Physics Formulae Thermal Physics Application Of First Law To Ideal Gas Laws

Subset – Definition and Properties

Our guide to the application of the First Law to Ideal Gas Laws helps students understand energy conservation. Learn how...
⚗️

Definition

The application of the First Law of Thermodynamics to ideal gas processes reveals how energy conservation manifests under different thermodynamic constraints. When specific properties are held constant (volume, pressure, or temperature), the first law simplifies to show clear relationships between heat, work, and internal energy changes. For ideal gases, internal energy depends only on temperature. This fundamental relationship, combined with the constraint conditions, allows us to determine exactly how energy flows in each process type. Understanding these applications is crucial for analyzing engines, air conditioning systems, and countless other technologies that rely on controlled thermodynamic processes.

\[ \Delta U = Q - W \]
First Law of Thermodynamics

Physical Properties

The First Law of Thermodynamics, ΔU = Q - W, is a statement of energy conservation. When applied to an ideal gas, its terms (change in internal energy ΔU, heat added Q, and work done by the gas W) are directly related to the gas's state variables like pressure, volume, and temperature.

PropertyDetails
Scalar/Vector NatureAll quantities in the equation (Heat, Work, Internal Energy) are scalar quantities. They have magnitude but no direction.
SI UnitsThe SI unit for all terms in the equation (ΔU, Q, and W) is the Joule (J).
Governing PrincipleThe formula is a direct application of the Law of Conservation of Energy, stating that energy cannot be created or destroyed, only transferred or transformed.
Key Variables<ul><li><strong>Q</strong>: Heat added to the system.</li><li><strong>W</strong>: Work done by the system on its surroundings.</li><li><strong>ΔU</strong>: Change in the internal energy of the system.</li></ul>
Dimensional FormulaAll terms have the dimension of energy, which is [M L^2 T^-2].
State vs. Path FunctionsChange in Internal Energy (ΔU) is a state function, depending only on the initial and final states. Heat (Q) and Work (W) are path functions, depending on the specific process taken between states.
📐

Diagram & Visualization

ΔU = Q - W Q W ΔU
The first law of thermodynamics applied to a gas in a piston: ΔU = Q - W.
🔑

Key Formulas

\[ Q = \Delta U \]
Isochoric Process (Constant Volume)
\[ W = p(V_2 - V_1) \]
Isobaric Process (Constant Pressure Work)
\[ Q = \Delta U + W \]
Isobaric Process (First Law)
\[ Q = W \]
Isothermal Process (Constant Temperature)
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Isothermal Process (Work Done)
\[ \Delta U = nC_V \Delta T \]
Internal Energy Change for an Ideal Gas
🔣

Variables

SymbolQuantitySI UnitDescription
QHeatJoule (J)Heat added to or removed from the system.
WWorkJoule (J)Work done by the system on its surroundings.
ΔUChange in Internal EnergyJoule (J)The change in the total energy contained within the system.
PPressurePascal (Pa)Force per unit area exerted by the gas.
VVolumeCubic meter (m³)The space occupied by the gas.
TAbsolute TemperatureKelvin (K)A measure of the average kinetic energy of the gas particles.
nAmount of substancemole (mol)The number of moles of gas.
RIdeal Gas ConstantJ/(mol·K)A proportionality constant in the ideal gas law.
CᵥMolar specific heat at constant volumeJ/(mol·K)The heat required to raise one mole of gas by 1 K at constant volume.
CₚMolar specific heat at constant pressureJ/(mol·K)The heat required to raise one mole of gas by 1 K at constant pressure.
ΔHChange in EnthalpyJoule (J)The heat absorbed or released in a process at constant pressure.
🔬

Derivation

The derivation for each process starts with the general form of the First Law of Thermodynamics, \( \Delta U = Q - W \), and the definition of internal energy change for an ideal gas, \( \Delta U = nC_V \Delta T \). Each process constraint simplifies these general equations.

