Comprehensive Thermodynamics: Energy Conservation, Fluid Statics, and Ideal Gas State Laws

Fundamentals of Thermodynamics and Thermodynamic Systems

• Thermodynamics Definition and Etymology:

  • Thermodynamics is defined as the science of energy.
  • Energy is defined as the ability to cause changes.
  • The word thermodynamics originates from the Greek words Therme (meaning heat) and Dynamis (meaning power).

• Thermodynamic System Classifications:

  • A thermodynamic system is defined as any specified portion of the universe that one wishes to study.
  • The system is separated from the surroundings (the rest of the universe) by a boundary.
  • Open System (Open Steady Flow–Steady State System): A system where both mass and energy cross the system boundary ($\dot{m}_{in}, \dot{m}_{out} \neq 0$ and $\dot{E}_{in}, \dot{E}_{out} \neq 0$). If no specific system type is stated in an engineering problem, the system is assumed by default to be an open system.

• Closed System (Non-Flow System): A system in which mass does not cross the boundary, meaning the mass remains entirely inside the system ($\dot{m}_{in} = \dot{m}_{out} = 0$), while energy is permitted to cross the system boundary ($\dot{E}_{in}, \dot{E}_{out} \neq 0$).

• Isolated System: A system where neither mass nor energy can cross the system boundary ($\dot{m} = 0$ and $\dot{E} = 0$).

• Boundary Classifications:
  • A boundary is the real or imaginary surface separating the system from its environment.
  • Fixed Boundary: A boundary that maintains an invariant spatial position and volume.
  • Moving Boundary: A boundary that changes spatial position or moves with the system (such as an expanding or compressing piston inside a cylinder).

Forms of Stored and Transitory Energy

• Classification of Energy Forms:

  • Stored Energy (Possessed Energy): Energy that is retained and stored within the system; its magnitude is intrinsically linked to the mass and internal condition of the system.
  • Transitory Energy (Energy in Transit / Motion): Energy crossing the boundaries of a system that loses its identity once absorbed or rejected; it is independent of a mass flow stream.

• Potential Energy ($PE$ or $P$):

  • Gravitational potential energy is the energy possessed by a body by virtue of its position or vertical elevation with respect to an arbitrary reference datum plane: $PE = F_g z = \frac{mgz}{k}$

• The change in gravitational potential energy between state 1 and state 2 is given by: $\Delta P = P_2 - P_1 = \frac{mg}{k}(z_2 - z_1)$

  • Variables and constants:
  • $m$: Mass of the body.
  • $g$: Acceleration due to gravity ($g = 9.8066\,\text{m/s}^2$ in SI units; $g = 32.174\,\text{ft/s}^2$ in English units).
  • $z$: Height or elevation above the datum plane.
  • $k$: Dimensional proportionality constant ($k = 1.0\,\frac{\text{kg}\cdot\text{m}}{\text{N}\cdot\text{s}^2}$ in SI units; $k = 32.174\,\frac{\text{lb}_m\cdot\text{ft}}{\text{lb}_f\cdot\text{s}^2}$ in English units).

• Kinetic Energy ($KE$ or $K$):

  • Kinetic energy is the energy or stored capacity for performing work possessed by a moving body by virtue of its velocity and momentum: $KE = \frac{m v^2}{2k}$

• The change in kinetic energy is expressed as: $\Delta K = K_2 - K_1 = \frac{m}{2k}(v_2^2 - v_1^2)$

  • Where $v$ represents velocity or speed relative to an inertial reference frame.

• Internal Energy ($U, u$):

  • Internal energy is the microscopic energy stored within a body or substance by virtue of the activity and configuration of its molecules, translational motions, and the rotational and vibrational movements of atoms within molecules.
  • Specific Internal Energy ($u$): Internal energy per unit mass ($\Delta u = u_2 - u_1$).
  • Total Internal Energy ($U$): Total energy for mass $m$, defined as $U = mu$ ($\Delta U = U_2 - U_1$).
  • Macroscopic Analysis: Approximated for ideal gases as $U = mc_v T$.
  • Microscopic Analysis: Expressed differentially as $du = c_v dT$, where the specific heat at constant volume $c_v$ is a function of absolute temperature $T$.

