Entropy Equations for Special Processes: Videos & Practice Problems

Entropy change can be calculated using different equations depending on the thermodynamic process. For isothermal processes, use Δs = q/T. For phase changes, apply Δs = ml/T. When temperature changes, use Δs = mc⁡ln(T_final / T_initial). In adiabatic processes, Δs = 0. Free expansion involves Δs = nR⁡ln(V_final / V_initial).

Entropy Equations for Special Thermodynamic Processes

Understanding entropy is crucial in thermodynamics, particularly when analyzing different processes. The change in entropy, denoted as ΔS, can be calculated using the equation ΔS = q/T, applicable in isothermal processes where temperature remains constant. However, in scenarios where temperature varies, alternative equations are necessary.

For phase changes, such as water transitioning to ice, the equation remains ΔS = q/T, but q is replaced with the heat of transformation, represented as q = mL, where L is the latent heat. This is because the temperature does not change during a phase transition.

When a substance changes temperature without a phase change, a different approach is required. The equation for this scenario is:

$$\Delta S = mc \ln\left(\frac{T_{final}}{T_{initial}}\right)$$

Here, m is the mass, c is the specific heat capacity, and T represents the temperatures in Kelvin. For example, to calculate the change in entropy for water warming from 20°C to 80°C, convert these temperatures to Kelvin (293 K and 353 K, respectively) and substitute into the equation along with the mass and specific heat (4186 J/kg·K). The result yields a positive change in entropy, indicating an increase in disorder as heat is added.

In adiabatic processes, where no heat is exchanged (q = 0), the change in entropy is zero (ΔS = 0). This indicates that the system remains unchanged in terms of entropy. Conversely, in a free expansion, where a gas expands into a larger volume without external pressure, the change in entropy can be calculated using:

$$\Delta S = nR \ln\left(\frac{V_{final}}{V_{initial}}\right)$$

In this case, n is the number of moles and R is the universal gas constant. The increase in volume leads to greater randomness and thus a positive change in entropy.

For processes occurring at constant volume or pressure (isovolumetric or isobaric), the equations are:

$$\Delta S = nC_V \ln\left(\frac{T_{final}}{T_{initial}}\right)$$

or

$$\Delta S = nC_P \ln\left(\frac{T_{final}}{T_{initial}}\right)$$

where C_V and C_P are the specific heat capacities at constant volume and pressure, respectively.

To summarize, recognizing the type of thermodynamic process is essential for selecting the correct equation to calculate entropy changes. The patterns in these equations often involve mass or moles, a specific heat or gas constant, and a logarithmic ratio of final to initial states, whether they be temperature or volume. This structured approach aids in understanding how energy disperses in various thermodynamic scenarios.

3 moles of an ideal gas are in the left side of an hourglass-shaped container, separated by a thin barrier. The right side is completely empty, but the volume of the left and right sides are equal. The barrier is suddenly removed, and the gas freely expands into the vacuum. What is the change in entropy?

• A. 17.3 J/K
• B. 0 J/K
• C. –17.3 J/K
• D. 49.9 J/K

Answer: (A)

Entropy & Calorimetry

In this problem, we explore the mixing of two bodies of water at different temperatures, a classic calorimetry scenario. We have 195 kilograms of water at 30°C and 5 kilograms of boiling water at 100°C. The goal is to determine the final equilibrium temperature of the mixture and the total change in entropy.

To begin, we convert the temperatures from Celsius to Kelvin: the colder water is at 303 K (30°C) and the hot water is at 373 K (100°C). The total mass of the water mixture is 200 kilograms. The equilibrium temperature can be calculated using the formula:

\[ T_f = \frac{m_c c_w T_c + m_h c_w T_h}{m_c c_w + m_h c_w} \]

Here, \(m_c\) and \(m_h\) are the masses of the cold and hot water, respectively, and \(c_w\) is the specific heat capacity of water, which is approximately 4186 J/(kg·K). Plugging in the values:

\[ T_f = \frac{195 \times 4186 \times 303 + 5 \times 4186 \times 373}{195 \times 4186 + 5 \times 4186} \]

After performing the calculations, the final equilibrium temperature is found to be approximately 304.8 K. This result aligns with expectations, as the final temperature is closer to that of the colder water due to its larger mass.