Isochoric Process (Constant Volume)

In an isochoric process, the volume of the system is held constant, so \( \Delta V = 0 \). Consequently, the differential volume change \( dV \) is zero.

\[ W = \int P \, dV = 0 \text{ (since dV = 0)} \]
Work Done at Constant Volume

Since no work is done, the First Law simplifies to show that all heat added to the system increases its internal energy.

\[ \Delta U = Q - W = Q - 0 \Rightarrow Q = \Delta U = nC_V \Delta T \]

Isobaric Process (Constant Pressure)

In an isobaric process, the pressure is constant. The work done can be calculated by integrating \( P \, dV \), where P can be taken out of the integral.

\[ W = \int_{V_1}^{V_2} P \, dV = P \int_{V_1}^{V_2} dV = P(V_2 - V_1) = P \Delta V \]
Work Done at Constant Pressure

Substituting this into the First Law, the heat added must account for both the change in internal energy and the work done by the gas.

\[ Q = \Delta U + W = nC_V \Delta T + P \Delta V \]

Isothermal Process (Constant Temperature)

For an isothermal process, the temperature is constant, so \( \Delta T = 0 \). For an ideal gas, internal energy is a function of temperature only, so the change in internal energy is zero.

\[ \Delta U = nC_V \Delta T = 0 \]

The First Law simplifies to show that all heat added is converted into work done by the system.

\[ 0 = Q - W \Rightarrow Q = W \]

To find the work done, we substitute \( P = nRT/V \) from the ideal gas law into the work integral.

\[ W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Work Done at Constant Temperature
📚

Types & Special Cases

The First Law of Thermodynamics simplifies in distinct ways for different thermodynamic processes, which are defined by holding a specific state variable constant.

Type / CaseDescriptionWhen to Use
Isochoric ProcessA process at constant volume (ΔV = 0). No work is done (W = 0), so the first law becomes <strong>ΔU = Q</strong>. All heat added increases the internal energy.When a gas is heated or cooled in a rigid, sealed container.
Isobaric ProcessA process at constant pressure (ΔP = 0). Work is done (W = PΔV), so the first law is <strong>ΔU = Q - PΔV</strong>. Heat added can increase internal energy and/or be used to do work.For systems with a freely moving piston under constant external pressure, such as a cylinder open to the atmosphere.
Isothermal ProcessA process at constant temperature (ΔT = 0). For an ideal gas, internal energy depends only on temperature, so ΔU = 0. The first law becomes <strong>Q = W</strong>. All heat added is converted into work done by the gas.For slow processes where the system is in continuous thermal contact with a large heat reservoir.
Adiabatic ProcessA process with no heat exchange (Q = 0). The first law becomes <strong>ΔU = -W</strong>. Work done by the gas is fueled entirely by its internal energy, causing its temperature to drop.For very rapid processes (e.g., compression in a diesel engine) or for systems that are thermally insulated from their surroundings.
🔥

Worked Example

2 moles of an ideal gas at 300 K are heated at constant volume until the temperature reaches 500 K. Given Cᵥ = 20.8 J/(mol·K), calculate the change in internal energy (ΔU), the heat required (Q), and the work done (W).
  1. Identify the process: Constant volume (isochoric).
  2. Calculate the change in internal energy: \( \Delta U = nC_V \Delta T = 2 \text{ mol} \times 20.8 \frac{\text{J}}{\text{mol·K}} \times (500 \text{ K} - 300 \text{ K}) = 2 \times 20.8 \times 200 = 8320 \text{ J} \).
  3. Determine the work done: For a constant volume process, \( \Delta V = 0 \), so \( W = \int P \, dV = 0 \text{ J} \).
  4. Apply the First Law to find the heat required: For an isochoric process, \( Q = \Delta U \). Therefore, \( Q = 8320 \text{ J} \).
  5. Verify the energy balance: \( \Delta U = Q - W \Rightarrow 8320 \text{ J} = 8320 \text{ J} - 0 \text{ J} \), which is consistent.
The change in internal energy is 8320 J, the heat required is 8320 J, and the work done is 0 J.
🧮

Try It

🏭

Applications

The principles of ideal gas processes are fundamental to the design and analysis of numerous thermal systems and engines.