• Flow Energy or Flow Work ($W_f$):

  • Flow work (also known as flow resistance energy) is the work done in pushing a fluid across an open system boundary against resisting pressure: $W_f = FL = (pA)L = pV = m(pv)$

• The net change in flow work between entrance (1) and discharge (2) is: $\Delta W_f = W_{f2} - W_{f1} = p_2 V_2 - p_1 V_1$

• Enthalpy ($H, h$):

  • Enthalpy (sometimes termed combination energy) is a composite thermodynamic property applicable to all fluids, defined as the sum of internal energy and flow energy: $h = u + pv$$H = mh = U + pV$
  • Differentially for ideal gases: $dh = c_p dT$, where $c_p$ is a function of temperature $T$.

• Other Stored Forms of Energy:

  • Chemical energy (released or absorbed during chemical reactions and bond alterations).
  • Nuclear energy (binding forces among protons and neutrons inside atomic nuclei).
  • Work performed in shearing a fluid.
  • Electromagnetic emanations.
  • Acoustic energy.
  • Energy stored via surface tension.

• Transitory Energy: Heat ($Q$):

  • Heat is energy in transition across a system boundary solely by virtue of a temperature difference between the systems or between a system and its surroundings.
  • Sign Convention: $Q$ is positive ($Q > 0$) when heat is added to the system; $Q$ is negative ($Q < 0$) when heat is rejected or transferred away from the system.
  • Sensible Heat ($Q_s$): The quantity of thermal energy required to alter the temperature of a substance without undergoing a phase change: $Q_s = mc\Delta T$
  • Example: Raising the temperature of $1\,\text{kg}$ of a material by $1\,^\circ\text{C}$ with a specific heat of $5\,\text{kJ}/(\text{kg}\cdot\text{K})$ requires exactly $5\,\text{kJ}$ of sensible heat.

• Latent Heat ($Q_L$): The heat required to alter the phase of a substance at constant temperature: $Q_L = m h_{fg}$

• Where $h_{fg}$ is the latent heat of fusion or latent heat of vaporization.

• Water Phase Transformations (at standard atmospheric pressure):

  • Latent heat of fusion (solid to liquid): Melting requires adding $80\,\text{cal/g}$; freezing requires removing $80\,\text{cal/g}$ ($Q = -80\,\text{cal/g}$).
  • Sensible heating/cooling of liquid water between $0\,^\circ\text{C}$ and $100\,^\circ\text{C}$ requires adding or removing $100\,\text{cal/g}$.
  • Latent heat of vaporization (liquid to gas): Vaporization requires adding $540\,\text{cal/g}$; condensation requires removing $540\,\text{cal/g}$ ($Q = -540\,\text{cal/g}$).

• Transitory Energy: Mechanical Work ($W$):
  • Work is the product of the displacement of a body and the force component aligned with the displacement ($dW = F_x dx = (pA)dL = p\,dV$). Work is transient and exists strictly while force acts through a distance.
  • Sign Convention: Work done by the system is positive ($W > 0$, representing an outflow of energy); work done on the system is negative ($W < 0$, representing an inflow of energy).
  • Nonflow Work ($W_{NF}$ or Boundary Work $W_b$): Work delivered by a closed system during a quasi-equilibrium expansion/compression process: $W_{NF} = \int_1^2 P\,dV$
• Mechanically, nonflow work represents the entire area bounded beneath the process path on the absolute pressure versus volume ($pV$) coordinate plane.

• Steady-Flow Work ($W_{SF}$ or $W_{rev}$): Reversible shaft work produced in an open steady flow system: $W_{SF} = -\int_1^2 V\,dP$
• Geometrically, steady-flow work corresponds to the area projected to the left ("at the back") of the process curve against the pressure axis on the $pV$ diagram.

