Internal Energy of Gases: Videos & Practice Problems

In thermodynamics, the total internal energy of an ideal gas is calculated using the equation E_internal = 3/2nRT for moles or E_internal_avg = nkT for particles. This energy reflects the microscopic kinetic and potential energies of gas particles, contrasting with macroscopic mechanical energy, which considers the whole object's motion. Understanding these concepts is crucial for analyzing gas behavior and energy transformations.

Internal Energy of Ideal Monoatomic Gases

In the study of ideal gases, understanding the distinction between average kinetic energy and total internal energy is crucial. The average kinetic energy per particle can be calculated using the formula:

\( \text{KE}_{\text{avg}} = \frac{3}{2} k_B T \)

where \( k_B \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \, \text{J/K} \)) and \( T \) is the temperature in Kelvin. For example, at a temperature of 300 K, the average kinetic energy for one particle would be:

\( \text{KE}_{\text{avg}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \, \text{J} \)

To find the total internal energy (\( E_{\text{internal}} \)) of a collection of particles, you multiply the average kinetic energy by the number of particles (\( n \)):

\( E_{\text{internal}} = n \times \text{KE}_{\text{avg}} \)

For instance, if there are 10 particles, the total internal energy would be:

\( E_{\text{internal}} = 10 \times 6.21 \times 10^{-21} = 62.1 \times 10^{-21} \, \text{J} \)

Alternatively, the total internal energy can also be expressed using the number of moles (\( n \)) of the gas with the equation:

\( E_{\text{internal}} = \frac{3}{2} n R T \)

Here, \( R \) is the universal gas constant (\( 8.314 \, \text{J/(mol K)} \)). This form is particularly useful when dealing with moles of gas rather than individual particles. It is important to note that these equations apply specifically to monoatomic gases, which consist of single atoms.

For example, if the total internal energy of a monoatomic gas is given as \( 2 \times 10^4 \, \text{J} \) at a temperature of 401 K, the number of moles can be calculated as follows:

\( n = \frac{E_{\text{internal}}}{\frac{3}{2} R T} = \frac{2 \times 10^4}{\frac{3}{2} \times 8.314 \times 401} \approx 4 \, \text{moles} \)

Understanding the difference between macroscopic and microscopic energy is also essential. Macroscopic energy refers to the total kinetic and potential energy of an entire object, while microscopic energy focuses on the kinetic and potential energy of individual particles within that object. In thermodynamics, total internal energy is a measure of the energy associated with the motion and interactions of gas particles, rather than the energy of the gas container or any external forces acting on it.

A container filled with 2 mol of an ideal, monoatomic gas is has a total internal energy equal to the kinetic energy of a 0.008kg bullet travelling at 700 m/s. What is the temperature of the gas in Kelvin?

• A. 123.2 K
• B. 78.6 K
• C. 235.7 K
• D. 11.2 K

Answer: (B)

Total Internal Energy

In this problem, we are tasked with calculating the total internal energy of an ideal monoatomic gas contained in a tank. The internal energy, denoted as \( E_{\text{internal}} \), can be determined using the equation:

\( E_{\text{internal}} = \frac{3}{2} nRT \)

Here, \( n \) represents the number of moles, \( R \) is the ideal gas constant (approximately \( 8.314 \, \text{J/(mol·K)} \)), and \( T \) is the temperature in Kelvin. To find \( E_{\text{internal}} \), we need to know two of the three variables in the equation. In this case, we are given \( n = 10 \) moles, but we need to determine the temperature \( T \).

Since the temperature is not provided directly, we can use the ideal gas law, expressed as:

\( PV = nRT \)

From this equation, we can solve for temperature \( T \) as follows:

\( T = \frac{PV}{nR} \)

Given the pressure \( P = 0.8 \, \text{atm} \) and volume \( V = 0.3 \, \text{m}^3 \), we first convert the pressure from atmospheres to Pascals using the conversion factor \( 1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa} \):

\( P = 0.8 \, \text{atm} \times 1.01 \times 10^5 \, \text{Pa/atm} = 8.08 \times 10^4 \, \text{Pa} \)

Now, substituting the values into the equation for temperature:

\( T = \frac{(8.08 \times 10^4 \, \text{Pa})(0.3 \, \text{m}^3)}{(10 \, \text{mol})(8.314 \, \text{J/(mol·K)})} \approx 291.6 \, \text{K} \)

With the temperature calculated, we can now substitute \( n \), \( R \), and \( T \) back into the internal energy equation:

\( E_{\text{internal}} = \frac{3}{2} (10)(8.314)(291.6) \approx 3.64 \times 10^4 \, \text{J} \)

Alternatively, we can derive the internal energy directly using the relationship between the ideal gas law and the internal energy equation. By substituting \( nRT \) with \( PV \) in the internal energy equation, we have:

\( E_{\text{internal}} = \frac{3}{2} PV \)

Substituting the known values:

\( E_{\text{internal}} = \frac{3}{2} (8.08 \times 10^4 \, \text{Pa})(0.3 \, \text{m}^3) \approx 3.64 \times 10^4 \, \text{J} \)

Both methods yield the same result, confirming the internal energy of the gas in the tank is approximately \( 3.64 \times 10^4 \, \text{J} \).

Here’s what students ask on this topic:

What is the total internal energy of an ideal gas?

The total internal energy of an ideal gas is the sum of the microscopic kinetic and potential energies of all the gas particles. For a monoatomic ideal gas, it can be calculated using the equation:

$E_{\mathrm{internal}}=\frac{3}{2}nRT$

where $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin. This equation is valid for monoatomic gases and reflects the energy due to the motion of the gas particles.

How do you calculate the average kinetic energy of a gas particle?

The average kinetic energy of a gas particle can be calculated using the equation:

$E_{\mathrm{avg}}=\frac{3}{2}{k}_{B}T$

where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin. This equation shows that the average kinetic energy of a gas particle is directly proportional to the temperature of the gas.

What is the difference between average kinetic energy and total internal energy of a gas?

The average kinetic energy of a gas refers to the energy per individual particle and is given by:

$E_{\mathrm{avg}}=\frac{3}{2}{k}_{B}T$

In contrast, the total internal energy of a gas is the sum of the kinetic energies of all particles in the gas. For a monoatomic ideal gas, it is given by:

$E_{\mathrm{internal}}=\frac{3}{2}nRT$

where $n$ is the number of moles. The total internal energy considers the entire collection of particles, making it a macroscopic property.

How does temperature affect the internal energy of an ideal gas?

The internal energy of an ideal gas is directly proportional to its temperature. For a monoatomic ideal gas, the internal energy is given by:

$E_{\mathrm{internal}}=\frac{3}{2}nRT$

where $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin. As the temperature increases, the internal energy increases proportionally, reflecting the increased kinetic energy of the gas particles.

What is the relationship between internal energy and the number of moles in an ideal gas?

The internal energy of an ideal gas is directly proportional to the number of moles. For a monoatomic ideal gas, the internal energy is given by:

$E_{\mathrm{internal}}=\frac{3}{2}nRT$

where $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin. This means that if the number of moles increases, the internal energy also increases proportionally, assuming constant temperature.