Problem 53 - University Physics with Modern Physics

Question

A Thermodymamic Process in a Liquid. A chemical engineer is studying the properties of liquid methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) She uses a stecl cylinder with a cross-sectional. The cylindor is and containing \(1.20 \times 10^{-2} \mathrm{m}^{3}\) of methanol. The cylinder is equipped with a tightly fitting piston that supports a load of \(3.00 \times 10^{4} \mathrm{N}\) . The temperature of the system is increased from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) . For methanol, the coefficient of volume expansion is \(1.20 \times 10^{-3} \mathrm{K}^{-1}\) , the density is 791 \(\mathrm{kg} / \mathrm{m}^{3}\) , and the specific heat capacity at constant pressure is \(c_{p}=2.51 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . You can ignore the expansion of the stecl cylinder. Find (a) the increase in volume of the methanol; (b) the mechanical work done by the methanol against the \(3.00 \times 10^{6} \mathrm{N}\) force; (c) the amount of heat added to the methanol; (d) the change in internal cnergy of the methanol. (e) Based on your results, explain whether there is any substantial difference between the specific heat capacities \(c_{p}\) (at constant pressure) and \(c_{V}\) (at constant volume) for methanol under these conditions.

Step-by-Step Solution

Answer

(a) \( 4.32 \times 10^{-4} \text{ m}^3 \); (b) 12.96 J; (c) 715636.14 J; (d) 715623.18 J; (e) \( c_p \) and \( c_V \) are similar due to small work done.

Step 1: Calculate Increase in Volume

First, use the formula for volume expansion: \( \Delta V = \beta V_0 \Delta T \), where \( \beta = 1.20 \times 10^{-3} \text{ K}^{-1} \) is the coefficient of volume expansion, \( V_0 = 1.20 \times 10^{-2} \text{ m}^3 \) is the initial volume, and \( \Delta T = 50 - 20 = 30 \text{ K} \).\[ \Delta V = 1.20 \times 10^{-3} \times 1.20 \times 10^{-2} \times 30 = 4.32 \times 10^{-4} \text{ m}^3 \].

Step 2: Calculate Mechanical Work Done

The work done by the methanol is given by \( W = P \Delta V \), where the pressure \( P = \frac{F}{A} \) and the force \( F = 3.00 \times 10^{4} \text{ N} \). Since the piston fits tightly, \( W = F \Delta V \).\[ W = 3.00 \times 10^4 \times 4.32 \times 10^{-4} = 12.96 \text{ J} \].

Step 3: Calculate the Amount of Heat Added

Use the formula \( Q = mc_p \Delta T \), where\( m = V_0 \times \text{density} \). Calculate \( m = 1.20 \times 10^{-2} \times 791 = 9.492 \text{ kg} \).Then calculate \( Q: \). \[ Q = 9.492 \times 2.51 \times 10^3 \times 30 = 715636.14 \text{ J} \].

Step 4: Calculate Change in Internal Energy

Use the first law of thermodynamics \( \Delta U = Q - W \), where \( Q = 715636.14 \text{ J} \) and \( W = 12.96 \text{ J} \).\[ \Delta U = 715636.14 - 12.96 = 715623.18 \text{ J} \].

Step 5: Discuss Specific Heat Capacities

The specific heat capacity at constant pressure \( c_p \) is used to find \( Q \). We notice that \( W \) is much smaller than \( Q \), indicating the volume work is quite small compared to energy added. Therefore, under these conditions, \( c_p \) and \( c_V \) are approximately the same, since the minor work does not significantly alter internal energy calculations.

Key Concepts

Volume Expansion

Volume expansion is a critical concept for understanding how liquids like methanol behave when heated. When the temperature of a liquid is increased, its volume typically increases too. This phenomenon is described using the coefficient of volume expansion, denoted by \( \beta \). This coefficient helps us quantify how much a liquid's volume changes per degree change in temperature.

The formula used to calculate the increase in volume is:

• \( \Delta V = \beta V_0 \Delta T \)

Where:

• \( \Delta V \) is the change in volume
• \( \beta \) is the coefficient of volume expansion
• \( V_0 \) is the original volume
• \( \Delta T \) is the change in temperature

In the example of methanol, we see an increase in temperature from 20.0°C to 50.0°C, leading to a volume increase, calculated by substituting the given values into the formula. Understanding this concept is pivotal, as it relates directly to other aspects of thermodynamic processes.

Specific Heat Capacity

Specific heat capacity is an essential property that tells us how much heat is required to change the temperature of a specific mass of a substance by one degree. It is often symbolized by \( c_p \) for constant pressure scenarios. In the context of heating methanol, knowing its specific heat capacity allows us to calculate the amount of heat energy it absorbs.

This property varies between substances and is defined by the formula:

• \( Q = mc_p \Delta T \)

Where:

• \( Q \) is the heat added
• \( m \) is the mass
• \( c_p \) is the specific heat capacity
• \( \Delta T \) is the temperature change

In the exercise, by calculating the mass of methanol contained in the steel cylinder, and using its specific heat capacity, we determine the amount of heat energy required to achieve the temperature increase. This concept is vital for designing energy-efficient systems and understanding heat transfer dynamics.

Internal Energy Change

The internal energy change of a system is central to thermodynamics, describing the system's total energy alterations. When methanol is heated, its internal energy increases as it takes in heat, indicated by the first law of thermodynamics.

This change in internal energy \( \Delta U \) can be calculated by:

• \( \Delta U = Q - W \)

Where:

• \( \Delta U \) is the change in internal energy
• \( Q \) is the heat added to the system
• \( W \) is the work done by the system

In the thermodynamic process discussed, methanol absorbs a substantial amount of heat, leading to a significant internal energy change. By subtracting the small amount of work done by the methanol, we obtain the net change in its internal energy. This concept highlights the balance between heat added, work done, and internal energy in physical processes.

Mechanical Work

Mechanical work within thermodynamics typically involves the energy transferred by the system when it expands or contracts against an external force. For methanol in this scenario, work arises as the liquid pushes against the force of the piston.

The work done by the expanding methanol is given by:

• \( W = P \Delta V \)

Since the piston area is fixed, work can also be expressed as:

• \( W = F \Delta V \)

Where:

• \( W \) is the mechanical work
• \( F \) is the force exerted by the piston or load
• \( \Delta V \) is the change in volume

In this study, the calculated work done by methanol was relatively small compared to the heat energy added, indicating that the mechanical work component does not dramatically affect energy balances when volume changes are minimal.

First Law of Thermodynamics

The first law of thermodynamics serves as a fundamental guideline, expressing the conservation of energy within a system. It states that the energy added to a system equates to the sum of the changes in internal energy and the work done by the system. Here, it supports the calculation of methanol's internal energy change.

The equation governing this principle is:

• \( \Delta U = Q - W \)

This law revolves around key concepts:

• \( \Delta U \), the change in internal energy
• \( Q \), the total heat added to the system
• \( W \), the work done by the system

The first law highlights how systems exchange energy with their surroundings, important for both theoretical studies and practical applications. Understanding this, in the context of methanol, emphasizes the negligible difference between specific heat capacities \( c_p \) and \( c_V \) due to minor work impacts in this reaction.