The Steady Flow Energy Equation (SFEE)

• Characteristics of a Steady Flow System:
  • Mass flow rate across every section is invariant; there is neither accumulation nor diminution of mass within the system ($\frac{dm_{CV}}{dt} = 0$).
  • Energy within the control volume remains constant; there is neither accumulation nor diminution of total energy ($\frac{dE_{CV}}{dt} = 0$).
  • The thermodynamic state and physical properties of the working substance at any specific spatial coordinate remain completely constant with respect to time.

• Steady Flow Energy Equation Derivation:

  • Equating total energy entering to total energy leaving the control volume: $P_1 + K_1 + W_{f1} + U_1 + Q = P_2 + K_2 + W_{f2} + U_2 + W$

• Regrouping in terms of changes: $Q = \Delta P + \Delta K + \Delta W_f + \Delta U + W$

• Combining internal energy and flow work into enthalpy ($H = U + W_f$): $P_1 + K_1 + H_1 + Q = P_2 + K_2 + H_2 + W$$Q = \Delta P + \Delta K + \Delta H + W$

• On a unit mass basis: $\frac{g z_1}{k} + \frac{v_1^2}{2k} + h_1 + q = \frac{g z_2}{k} + \frac{v_2^2}{2k} + h_2 + w$

• Fully Worked Steady-Flow Examples:

  • Example 1: Open System with Unknown Work:
  • Given Parameters: Steady flow process with working substance dropping from $p_1 = 200\,\text{psia}$ to $p_2 = 20\,\text{psia}$; speed rises from $v_1 = 200\,\text{fps}$ to $v_2 = 1000\,\text{fps}$; specific internal energy decreases by $25\,\text{Btu/lb}$ ($\Delta u = -25\,\text{Btu/lb}$); specific volume expands from $v_1 = 1\,\text{ft}^3/\text{lb}$ to $v_2 = 8\,\text{ft}^3/\text{lb}$; heat transfer $Q = 0$. Mass flow is $10\,\text{lb/min}$.
  • Kinetic Energy Computations (Basis $1\,\text{lb}_m$): $K_1 = \frac{v_1^2}{2k} = \frac{(200)^2}{(2)(32.174)(778)} = 0.80\,\text{Btu/lb}_m$$K_2 = \frac{v_2^2}{2k} = \frac{(1000)^2}{(2)(32.174)(778)} = 19.97\,\text{Btu/lb}_m$

• Flow Work Computations: $W_{f1} = p_1 v_1 = \frac{(200)(144)(1)}{778} = 37.02\,\text{Btu/lb}_m$$W_{f2} = p_2 v_2 = \frac{(20)(144)(8)}{778} = 29.61\,\text{Btu/lb}_m$

• Energy Balance: $K_1 + W_{f1} = K_2 + W_{f2} + \Delta u + W$$0.80 + 37.02 = 19.97 + 29.61 - 25 + W$$37.82 = 24.58 + W \implies W = 13.24\,\text{Btu/lb}_m$

• Because $W > 0$, work is done by the substance.

• Power Calculation ($1\,\text{hp} = 42.4\,\text{Btu/min}$): $\text{Power} = \frac{(13.24\,\text{Btu/lb}_m)(10\,\text{lb}_m/\text{min})}{42.4\,\text{Btu}/(\text{min}\cdot\text{hp})} = 3.12\,\text{hp}$

• Example 2: 100-hp Steam Turbine:

  • Given Parameters: Turbine delivers $100\,\text{hp}$. Inlet state: $p_1 = 200\,\text{psia}$, $u_1 = 1163.3\,\text{Btu/lb}$, $v_1 = 2.65\,\text{ft}^3/\text{lb}$, $v_1 = 400\,\text{fps}$. Exhaust state: $p_2 = 1\,\text{psia}$, $u_2 = 925\,\text{Btu/lb}$, $v_2 = 294\,\text{ft}^3/\text{lb}$, $v_2 = 1100\,\text{fps}$. Heat loss: $Q = -10\,\text{Btu/lb}$. Neglect potential energy changes.
  • Kinetic Energy Evaluations: $K_1 = \frac{(400)^2}{(2)(32.174)(778)} = 3.20\,\text{Btu/lb}_m$$K_2 = \frac{(1100)^2}{(2)(32.174)(778)} = 24.17\,\text{Btu/lb}_m$

