Heat Equations for Special Processes & Molar Specific Heats: Videos & Practice Problems

In isobaric and isovolumetric processes, the change in internal energy can be calculated using heat transfer equations tailored for gases. For isobaric processes, use q = nCPmΔT, and for isovolumetric, q = nCVmΔT. The specific heat values depend on whether the gas is monoatomic or diatomic, simplifying calculations. Understanding these concepts is crucial for analyzing thermodynamic systems.

Heat Equations for Isobaric & Isovolumetric Processes

In thermodynamics, understanding isobaric and isovolumetric processes is crucial for calculating changes in internal energy, especially when heat transfer data is not provided. In these scenarios, we can utilize specific heat equations tailored for gases, which differ from those used for solids and liquids due to the absence of mass in kilograms. Instead of the familiar equation q = mcΔT, we employ q = nCΔT, where n represents the number of moles and C is the molar specific heat.

For isobaric processes (constant pressure), the molar specific heat is denoted as C_P, while for isovolumetric processes (constant volume), it is C_V. The values for these specific heats depend on the type of gas involved, categorized as monoatomic or diatomic. For monoatomic gases, C_V = \frac{3}{2}R and C_P = \frac{5}{2}R, while for diatomic gases, C_V = \frac{5}{2}R and C_P = \frac{7}{2}R, where R is the universal gas constant (approximately 8.314 J/(mol·K)).

To find the change in internal energy (ΔE), we can apply the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system: ΔE = q - W. In isovolumetric processes, the work done is zero because there is no change in volume, simplifying our equation to ΔE = q. Thus, we can calculate the change in internal energy directly from the heat transfer equation: ΔE = nC_VΔT.

For example, if we have 3 moles of a monoatomic gas undergoing an isovolumetric process with a temperature change from 300 K to 350 K, we can calculate the change in internal energy. First, we determine C_V = \frac{3}{2}R, and then we find ΔT = 350 K - 300 K = 50 K. Plugging these values into the equation gives us:

ΔE = nC_VΔT = 3 \cdot \frac{3}{2}R \cdot 50

Substituting R = 8.314 J/(mol·K) results in a change in internal energy of approximately 1,871 joules. This value also represents the heat transfer in this isovolumetric process, illustrating the direct relationship between heat transfer and internal energy change when no work is done.

In summary, mastering the application of these equations and understanding the specific heat values for different gases allows for effective calculations of internal energy changes in thermodynamic processes.

How much heat energy is needed to increase the temperature of 5 mol of an ideal diatomic gas by from 273K to 300K if the a) pressure is held constant; b) the volume is held constant?

• A. a) 3928 Jb) 2806 J
• B. a) 3928 Jb) 1683 J
• C. a) 135 Jb) 135 J
• D. a) 3.97 × 10⁴ Jb) 4.36 × 10⁴ J

Answer: (A)

Calculating Temperature Changes

In this scenario, we analyze the behavior of a monoatomic gas undergoing two distinct thermodynamic processes: an isobaric process followed by an isovolumetric process. Initially, the gas is at a temperature of 293 Kelvin. The first process involves adding 2,000 joules of heat at constant pressure, which results in a temperature increase. The second process entails removing the same amount of heat at constant volume, leading to a temperature decrease. Our goal is to determine the final temperature of the gas after these processes.

To visualize these processes, we can represent them on a PV diagram. The transition from point A to point B, where heat is added at constant pressure, appears as a horizontal line (isobaric process). The transition from point B to point C, where heat is removed at constant volume, is depicted as a vertical line moving downward (isovolumetric process).

We denote the initial temperature at point A as \( T_A = 293 \, \text{K} \). The final temperature at point C, \( T_C \), can be expressed as:

\[ T_C = T_A + \Delta T_{AB} + \Delta T_{BC} \]

Here, \( \Delta T_{AB} \) represents the temperature change during the isobaric process, and \( \Delta T_{BC} \) represents the temperature change during the isovolumetric process.

