Thermodynamic Laws and Ideal Gas Calculations

Systems, properties and thermodynamic laws

A system is the matter or region selected for an energy balance. A closed system exchanges energy but not mass; an open control volume can exchange both. A property depends on state, while heat and work describe transfers across the boundary. Use the zeroth law to explain thermal equilibrium and temperature measurement. The first law expresses energy conservation: for a closed system, ΔU = Q − W when heat into the system and work done by the system are positive. State the convention before using it.

If a closed cylinder receives 500 kJ of heat and does 180 kJ of work, its internal energy increases by 320 kJ. For a steady boiler control volume, mass-flow enthalpy carries flow work; neglecting kinetic/potential changes and shaft work gives heat input approximately equal to mass flow times the outlet-minus-inlet enthalpy. Do not add flow work again after using enthalpy.

The second law identifies direction and limitations. Heat does not spontaneously flow from a cold body to a hot body without another effect, and a cyclic heat engine cannot convert all heat received from a single reservoir into work. A reversible engine between reservoirs at 600 K and 300 K has maximum efficiency 1 − 300/600 = 50%. A real engine between those same reservoirs is less efficient. The Carnot expression is a limiting result, not a Rankine-cycle efficiency formula. Entropy generation indicates irreversibility; it cannot be negative for an isolated overall system. The third law concerns the approach to absolute zero and the entropy of a perfect crystal.

Ideal Gas Laws & The Characteristic Equation of State

Air, nitrogen, oxygen, and flue gas mixtures behave in close conformity with the ideal gas laws at the moderate pressures and high temperatures typical of power plant draft systems, compressors, and pneumatic controls.

The Empirical Gas Laws

                       SUMMARY OF CLASSICAL GAS LAWS
Boyle's Law       : P₁ · V₁ = P₂ · V₂          (Constant Temperature)
Charles's Law     : V₁ / T₁ = V₂ / T₂          (Constant Pressure)
Gay-Lussac's Law  : P₁ / T₁ = P₂ / T₂          (Constant Volume)
Combined Gas Law  : (P₁ · V₁) / T₁ = (P₂ · V₂) / T₂

1. Boyle's Law (Robert Boyle, 1662): At constant temperature, the volume of a given mass of gas is inversely proportional to its absolute pressure:

$$P_1 V_1 = P_2 V_2 = \text{constant}$$

Example: If an air compressor halves the volume of air isothermally, the absolute pressure doubles. 2. Charles's Law (Jacques Charles, 1787): At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2} = \text{constant}$$

Example: When combustion air is heated across an air preheater, its volume expands, requiring larger ductwork and variable-speed draft fans. 3. Gay-Lussac's Law (Joseph Louis Gay-Lussac, 1802): At constant volume, the absolute pressure of a given mass of gas is directly proportional to its absolute temperature:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2} = \text{constant}$$

Safety Implication: If an isolated compressed nitrogen cylinder or air receiver is exposed to plant fire, its absolute pressure increases in direct proportion to absolute temperature until the vessel ruptures or its relief valve opens. 4. General Combined Gas Law: Combining Boyle's and Charles's laws for a constant mass of gas undergoing simultaneous changes in pressure, volume, and temperature:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

The Characteristic Ideal Gas Equation ($PV = mRT$)

When gas mass ($m$) is considered, the thermodynamic state is defined by the characteristic gas equation:

$$P V = m R T$$

Where:

• $P$ = absolute pressure in kilopascals ($\text{kPa}_{\text{abs}}$)
• $V$ = volume of gas in cubic metres ($\text{m}^3$)
• $m$ = mass of gas in kilograms ($\text{kg}$)
• $R$ = characteristic (specific) gas constant ($[\text{kJ}/(\text{kg}\cdot\text{K})]$ or $[\text{J}/(\text{kg}\cdot\text{K})]$)
• $T$ = absolute temperature in Kelvin ($\text{K} = ^\circ\text{C} + 273.15$)

Characteristic Gas Constant ($R$)

The specific gas constant ($R$) is derived by dividing the Universal Gas Constant ($R_u = 8.31447\text{ kJ/(kmol}\cdot\text{K)}$) by the molar molecular mass ($M$ in $\text{kg/kmol}$) of the gas:

$$R = \frac{R_u}{M}$$

For dry atmospheric air (molecular mass $M \approx 28.97\text{ kg/kmol}$):

$$R_{\text{air}} = \frac{8.31447\text{ kJ/(kmol}\cdot\text{K)}}{28.97\text{ kg/kmol}} = 0.287\text{ kJ/(kg}\cdot\text{K)} = 287\text{ J/(kg}\cdot\text{K)}$$

The SOPEEC Exam Trap: Absolute Units Imperative

CRITICAL EXAM WARNING: The single most common error on power engineering gas calculations is substituting gauge pressure ($P_g$) or Celsius ($T_C$) directly into gas equations. Gas laws are derived strictly from molecular kinetic theory referenced to absolute zero and absolute vacuum.