• Flow Work Evaluations: $W_{f1} = \frac{(200)(144)(2.65)}{778} = 98.10\,\text{Btu/lb}_m$$W_{f2} = \frac{(1)(144)(294)}{778} = 54.42\,\text{Btu/lb}_m$

• Energy Balance for Work per lb: $K_1 + W_{f1} + u_1 + Q = K_2 + W_{f2} + u_2 + W$$3.20 + 98.10 + 1163.3 + (-10) = 24.17 + 54.42 + 925 + W$$1254.60 = 1003.59 + W \implies W = 251\,\text{Btu/lb}_m$

• Steam Flow Rate Determination ($1\,\text{hp}\cdot\text{h} = 2544\,\text{Btu}$): $\dot{m} = \frac{(100\,\text{hp})(2544\,\text{Btu}/(\text{hp}\cdot\text{h}))}{251\,\text{Btu/lb}_m} = 1014\,\text{lb}_m/\text{h}$

• Example 3: Steady-Flow Air Compressor:

  • Given Parameters: Air compressor flows $272\,\text{kg/min}$. Inlet: $p_1 = 99.29\,\text{kPa}$, $v_1 = 0.026\,\text{m}^3/\text{kg}$, $u_1 = 1594\,\text{J/kg}$. Discharge: $p_2 = 689.5\,\text{kPa}$, $v_2 = 0.0051\,\text{m}^3/\text{kg}$, $u_2 = 6241\,\text{J/kg}$. Jacket cooling water extracts $4383\,\text{J/kg}$ ($Q = -4.383\,\text{kJ/kg}$). Kinetic energy change: $\Delta K = +896\,\text{J/kg} = +0.896\,\text{kJ/kg}$.
  • Flow Work: $W_{f1} = p_1 v_1 = (99.29\,\text{kN/m}^2)(0.026\,\text{m}^3/\text{kg}) = 2.582\,\text{kJ/kg}$$W_{f2} = p_2 v_2 = (689.5\,\text{kN/m}^2)(0.0051\,\text{m}^3/\text{kg}) = 3.516\,\text{kJ/kg}$

• Energy Balance: $W_{f1} + u_1 + Q = \Delta K + W_{f2} + u_2 + W$$2.582 + 1.594 - 4.383 = 0.896 + 3.516 + 6.241 + W$$-0.207 = 10.653 + W \implies W = -10.86\,\text{kJ/kg}$

• Total Shaft Work Rate: $W = (-10.86\,\text{kJ/kg})(272\,\text{kg/min}) = -2954\,\text{kJ/min}$

• Example 4: Steady-Flow Centrifugal Water Pump:

  • Given Parameters: Water delivery rate $\dot{m} = 2270\,\text{kg/min}$. Inlet pressure $p_1 = 82740\,\text{Pa}$, discharge pressure $p_2 = 275800\,\text{Pa}$. Inlet pipe diameter $d_1 = 15.24\,\text{cm} = 0.1524\,\text{m}$, discharge pipe diameter $d_2 = 10.16\,\text{cm} = 0.1016\,\text{m}$. Fluid density $\rho = 1000\,\text{kg/m}^3$.

• Cross-Sectional Areas: $A_1 = \frac{\pi}{4}(0.1524)^2 = 0.01824\,\text{m}^2$$A_2 = \frac{\pi}{4}(0.1016)^2 = 0.008107\,\text{m}^2$

• Velocities: $v_1 = \frac{\dot{m}}{\rho_1 A_1} = \frac{2270/60}{(1000)(0.01824)} = 2.074\,\text{m/s}$$v_2 = \frac{\dot{m}}{\rho_2 A_2} = \frac{2270/60}{(1000)(0.008107)} = 4.667\,\text{m/s}$

