problem, we're interested in understanding whether the gas molecules are linear or non-linear. This is determined using the concept of degrees of freedom, influencing how a molecule can store energy:
- Linear molecules typically have 5 degrees of freedom.
- Non-linear molecules usually have 6 degrees.
Degrees of Freedom
Degrees of freedom in thermodynamics and molecular physics refer to the independent ways in which a system can possess energy. For gas molecules, degrees of freedom are vital in determining how energy is divided among its motions. These motions include:
- Translational (movement through space).
- Rotational (spinning around an axis).
- Vibrational (bond stretching and contracting).
Universal Gas Constant
The Universal Gas Constant, \(R\), is a fundamental constant that appears in equations describing the behavior of gases. It relates to the Ideal Gas Law,\(PV = nRT\), connecting pressure (P), volume (V), and temperature (T) for a given amount of gas (n). The value of \(R\) is roughly 8.314 J/mol·K.
The constant's versatility lets it be expressed in different units, including kcal/mol·K, to suit specific problem needs. In the problem, \(R\) is recalibrated to match caloric units, such as converting 8.314 J to 0.00198 kcal for ease of calculation.
This conversion is crucial for finding other properties, like specific heats, which relate via \(C_P = C_V + r\). Here, \(r\) computes from \(R\) and molecular mass, indicating how energy changes influence pressure and volume transformations in gases.
The constant's versatility lets it be expressed in different units, including kcal/mol·K, to suit specific problem needs. In the problem, \(R\) is recalibrated to match caloric units, such as converting 8.314 J to 0.00198 kcal for ease of calculation.
This conversion is crucial for finding other properties, like specific heats, which relate via \(C_P = C_V + r\). Here, \(r\) computes from \(R\) and molecular mass, indicating how energy changes influence pressure and volume transformations in gases.