Next, we calculate the total change in entropy (\(\Delta S\)) for the system. The change in entropy for each body of water can be expressed using the formula:

\[ \Delta S = m c \ln\left(\frac{T_f}{T_i}\right) \]

For the hot water, the change in entropy is:

\[ \Delta S_h = m_h c_w \ln\left(\frac{T_f}{T_h}\right) = 5 \times 4186 \times \ln\left(\frac{304.8}{373}\right) \]

For the cold water, the change in entropy is:

\[ \Delta S_c = m_c c_w \ln\left(\frac{T_f}{T_c}\right) = 195 \times 4186 \times \ln\left(\frac{304.8}{303}\right) \]

Calculating these values yields:

\[ \Delta S_h \approx -4222 \text{ J/K} \]

\[ \Delta S_c \approx 4835 \text{ J/K} \]

Adding these together gives the total change in entropy:

\[ \Delta S_{total} = \Delta S_h + \Delta S_c \approx 609 \text{ J/K} \]

This positive change in entropy confirms the second law of thermodynamics, indicating that the overall entropy of the universe increases during this process. The hot water loses entropy while the cold water gains more, resulting in a net increase in entropy.

Entropy & Ideal Gas Processes

In this analysis of an ideal monoatomic gas undergoing a two-step process, we aim to calculate the total change in entropy, denoted as ΔS_total. This process consists of two segments: from point A to point B and from point B to point C. The total change in entropy can be expressed as the sum of the changes in entropy for each segment, specifically ΔS_total = ΔS_AB + ΔS_BC.

For the first segment (A to B), we identify the nature of the process. It is determined that this process is adiabatic, meaning there is no heat transfer (Q = 0). Consequently, the change in entropy for this segment is ΔS_AB = 0. This simplifies our calculation of total entropy change to just the change from the second segment.

In the second segment (B to C), we observe that this process is isobaric, characterized by constant pressure. To calculate the change in entropy for an isobaric process involving an ideal gas, we use the formula:

ΔS_BC = nC_P ln(T_C/T_B)

where n is the number of moles, C_P is the molar heat capacity at constant pressure, T_C is the final temperature, and T_B is the initial temperature.

To find the temperatures T_C and T_B, we apply the ideal gas law, PV = nRT, rearranging it to solve for temperature:

T = \frac{PV}{nR}

For T_A (which equals T_C since both points lie on the same isotherm), we substitute the known values:

T_A = \frac{P_AV_A}{nR} = \frac{(1.5 \times 10^{5})(0.022)}{(0.8)(8.314)} = 496 \text{ K}

For T_B, we similarly calculate:

T_B = \frac{P_BV_B}{nR} = \frac{(0.5 \times 10^{5})(0.042)}{(0.8)(8.314)} = 316 \text{ K}

With T_C = 496 K and T_B = 316 K, we can now substitute these values into the entropy change equation for the isobaric process:

ΔS_BC = (0.8) \left(\frac{5}{2} \times 8.314\right) \ln\left(\frac{496}{316}\right)

Upon calculating, we find that the total change in entropy for the entire process is:

ΔS_total = ΔS_AB + ΔS_BC = 0 + 7.5 \text{ J/K} = 7.5 \text{ J/K}

This result indicates that the total change in entropy for the gas during this two-step process is 7.5 joules per Kelvin, reflecting the increase in disorder as the gas transitions through the defined states.

Here’s what students ask on this topic:

What is the entropy change equation for an isothermal process?

For an isothermal process, where the temperature remains constant, the change in entropy (ΔS) can be calculated using the equation:

$\mathrm{\Delta S}=\frac{q}{T}$

Here, q is the heat added to the system, and T is the absolute temperature in Kelvin. This equation is straightforward because the temperature does not change during the process.

How do you calculate the change in entropy for a phase change?

During a phase change, the temperature remains constant, so the change in entropy (ΔS) can be calculated using:

$\mathrm{\Delta S}=\frac{\mathrm{ml}}{T}$

Here, m is the mass of the substance, l is the latent heat of the phase change, and T is the absolute temperature in Kelvin. This equation is derived from the fact that the heat added or removed during a phase change is used to change the phase, not the temperature.

What is the entropy change equation for a substance changing temperature?

When a substance changes temperature without a phase change, the change in entropy (ΔS) is given by:

$\mathrm{\Delta S}=\mathrm{mc}ln(\frac{T}{}$_finalT_initial)

Here, m is the mass of the substance, c is the specific heat capacity, T_final is the final temperature, and T_initial is the initial temperature. This equation accounts for the fact that the temperature is not constant during the process.

How is the change in entropy calculated for an adiabatic process?

In an adiabatic process, there is no heat transfer (q = 0). Therefore, the change in entropy (ΔS) is zero:

$\mathrm{\Delta S}=0$

This means that the entropy of the system does not change during an adiabatic process, as there is no exchange of heat with the surroundings.

What is the entropy change equation for a free expansion process?

For a free expansion process, where a gas expands to a larger volume without doing work, the change in entropy (ΔS) is given by:

$\mathrm{\Delta S}=\mathrm{nR}ln(\frac{V}{}$_finalV_initial)

Here, n is the number of moles of the gas, R is the universal gas constant, V_final is the final volume, and V_initial is the initial volume. This equation reflects the increase in randomness and disorder as the gas expands.