First Law of Thermodynamics: Videos & Practice Problems

The first law of thermodynamics relates internal energy change (ΔE) to heat (q) and work (w) through the equation ${\mathrm{\Delta E}}^{=}(q-w)$. Here, q represents heat added to the system, while w denotes work done by the system. Understanding the signs of q and w is crucial: adding heat increases internal energy, while work done by the system decreases it. This foundational concept is essential for analyzing thermodynamic processes.

The First Law of Thermodynamics

In physics, the concept of work is defined as the transfer of mechanical energy, while heat represents the transfer of thermal energy. The first law of thermodynamics connects these two ideas through an important equation that relates heat, work, and the internal energy of a system. This foundational principle is expressed as:

$$\Delta E = q - w$$

In this equation, ΔE signifies the change in internal energy of the system, which is often a gas in many problems. The variable q represents the heat added to the system, while w denotes the work done by the system. It is crucial to understand these definitions, as they will guide you in solving thermodynamic problems.

For example, if a gas receives 500 joules of heat and does 200 joules of work on its surroundings, the change in internal energy can be calculated as follows:

$$\Delta E = 500 \, \text{J} - 200 \, \text{J} = 300 \, \text{J}$$

This indicates that the internal energy of the gas has increased by 300 joules due to the heat added, despite the work done on the environment.

Understanding how heat and work affect internal energy is essential. When heat is added to a system, it is considered positive, leading to an increase in internal energy. Conversely, if heat is removed, it is negative, resulting in a decrease in internal energy. Similarly, when a gas expands and does work on its surroundings, the work is positive, which decreases internal energy. If the gas is compressed, the work is negative, leading to an increase in internal energy.

To illustrate this further, consider a scenario where a gas in a movable piston loses 240 joules of heat while expanding from a volume of 1 to 3 cubic meters at a constant pressure of 100 pascals. The work done by the gas can be calculated using the formula:

$$w = P \cdot \Delta V$$

Here, P is the pressure and ΔV is the change in volume. The change in volume is:

$$\Delta V = V_{final} - V_{initial} = 3 \, \text{m}^3 - 1 \, \text{m}^3 = 2 \, \text{m}^3$$

Thus, the work done by the gas is:

$$w = 100 \, \text{Pa} \cdot 2 \, \text{m}^3 = 200 \, \text{J}$$

Substituting these values into the first law equation gives:

$$\Delta E = -240 \, \text{J} - 200 \, \text{J} = -440 \, \text{J}$$

This result indicates that the internal energy of the gas has decreased by 440 joules, reflecting the removal of heat and the work done by the gas during expansion.

In summary, the first law of thermodynamics provides a framework for understanding how energy is conserved and transformed within a system, emphasizing the interplay between heat, work, and internal energy.

A gas in a cylinder held at a constant pressure 1.80×10⁵ Pa expands from a volume of 1.2 m³ to 1.6 m³. The internal energy of the gas decreases from 4.40×10⁵ J to 3×10⁵ J. How much heat was transferred to the gas?

• A. –6.8 × 10⁴ J
• B. 2.12 × 10⁵ J
• C. 4.72 × 10⁵ J
• D. 2.28 × 10⁵ J

Answer: (A)

Alternate Equation of the First Law of Thermodynamics

The first law of thermodynamics is a fundamental principle that describes the relationship between heat, work, and internal energy in a system. It can be expressed in two forms: ΔE = q - w and ΔE = q + w. The key difference between these two equations lies in the definition of work (w). In the first form, work is defined as the work done by the system, while in the second form, it refers to the work done on the system. Understanding this distinction is crucial for correctly applying the first law in various scenarios.

To illustrate this, consider a scenario where a gas is contained in a piston. If 300 joules of heat is added to the gas at a constant pressure of 100 pascals, and the gas is compressed from 5 to 3 cubic meters, we can analyze the work done. The work done by the gas on the piston can be calculated using the formula:

W = P \cdot \Delta V

Here, P is the pressure (100 pascals) and ΔV is the change in volume, which is negative 2 cubic meters (from 5 to 3). Thus, the work done by the gas is:

W = 100 \, \text{Pa} \cdot (-2 \, \text{m}^3) = -200 \, \text{J}

This negative value indicates that work is done by the gas on the surroundings. Conversely, the work done on the gas by the piston is equal in magnitude but opposite in sign, resulting in:

W_{on} = +200 \, \text{J}

Next, to find the change in internal energy (ΔE), we can use both forms of the first law. Using the first form:

ΔE = q - W_{by}

Substituting the values, we have:

ΔE = 300 \, \text{J} - (-200 \, \text{J}) = 300 \, \text{J} + 200 \, \text{J} = 500 \, \text{J}

Using the second form:

ΔE = q + W_{on}

Again substituting the values gives:

ΔE = 300 \, \text{J} + 200 \, \text{J} = 500 \, \text{J}

Both forms yield the same result, confirming the consistency of the first law of thermodynamics. This principle emphasizes that energy is conserved in a closed system, and understanding the signs associated with work is essential for accurate calculations. Whether using the equation with a minus or plus sign for work, the outcome should remain consistent, reflecting the fundamental nature of energy transfer in thermodynamic processes.

