Problem 55 - University Physics with Modern Physics

Question

CP A Thermodynamic Process in a Liquid. A chemical engineer is studying the properties of liquid methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) . She uses a steel cylinder with a cross-sectional area of 0.0200 \(\mathrm{m}^{2}\) 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 \(79 \mathrm{I} \mathrm{kg} / \mathrm{m}^{3},\) and the specific heat 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 steel cylinder. Find (a) the increase in volume of the methanol; (b) the mechanical work done by the methanol against the \(3.00 \times 10^{4}\) N force; (c) the amount of heat added to the methanol; (d) the change in internal energy of the methanol. (e) Based on your results, explain whether there is any substantial difference between the specific heats \(c_{p}\) (at constant pressure) and \(c_{V}\) (at constant volume) for methanol under these conditions.

Step-by-Step Solution

Answer

(a) \(4.32 \times 10^{-4} \mathrm{m}^{3}\), (b) \(648 \mathrm{J}\), (c) \(714178.8 \mathrm{J}\), (d) \(713530.8 \mathrm{J}\); \(c_p \approx c_V\).

Step 1: Calculate the Increase in Volume

First, we need to find the increase in volume of methanol due to the temperature increase using the coefficient of volume expansion formula:\[\Delta V = \beta V_0 \Delta T\]where \(\beta = 1.20 \times 10^{-3} \mathrm{K}^{-1}\), \(V_0 = 1.20 \times 10^{-2} \mathrm{m}^{3}\), and \(\Delta T = 50.0 - 20.0 = 30.0^{\circ} \mathrm{C}\). Calculating gives:\[\Delta V = 1.20 \times 10^{-3} \times 1.20 \times 10^{-2} \times 30.0 = 4.32 \times 10^{-4} \mathrm{m}^{3}\]

Step 2: Calculate the Mechanical Work Done

The work done by the methanol can be calculated using the formula:\[W = F \times \Delta h = F \times \frac{\Delta V}{A}\]where \(F = 3.00 \times 10^{4} \mathrm{N}\) and \(A = 0.0200 \mathrm{m}^{2}\). Thus, \(\Delta h = \frac{4.32 \times 10^{-4}}{0.0200} = 0.0216 \mathrm{m}\). Substituting the values gives:\[W = 3.00 \times 10^{4} \times 0.0216 = 648 \mathrm{J}\]

Step 3: Calculate the Heat Added

To find the heat added, we use the formula for heat transfer:\[Q = m c_p \Delta T\]where density \(\rho = 791 \mathrm{kg} / \mathrm{m}^{3}\), \(m = \rho V_0 = 791 \times 1.20 \times 10^{-2} = 9.492 \mathrm{kg}\), and \(c_p = 2.51 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\). Calculating gives:\[Q = 9.492 \times 2.51 \times 10^{3} \times 30.0 = 714178.8 \mathrm{J}\]

Step 4: Calculate the Change in Internal Energy

The change in internal energy can be found using the first law of thermodynamics;\[\Delta U = Q - W\]Substituting from previous steps:\[Q = 714178.8 \mathrm{J}, \quad W = 648 \mathrm{J}\]Thus;\[\Delta U = 714178.8 - 648 = 713530.8 \mathrm{J}\]

Step 5: Differentiate Specific Heats at Constant Pressure and Volume

Since the work done \(W\) is quite small compared to the heat added \(Q\), the specific heat at constant volume \(c_V\) is nearly equal to \(c_p\). Hence, under these conditions, there isn't a substantial difference between \(c_p\) and \(c_V\) for methanol.

Key Concepts

Volume Expansion

When a liquid like methanol is heated, its volume increases. This phenomenon is known as volume expansion. Volume expansion is quantified by the coefficient of volume expansion, denoted as \( \beta \). This coefficient measures how much a unit volume of the substance expands per degree change in temperature.

The formula for calculating the increase in volume \( \Delta V \) is:

• \( \Delta V = \beta V_0 \Delta T \)

Where:

• \( \beta \) is the coefficient of volume expansion,
• \( V_0 \) is the initial volume,
• \( \Delta T \) is the change in temperature.

Using this formula, you can determine the volume increase of methanol when its temperature shifts from 20°C to 50°C.

Understanding volume expansion is crucial in thermodynamics, as it affects how the liquid interacts with its container and the surrounding environment.

Mechanical Work

In thermodynamics, work is done when a force causes a displacement. When methanol expands, it pushes against the load, exerting a force. This work done by methanol is known as mechanical work.

Mechanical work \( W \) is calculated using the formula:

• \( W = F \times \Delta h \)

Where:

• \( F \) is the force exerted by the methanol,
• \( \Delta h \) is the change in height caused by the volume expansion.

If the cross-sectional area \( A \) of the cylinder is known, \( \Delta h \) can be expressed as:

• \( \Delta h = \frac{\Delta V}{A} \)

Through this method, we can determine the mechanical work done by methanol against the force during the expansion process.

Heat Transfer

Heat transfer in thermodynamics involves the movement of thermal energy from a hotter object to a cooler one. When it comes to methanol being heated, the heat is absorbed, causing the molecules in the liquid methanol to move faster and lead to an increase in temperature.

The amount of heat \( Q \) added to methanol can be found using the equation:

• \( Q = m c_p \Delta T \)

Where:

• \( m \) is the mass of the methanol,
• \( c_p \) is the specific heat at constant pressure,
• \( \Delta T \) is the temperature change.

Calculating heat transfer accurately helps in understanding how energy is exchanged within a system, influencing changes in state and even chemical reactions.

Internal Energy

Internal energy is the total energy contained within a system that accounts for both kinetic and potential energies of the particles. When methanol undergoes a temperature change, its internal energy also changes.

According to the first law of thermodynamics, the change in internal energy \( \Delta U \) can be calculated by:

• \( \Delta U = Q - W \)

Where:

• \( Q \) is the heat added to the system,
• \( W \) is the work done by the system.

This principle underlines how energy is conserved within the system, being either transformed or transferred. The change in internal energy helps in understanding how a system behaves under different conditions and how energy is balanced during processes.

Specific Heat

Specific heat is an intrinsic property that defines how much heat energy is needed to change a substance's temperature by a specified amount. Methanol, like all substances, has a specific heat at constant pressure \( c_p \) and at constant volume \( c_v \).

In simple terms, specific heat at constant pressure means the energy required to increase the temperature while the system is allowed to expand, whereas specific heat at constant volume keeps the volume unchanged.

For methanol, the difference between \( c_p \) and \( c_v \) isn't substantial due to the low work done compared to the heat added. Thus, in many practical scenarios, both can be considered nearly equal, simplifying calculations and improving analytical efficiency.