• Kinetic Energy and Flow Work per Unit Mass ($1\,\text{kg}_m$ basis): $K_1 = \frac{v_1^2}{2k} = \frac{(2.074)^2}{(2)(1)} = 2.151\,\text{N}\cdot\text{m/kg}_m$$K_2 = \frac{v_2^2}{2k} = \frac{(4.667)^2}{(2)(1)} = 10.89\,\text{N}\cdot\text{m/kg}_m$$W_{f1} = \frac{p_1}{\rho_1} = \frac{82740}{1000} = 82.74\,\text{N}\cdot\text{m/kg}_m$$W_{f2} = \frac{p_2}{\rho_2} = \frac{275800}{1000} = 275.8\,\text{N}\cdot\text{m/kg}_m$

• Pump Work Calculation: $K_1 + W_{f1} = K_2 + W_{f2} + W$$2.151 + 82.74 = 10.89 + 275.8 + W$$84.891 = 286.69 + W \implies W = -201.8\,\text{N}\cdot\text{m/kg}_m$

• Total Work Per Minute: $W = (-201.8\,\text{N}\cdot\text{m/kg}_m)(2270\,\text{kg/min}) = -458086\,\text{J/min} = -458.1\,\text{kJ/min}$

• Example 5: Steady-Flow Steam Turbine with Elevation Change:

  • Given Parameters: Steam enters at $p_1 = 1200\,\text{kPa}$, $T_1 = 188\,^\circ\text{C}$, $h_1 = 2785\,\text{kJ/kg}$, $v_1 = 33.3\,\text{m/s}$, $z_1 = 3\,\text{m}$. Leaves at $p_2 = 20\,\text{kPa}$, $h_2 = 2512\,\text{kJ/kg}$, $v_2 = 100\,\text{m/s}$, $z_2 = 0\,\text{m}$. Heat loss to surroundings is $\dot{Q} = -0.29\,\text{kJ/s}$. Steam mass flow rate is $\dot{m} = 0.42\,\text{kg/s}$.
  • Energy Term Evaluations per kg: $P_1 = \frac{g z_1}{k} = \frac{(9.8066\,\text{m/s}^2)(3\,\text{m})}{(1\,\text{kg}\cdot\text{m}/(\text{N}\cdot\text{s}^2))(1000\,\text{J/kJ})} = 0.0294\,\text{kJ/kg}$$K_1 = \frac{v_1^2}{2k} = \frac{(33.3)^2}{2(1)(1000)} = 0.5544\,\text{kJ/kg}$$K_2 = \frac{v_2^2}{2k} = \frac{(100)^2}{2(1)(1000)} = 5.000\,\text{kJ/kg}$$q = \frac{\dot{Q}}{\dot{m}} = \frac{-0.29\,\text{kJ/s}}{0.42\,\text{kg/s}} = -0.6905\,\text{kJ/kg}$

• Energy Balance: $P_1 + K_1 + h_1 + q = K_2 + h_2 + w$$0.0294 + 0.5544 + 2785 + (-0.6905) = 5.000 + 2512 + w$$2784.8933 = 2517.000 + w \implies w = 267.893\,\text{kJ/kg}$

• Power Output in kW: $\text{Power} = w \dot{m} = (267.9\,\text{kJ/kg})(0.42\,\text{kg/s}) = 112.52\,\text{kW}$

• Comprehensive Practice and Review Problems:

  • Problem 1: A $3220\,\text{lb}$ body falls $778\,\text{ft}$ from rest under zero heat transfer and zero friction. Applying the steady flow equation deleting irrelevant energy terms gives a terminal speed of:
    $\text{Answer: } 224\,\text{fps}$
  • Problem 2: A reciprocating compressor draws in $500\,\text{ft}^3/\text{min}$ of air with density $\rho_1 = 0.079\,\text{lb/ft}^3$ at $p_1 = 15\,\text{psia}$ and discharges it at $\rho_2 = 0.304\,\text{lb/ft}^3$ and $p_2 = 80\,\text{psia}$. The specific internal energy increases by $33.8\,\text{Btu/lb}$ and cooling heat rejection is $13\,\text{Btu/lb}$. Neglecting changes in kinetic energy, the work on the air is:
    $\text{Answer: } 56.25\,\text{hp}$
  • Problem 3: Steam enters a turbine with $h_1 = 1292\,\text{Btu/lb}$ and exhausts with $h_2 = 1098\,\text{Btu/lb}$. Heat loss is $13\,\text{Btu/lb}$. For a steam flow of $2\,\text{lb/sec}$, the resulting work output is:
    $\text{Answer: } 512.3\,\text{hp}$
  • Problem 4: A steady flow system receives $4.56\,\text{kg/min}$ of fluid at $p_1 = 137.90\,\text{kPa}$, $v_1 = 0.0388\,\text{m}^3/\text{kg}$, $v_1 = 122\,\text{m/s}$, and $u_1 = 17.16\,\text{kJ/kg}$. It discharges at $p_2 = 551.6\,\text{kPa}$, $v_2 = 0.193\,\text{m}^3/\text{kg}$, $v_2 = 183\,\text{m/s}$, and $u_2 = 52.80\,\text{kJ/kg}$. During transit, the fluid receives $3000\,\text{J/s}$ of heat. The net work rate is:
    $\text{Answer: } -486\,\text{kJ/min}$
  • Problem 5: Air flows steadily at $0.5\,\text{kg/s}$ through a compressor, entering at $v_1 = 7\,\text{m/s}$, $p_1 = 100\,\text{kPa}$, and $v_1 = 0.95\,\text{m}^3/\text{kg}$, and leaving at $v_2 = 5\,\text{m/s}$, $p_2 = 700\,\text{kPa}$, and $v_2 = 0.19\,\text{m}^3/\text{kg}$. The internal energy of the discharged air is $90\,\text{kJ/kg}$ greater than inlet air. Compressor cooling jackets absorb heat at $58\,\text{kW}$. The work input is:
    $\text{Answer: } -122\,\text{kW}$
  • Problem 6: An apparatus performs $135\,\text{kJ}$ of work per kilogram of fluid flowing through it. Inlet conditions: $v_1 = 0.37\,\text{m}^3/\text{kg}$, $p_1 = 600\,\text{kPa}$, $v_1 = 16\,\text{m/s}$, and elevation $z_1 = 32\,\text{m}$. Discharge conditions: $v_2 = 0.62\,\text{m}^3/\text{kg}$, $p_2 = 100\,\text{kPa}$, $v_2 = 270\,\text{m/s}$, and elevation $z_2 = 0\,\text{m}$. Heat loss is $9\,\text{kJ/kg}$. The specific internal energy change is:
    $\text{Answer: } \Delta u = -20.01\,\text{kJ/kg}$
  • Problem 7: Steam enters a turbine stage with an enthalpy of $3628\,\text{kJ/kg}$ at $70\,\text{m/s}$ and exits the same stage with $h_2 = 2846\,\text{kJ/kg}$ and velocity $124\,\text{m/s}$. The work done by the steam is:
    $\text{Answer: } 776.8\,\text{kJ/kg}$

Conservation of Mass and the Continuity Equation

• Continuity Equation Formulations:

  • Mass flow rate ($\dot{m}$) is defined as the mass passing through a normal cross-section per unit time: $\dot{m} = \int_{A_c} \rho v_n\,dA_c = \rho v_{avg} A_c = \rho v A = \frac{m}{t}$

• Volume flow rate ($\dot{V}$) is defined as the volumetric displacement passing through a cross-section per unit time. The initial formulation of volumetric conservation was published in 1628 by the Italian monk Benedetto Castelli (circa 1577–1644): $\dot{V} = \int_{A_c} v_n\,dA_c = v A = \frac{V}{t}$

• Relation between mass and volume flow rate via specific volume ($v$) and density ($\rho$): $\dot{m} = \rho \dot{V} = \frac{\dot{V}}{v}$

• Conservation of Mass Principle:

  • Mass is strictly indestructible.
  • For any control volume (CV) over a finite time interval $\Delta t$: $(\text{Total mass entering the CV}) - (\text{Total mass leaving the CV}) = (\text{Net change of mass inside the CV})$