For the isobaric process (A to B), we use the equation:

\[ q = n C_P \Delta T_{AB} \]

Where \( q \) is the heat added, \( n \) is the number of moles, and \( C_P \) is the molar specific heat at constant pressure. Given that \( q = 2000 \, \text{J} \) and \( n = 3 \) moles, we can calculate \( C_P \) for a monoatomic gas as:

\[ C_P = \frac{5}{2} R \]

Substituting \( R = 8.314 \, \text{J/(mol K)} \), we find:

\[ C_P = \frac{5}{2} \times 8.314 = 20.785 \, \text{J/(mol K)} \]

Now, rearranging the equation to solve for \( \Delta T_{AB} \):

\[ \Delta T_{AB} = \frac{q}{n C_P} = \frac{2000}{3 \times 20.785} \approx 32.1 \, \text{K} \]

Next, for the isovolumetric process (B to C), we use the equation:

\[ q = n C_V \Delta T_{BC} \]

Here, \( C_V \) is the molar specific heat at constant volume, calculated as:

\[ C_V = \frac{3}{2} R \]

Substituting \( R \) gives:

\[ C_V = \frac{3}{2} \times 8.314 = 12.471 \, \text{J/(mol K)} \]

Since we are removing heat, \( q \) is negative, so:

\[ \Delta T_{BC} = \frac{-2000}{3 \times 12.471} \approx -53.5 \, \text{K} \]

Now, we can calculate the final temperature:

\[ T_C = 293 + 32.1 - 53.5 \approx 271.6 \, \text{K} \]

This result indicates that the final temperature of the gas after both processes is 271.6 Kelvin. The analysis highlights how the specific heat capacities influence temperature changes during different thermodynamic processes, with the isovolumetric process resulting in a greater temperature change for the same amount of heat transfer compared to the isobaric process.

Here’s what students ask on this topic:

What is the difference between isobaric and isovolumetric processes in thermodynamics?

In thermodynamics, isobaric and isovolumetric processes refer to specific types of processes involving gases. An isobaric process occurs at constant pressure, meaning the pressure of the gas does not change during the process. The heat transfer equation for an isobaric process is given by $q=n{C}_P∆T$, where $n$ is the number of moles, $C_P$ is the molar specific heat at constant pressure, and $∆T$ is the change in temperature. An isovolumetric process, on the other hand, occurs at constant volume, meaning the volume of the gas does not change. The heat transfer equation for an isovolumetric process is $q=n{C}_V∆T$, where $C_V$ is the molar specific heat at constant volume. Understanding these processes is crucial for analyzing thermodynamic systems.

How do you calculate the change in internal energy for an isovolumetric process?

To calculate the change in internal energy for an isovolumetric process, you can use the heat transfer equation tailored for constant volume processes. The equation is $\mathrm{∆E}=q=n{C}_V∆T$, where $\mathrm{∆E}$ is the change in internal energy, $n$ is the number of moles of the gas, $C_V$ is the molar specific heat at constant volume, and $∆T$ is the change in temperature. Since the volume is constant, no work is done by the gas, and the entire heat transfer contributes to the change in internal energy. For example, if you have 3 moles of a monoatomic gas with a temperature change from 300 K to 350 K, you would use $C_V=\frac{3}{2}R$ (where $R$ is the gas constant) to find the change in internal energy.

What are the specific heat values for monoatomic and diatomic gases in isobaric and isovolumetric processes?

The specific heat values for monoatomic and diatomic gases differ based on whether the process is isobaric (constant pressure) or isovolumetric (constant volume). For monoatomic gases, the molar specific heat at constant volume $C_V$ is $\frac{3}{2}R$ and at constant pressure $C_P$ is $\frac{5}{2}R$. For diatomic gases, the molar specific heat at constant volume $C_V$ is $\frac{5}{2}R$ and at constant pressure $C_P$ is $\frac{7}{2}R$. These values simplify calculations as they depend only on the type of gas (monoatomic or diatomic) and not on the specific gas itself.

How do you determine the heat transfer in an isobaric process?

To determine the heat transfer in an isobaric process, you use the heat transfer equation for constant pressure processes: $q=n{C}_P∆T$. Here, $q$ is the heat transfer, $n$ is the number of moles of the gas, $C_P$ is the molar specific heat at constant pressure, and $∆T$ is the change in temperature. For example, if you have 2 moles of a diatomic gas with a temperature change from 250 K to 300 K, you would use $C_P=\frac{7}{2}R$ (where $R$ is the gas constant) to find the heat transfer. Plugging in the values, you get $q=2\times \frac{7}{2}\times 8.314\times 50$, which gives the heat transfer.

Why is the specific heat per mole used instead of specific heat per kilogram for gases?

The specific heat per mole is used instead of specific heat per kilogram for gases because gas problems typically provide the amount of gas in moles rather than mass. The specific heat per mole, also known as molar specific heat, simplifies calculations as it directly relates to the number of moles of gas. The equations for heat transfer in gases are $q=n{C}_P∆T$ for isobaric processes and $q=n{C}_V∆T$ for isovolumetric processes, where $n$ is the number of moles. This approach is more practical and aligns with the way gas quantities are usually measured and provided in thermodynamic problems.