1. Always Convert Pressure to Absolute:

$$P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atmospheric}}$$

(Unless specified otherwise on the exam, use standard sea-level atmospheric pressure $P_{\text{atm}} = 101.325\text{ kPa}$ or $101.3\text{ kPa}$). 2. Always Convert Temperature to Kelvin:

$$T_K = T_C + 273.15$$

Worked Numerical Examples

Example 1: Mass of Air in a Compressed Air Receiver

Problem: A plant air receiver tank has an internal volumetric capacity of $4.0\text{ m}^3$. The tank pressure gauge indicates $720\text{ kPa}$, and the air temperature inside the vessel is $26^\circ\text{C}$. Assuming atmospheric pressure is $101.325\text{ kPa}$ and the specific gas constant for air is $R = 0.287\text{ kJ/(kg}\cdot\text{K)}$, determine the mass of air stored within the receiver.

Step-by-Step Solution:

1. Convert gauge pressure to absolute pressure:

$$P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} = 720\text{ kPa} + 101.325\text{ kPa} = 821.325\text{ kPa}$$

1. Convert Celsius temperature to Kelvin:

$$T_K = 26^\circ\text{C} + 273.15 = 299.15\text{ K}$$

1. Rearrange $P V = m R T$ to solve for mass ($m$):

$$m = \frac{P V}{R T}$$

1. Substitute the thermodynamic values (noting that $1\text{ kPa} \times 1\text{ m}^3 = 1\text{ kJ}$):

$$m = \frac{(821.325\text{ kPa}) \times (4.0\text{ m}^3)}{(0.287\text{ kJ/(kg}\cdot\text{K)}) \times (299.15\text{ K})}$$ $$m = \frac{3285.3}{85.85605} = 38.27\text{ kg}$$

Result: The air receiver stores $38.27\text{ kg}$ of compressed air.

Example 2: Combined Gas Law Compression Cycle

Problem: An air compressor draws in $15.0\text{ m}^3$ of atmospheric combustion air at $101.3\text{ kPa}_{\text{abs}}$ and $20^\circ\text{C}$. The air is compressed into a storage reservoir until its final volume is $2.2\text{ m}^3$ and its final temperature rises to $165^\circ\text{C}$. Calculate:

1. The final absolute pressure of the compressed air ($P_2$) in $\text{kPa}$.
2. The final gauge pressure displayed on the reservoir pressure gauge.

Step-by-Step Solution:

1. Establish State 1 (Initial Conditions):
   • $P_1 = 101.3\text{ kPa}_{\text{abs}}$
   • $V_1 = 15.0\text{ m}^3$
   • $T_1 = 20^\circ\text{C} + 273.15 = 293.15\text{ K}$
2. Establish State 2 (Final Conditions):
   • $V_2 = 2.2\text{ m}^3$
   • $T_2 = 165^\circ\text{C} + 273.15 = 438.15\text{ K}$
3. Apply the General Combined Gas Law:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

1. Rearrange to solve for $P_2$:

$$P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)$$ $$P_2 = 101.3\text{ kPa} \times \left(\frac{15.0\text{ m}^3}{2.2\text{ m}^3}\right) \times \left(\frac{438.15\text{ K}}{293.15\text{ K}}\right)$$ $$P_2 = 101.3 \times 6.81818 \times 1.494627 = 101.3 \times 10.190641 = 1032.31\text{ kPa}_{\text{abs}}$$

1. Determine final gauge pressure:

$$P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}} = 1032.31\text{ kPa} - 101.3\text{ kPa} = 931.01\text{ kPa}$$

Result: The final absolute pressure is $1032.31\text{ kPa}$ and the gauge reads $931.01\text{ kPa}$.

Source notes

Official scope and technical references checked 2026-10-17: Reference 1; Reference 2. Archived DOE handbooks are used for physical principles, not current Canadian legal requirements.