  • Internal Combustion Engines: The Otto cycle, which models gasoline engines, involves isochoric (constant volume) heat addition and rejection steps. The Diesel cycle involves an isobaric (constant pressure) heat addition step.
  • Gas Turbines & Jet Engines: The Brayton cycle, which models these engines, uses isobaric combustion and heat rejection processes.
  • Refrigeration and Air Conditioning: These systems rely on the Rankine cycle, where refrigerants undergo phase changes (evaporation and condensation) at constant pressure.
  • Steam Power Plants: The Rankine cycle also describes steam power plants, involving isobaric boiling and condensation of water to drive turbines.
  • Industrial Compressors: Gas compression is often modeled as an isothermal or adiabatic process, with interstage cooling designed to approximate isothermal conditions for higher efficiency.
🌍

Real-World Examples

A pressure cooker with a volume of 6 L contains air at 1 atm and 25°C. It is heated on a stove until the pressure reaches 2 atm. Assuming the volume is constant and air is an ideal gas with Cᵥ = 20.8 J/(mol·K), how much heat was added to the air?
  1. Identify the process: Constant volume (isochoric), so W = 0 and Q = ΔU.
  2. Find the initial number of moles using PV=nRT: \( n = \frac{P_1 V}{RT_1} = \frac{(101325 \text{ Pa})(0.006 \text{ m}^3)}{(8.314 \text{ J/mol·K})(298.15 \text{ K})} \approx 0.245 \text{ mol} \).
  3. Find the final temperature using the pressure-temperature relation for an isochoric process: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \Rightarrow T_2 = T_1 \frac{P_2}{P_1} = (298.15 \text{ K}) \frac{2 \text{ atm}}{1 \text{ atm}} = 596.3 \text{ K} \).
  4. Calculate the change in internal energy: \( Q = \Delta U = nC_V \Delta T = (0.245 \text{ mol})(20.8 \text{ J/mol·K})(596.3 \text{ K} - 298.15 \text{ K}) \approx 1523 \text{ J} \).
Approximately 1523 J of heat was added to the air in the pressure cooker.
A hot air balloon expands from 2800 m³ to 2900 m³ at a constant atmospheric pressure of 101 kPa as the air inside is heated. How much work is done by the air in the balloon on the surrounding atmosphere?
  1. Identify the process: Constant pressure (isobaric).
  2. Use the formula for work done in an isobaric process: \( W = P \Delta V \).
  3. Calculate the change in volume: \( \Delta V = V_2 - V_1 = 2900 \text{ m}^3 - 2800 \text{ m}^3 = 100 \text{ m}^3 \).
  4. Calculate the work done: \( W = (101 \times 10^3 \text{ Pa}) (100 \text{ m}^3) = 10.1 \times 10^6 \text{ J} \) or 10.1 MJ.
The air in the balloon does 10.1 MJ of work on the atmosphere.
🏞️

Real-World Scenarios

V=const Q
Pressure Cooker
In a sealed pressure cooker, volume is constant (isochoric). Added heat (Q) directly increases the internal energy (ΔU), raising temperature and pressure.
P=const W Q
Boiling Water
Boiling water in an open pot occurs at constant pressure (isobaric). Heat added (Q) increases internal energy (ΔU) and does work (W) as steam expands.
T=const Q W
Inflating a Tire
When a tire is inflated slowly, temperature remains constant (isothermal). Work (W) done compressing the air dissipates as heat (Q) into the surroundings.

Pressure Cooker: When a sealed pressure cooker is heated, its volume remains essentially constant. The added heat increases the internal energy of the steam and air inside, causing a proportional rise in both temperature and pressure. This is a clear example of an isochoric process where Q = ΔU.

Boiling Water in an Open Pot: As water boils in a pot open to the atmosphere, the steam is produced at a constant pressure (atmospheric pressure). The heat added from the stove not only increases the internal energy of the water to turn it into steam (phase change) but also does work by pushing the surrounding air away. This is an isobaric process.

Slowly Inflating a Tire: When a tire is inflated very slowly, the heat generated by the compression of the air has time to dissipate into the environment, keeping the air's temperature nearly constant. All the work done on the air by the pump is converted into heat that flows out, making this an approximation of an isothermal process where Q ≈ W.