• Rate equation: $\dot{m}_{in} - \dot{m}_{out} = \frac{dm_{CV}}{dt}$

• For any steady-flow process, the time rate of mass accumulation within the control volume is zero ($\frac{dm_{CV}}{dt} = 0$), dictating: $\sum \dot{m}_{in} = \sum \dot{m}_{out}$

• Single-Stream Steady Flow: $\dot{m}_1 = \dot{m}_2 \implies \rho_1 v_1 A_1 = \rho_2 v_2 A_2 \implies \frac{v_1 A_1}{v_1} = \frac{v_2 A_2}{v_2}$

• Incompressible Steady Flow ($\rho_1 = \rho_2 = \text{constant}$): $\sum \dot{V}_{in} = \sum \dot{V}_{out}$$v_1 A_1 = v_2 A_2$

• Worked Continuity Equation Examples:

  • Example 1: Air Flow in a Pipeline:
  • Given Parameters: Fluid moves in steady flow between two sections. Section 1: $A_1 = 1\,\text{ft}^2$, $v_1 = 1000\,\text{ft/min}$, $v_1 = 4\,\text{ft}^3/\text{lbm}$. Section 2: $A_2 = 2\,\text{ft}^2$, $\rho_2 = 0.20\,\text{lbm/ft}^3$.
  • Mass Flow Rate Calculation: $\dot{m} = \frac{A_1 v_1}{v_1} = \frac{(1\,\text{ft}^2)(1000\,\text{ft/min})}{4\,\text{ft}^3/\text{lbm}} = 250\,\text{lbm/min}$$\dot{m} = (250\,\text{lbm/min})(60\,\text{min/hr}) = 15000\,\text{lbm/hr}$

• Discharge Velocity Evaluation: $\dot{m} = \rho_2 A_2 v_2 \implies 250\,\text{lbm/min} = (0.20\,\text{lbm/ft}^3)(2\,\text{ft}^2)(v_2)$$v_2 = \frac{250}{0.40} = 625\,\text{ft/min} = \frac{625}{60} = 10.42\,\text{ft/s}$

• Example 2: Combining Tube with Two Inlet Streams:

  • Given Parameters: Two gaseous streams mix in a combining tube to form a single exit stream.
  • Stream A: $A_A = 75\,\text{in}^2$, $v_A = 500\,\text{fps}$, $v_A = 10\,\text{ft}^3/\text{lb}$.
  • Stream B: $A_B = 50\,\text{in}^2$, $\dot{m}_B = 20\,\text{lb/sec}$, $\rho_B = 0.12\,\text{lb/ft}^3$.
  • Exit Mixture: $v_3 = 250\,\text{fps}$, $v_3 = 5\,\text{ft}^3/\text{lb}$.
  • Inlet Mass Flow Rates: $\dot{m}_A = \frac{A_A v_A}{v_A} = \frac{\left(\frac{75}{144}\,\text{ft}^2\right)(500\,\text{ft/s})}{10\,\text{ft}^3/\text{lb}} = \frac{260.417}{10} = 26.042\,\text{lb/sec}$$\dot{m}_B = 20\,\text{lb/sec}$

• Exit Mass Flow Rate: $\dot{m}_{exit} = \dot{m}_A + \dot{m}_B = 26.042 + 20 = 46.042\,\text{lb/sec}$

• Example 3: Filling a 10-Gallon Bucket via Garden Hose and Nozzle:

  • Given Parameters: A $10\,\text{gal}$ bucket is filled in $50\,\text{s}$. Hose inner diameter is $D_1 = 2\,\text{cm} = 0.02\,\text{m}$; nozzle exit diameter is $D_2 = 0.8\,\text{cm} = 0.008\,\text{m}$. Conversion: $1\,\text{gal} = 0.00378541\,\text{m}^3$.
  • Total Volume: $V = (10)(0.00378541) = 0.037854\,\text{m}^3 = 37.854\,\text{L}$