The internal energy of a system decreases by 500 J, and 230 J of work is done on the system. What is the heat transfer into or out of this system?

• A. 730 J
• B. 270 J
• C. – 270 J
• D. – 730 J

Answer: (D)

Calculating Work Done on Monoatomic Gas

The internal energy of a monoatomic ideal gas can be expressed using the formula:

$$ U = \frac{3}{2} nRT $$

In this equation, \( n \) represents the number of moles of the gas, \( R \) is the ideal gas constant (approximately 8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin. In a scenario where 1300 joules of heat is added to the gas, causing its temperature to rise from 270 K to 320 K, we can determine the work done by the gas using the first law of thermodynamics.

The first law of thermodynamics states that the change in internal energy (\( \Delta U \)) of a system is equal to the heat added to the system (\( Q \)) minus the work done by the system (\( W \)). This can be mathematically represented as:

$$ \Delta U = Q - W $$

To find the work done by the gas, we can rearrange this equation to:

$$ W = Q - \Delta U $$

First, we need to calculate the change in internal energy. The change in internal energy can be calculated as:

$$ \Delta U = U_{\text{final}} - U_{\text{initial}} = \frac{3}{2} nR T_{\text{final}} - \frac{3}{2} nR T_{\text{initial}} $$

Substituting the values for \( T_{\text{final}} = 320 \, \text{K} \) and \( T_{\text{initial}} = 270 \, \text{K} \), we have:

$$ \Delta U = \frac{3}{2} nR (T_{\text{final}} - T_{\text{initial}}) $$

Now, substituting the known values:

$$ \Delta U = \frac{3}{2} \cdot 1.5 \cdot 8.314 \cdot (320 - 270) $$

Calculating this gives:

$$ \Delta U = 935 \, \text{J} $$

Now that we have the change in internal energy, we can substitute back into the rearranged first law equation:

$$ W = Q - \Delta U $$

Substituting \( Q = 1300 \, \text{J} \) and \( \Delta U = 935 \, \text{J} \):

$$ W = 1300 - 935 $$

Thus, the work done by the gas is:

$$ W = 365 \, \text{J} $$

This calculation illustrates the relationship between heat transfer, internal energy change, and work done in thermodynamic processes, highlighting the application of the first law of thermodynamics in practical scenarios.

Here’s what students ask on this topic:

What is the first law of thermodynamics?

The first law of thermodynamics states that the change in internal energy of a system (ΔE) is equal to the heat added to the system (q) minus the work done by the system (w). Mathematically, it is expressed as:

$\mathrm{\Delta E}=q-w$

This law is a statement of the conservation of energy, indicating that energy can neither be created nor destroyed, only transferred or converted from one form to another. In thermodynamic processes, understanding the signs of q and w is crucial: adding heat increases internal energy, while work done by the system decreases it.

How do you calculate the change in internal energy using the first law of thermodynamics?

To calculate the change in internal energy (ΔE) using the first law of thermodynamics, you use the equation:

$\mathrm{\Delta E}=q-w$

Here, q represents the heat added to the system, and w represents the work done by the system. For example, if 500 joules of heat are added to a gas and the gas does 200 joules of work, the change in internal energy would be:

$\mathrm{\Delta E}=500-200=300$ joules

This means the internal energy of the gas increases by 300 joules.

What is the difference between q-w and q+w in the first law of thermodynamics?

The difference between the forms q-w and q+w in the first law of thermodynamics lies in the definition of work (w). In the equation ΔE = q - w, w represents the work done by the system. In the equation ΔE = q + w, w represents the work done on the system. The choice of form depends on the convention used in the textbook or course. Both forms are correct but require consistent use of the definitions of work.

How does adding or removing heat affect the internal energy of a system?

Adding heat to a system increases its internal energy, while removing heat decreases it. In the first law of thermodynamics, this is represented by the term q in the equation ΔE = q - w. If q is positive (heat added), the internal energy (ΔE) increases. If q is negative (heat removed), the internal energy decreases. For example, adding 300 joules of heat to a system increases its internal energy by 300 joules, while removing 300 joules of heat decreases its internal energy by 300 joules.

What is the significance of the first law of thermodynamics in thermodynamic processes?

The first law of thermodynamics is significant because it establishes the principle of energy conservation in thermodynamic processes. It allows us to analyze how energy is transferred and transformed within a system. By relating internal energy change (ΔE) to heat (q) and work (w), it provides a framework for understanding how different processes, such as heating, cooling, expansion, and compression, affect the energy state of a system. This foundational concept is essential for studying and solving problems in thermodynamics.