⚠️

Limitations

⚠️ The formulas are derived assuming the substance is an ideal gas. Real gases deviate from ideal behavior, especially at high pressures and low temperatures, due to intermolecular forces and the finite volume of gas molecules. In these cases, ΔU also depends weakly on volume.
⚠️ The processes described (isochoric, isobaric, isothermal) are idealizations. Real-world processes are often quasi-static approximations and may involve simultaneous changes in pressure, volume, and temperature. For example, rapid compression is closer to an adiabatic (no heat exchange) process than an isothermal one.

Common Mistakes

⚠️ Using the wrong heat capacity. Remember to use Cᵥ (constant volume) for calculating ΔU in any ideal gas process and for calculating Q in an isochoric process. Use Cₚ (constant pressure) only for calculating Q in an isobaric process, where Q = nCₚΔT.
⚠️ Incorrectly calculating work. A common error is to use W = PΔV for non-isobaric processes. This formula is only valid when pressure is constant. For isothermal processes, the integral form W = nRT ln(V₂/V₁) must be used, and for isochoric processes, W is always zero.
⚠️ Confusing signs for work and heat. The convention used here is W is work done *by* the system (positive for expansion) and Q is heat added *to* the system (positive for heating). Be consistent. If work is done *on* the system (compression), W is negative.
📏

Units and Dimensions

QuantitySymbolSI UnitDimensional Formula
Energy (Heat, Work, Internal Energy)Q, W, UJoule (J)[M L² T⁻²]
PressurePPascal (Pa = N/m²)[M L⁻¹ T⁻²]
VolumeVCubic meter (m³)[L³]
TemperatureTKelvin (K)[Θ]
Amount of substancenmole (mol)[N]
Ideal Gas ConstantRJoule per mole-kelvin (J/(mol·K))[M L² T⁻² Θ⁻¹ N⁻¹]
Molar Heat CapacityCᵥ, CₚJoule per mole-kelvin (J/(mol·K))[M L² T⁻² Θ⁻¹ N⁻¹]
🎯

Study Strategy

1 🧠 Grasp the Fundamentals
  • Thoroughly review the first law of thermodynamics: ΔU = Q - W, defining each term: internal energy, heat, and work.
  • Read the 'DEFINITION' section, focusing on how the first law simplifies for ideal gases under specific constraints (constant V, P, T).
  • Internalize the key concept for ideal gases: the change in internal energy (ΔU) depends only on the change in temperature (ΔT).
  • Create a concept map of the four main processes: isobaric, isochoric, isothermal, and adiabatic, noting which variable is held constant in each.
2 📝 Commit the Formula to Memory
  • Memorize the universal formula for an ideal gas's internal energy change: ΔU = nCᵥΔT. This applies to ALL processes.
  • Learn the specific formulas for work (W) in different processes, especially W = PΔV for isobaric (constant pressure) processes.
  • Memorize the formulas for heat (Q). For an isochoric process, Q = ΔU = nCᵥΔT. For an isobaric process, Q = nCₚΔT.
  • Create flashcards for each process, listing the constant variable, the simplified first law, and the specific formulas for Q, W, and ΔU.
3 ✍️ Practice with Problems
  • Start with single-step problems for each process type (e.g., find the heat added during an isochoric temperature increase).
  • Carefully study the 'COMMON_MISTAKES' section. Actively check your work to ensure you're using Cᵥ for ΔU and the correct formula for work.
  • Practice calculating work from P-V diagrams. Remember that work is the area under the curve and depends on the path taken.
  • Move on to multi-step problems involving thermodynamic cycles, calculating the net work, heat transfer, and change in internal energy for a full cycle.
4 🌍 Connect to Real-World Physics
  • Read the 'APPLICATIONS' section and explain to a friend how the Otto cycle in a car engine uses an isochoric process for heat addition.
  • Research the Brayton cycle mentioned in 'APPLICATIONS'. Sketch its P-V diagram and identify the isobaric heat addition and rejection steps.
  • Consider an aerosol can getting cold during use. Explain this phenomenon as a nearly adiabatic expansion where the gas does work, decreasing its internal energy.
  • Analyze everyday scenarios. Is pumping a bicycle tire quickly an adiabatic or isothermal process? Justify your answer using the first law.
Master energy conservation in gases by first understanding the core law, then specializing for each process, practicing diligently, and finally seeing it power our world.

Frequently Asked Questions

×

×