• Flow Rates: $\dot{V} = \frac{V}{t} = \frac{0.037854\,\text{m}^3}{50\,\text{s}} = 7.571 \times 10^{-4}\,\text{m}^3/\text{s} = 0.757\,\text{L/s}$$\dot{m} = \rho \dot{V} = (1000\,\text{kg/m}^3)(7.571 \times 10^{-4}\,\text{m}^3/\text{s}) = 0.7571\,\text{kg/s}$

• Velocity at Nozzle Exit: $A_2 = \frac{\pi}{4}(0.008\,\text{m})^2 = 5.0265 \times 10^{-5}\,\text{m}^2$$v_2 = \frac{\dot{V}}{A_2} = \frac{7.571 \times 10^{-4}\,\text{m}^3/\text{s}}{5.0265 \times 10^{-5}\,\text{m}^2} = 15.06\,\text{m/s}$

Fluid Statics, Pascal's Law, and Pressure Variations

• Pascal's Law:
  • Formulated by the French mathematician and physicist Blaise Pascal.
  • Pascal's law states that an external pressure applied to a confined fluid is transmitted equally, undiminished, in all directions and to all parts of the container.
  • When liquid flows into communicating containers, pressure forces the liquid upward to the exact same vertical elevation regardless of the shape, volume, cross-sectional area, or inclination angle of the tubes.

• Pascal's Principle Applied to Fluids:
  • For Liquids: Homogeneous liquids at the same vertical depth possess identical pressures, irrespective of container geometry: $P_1 = P_2 = P_3 = P_4 = \gamma h$
  • Unit dimensional validation: $\left[\frac{\text{kN}}{\text{m}^3}\right] [\text{m}] = \frac{\text{kN}}{\text{m}^2} = \text{kPa}$.
• For Confined Gases: The pressure everywhere within a closed vessel containing a gas is uniform and acts equally in all directions: $P_1 = P_2 = P_3 = P_4 = P_g$
• Siphon Mechanics: Hydrostatic pressure forces fluid to climb over an elevation boundary until the fluid source is exhausted or the vacuum suction is severed, enabling the draining of vessels without tipping.

• Hydrostatic Pressure Variation Derivation:
  • Consider an elementary prism of fluid at rest aligned at an arbitrary angle $\theta$ to the horizontal, having a cross-sectional area $a$, length $L$, and elevation difference $h = L\sin(\theta)$.

• Total weight of the prism is $W = \gamma a L$.

• Summing forces along the longitudinal axis in static equilibrium ($\sum F_x = 0$): $P_2 a - P_1 a - W \sin(\theta) = 0$$P_2 a - P_1 a - (\gamma a L)\left(\frac{h}{L}\right) = 0$$(P_2 - P_1)a = \gamma a h \implies P_2 - P_1 = \gamma h$

• Consequences of Hydrostatic Equilibrium:

  • In a homogeneous fluid at rest, the difference in pressure between two points depends solely on the specific weight ($\gamma$) and vertical distance ($h$).
  • Points lying along the exact same horizontal plane ($h = 0$) possess identical pressure ($P_1 = P_2$).
  • If point 1 is at a free liquid surface exposed to atmospheric pressure, gage pressure $P_1 = 0$, yielding the standard gage equation $P = \gamma h$.

• Multiple Fluid Stratification:

  • For a closed container holding stacked layers of immiscible liquids with an overhead gas under gage pressure $P_a$: $P_{bottom} = P_a + \sum_{i=1}^n \gamma_i h_i = P_a + \sum_{i=1}^n \rho_i g h_i$

• Barometric Instruments:

  • Mercury Barometer: Invented to determine atmospheric pressure. At sea level, atmospheric pressure supports a column of mercury of vertical height $760\,\text{mm}\,\text{Hg}$ ($29.9\,\text{in}\,\text{Hg}$). Greater altitudes produce lower barometric heights due to reduced atmospheric density and pressure.

• Aneroid Barometer: Operates using a partially evacuated, corrugated metallic capsule (drum) that expands or contracts in response to external barometric variations. A linkage system of levers, hairsprings, and chains magnifies this displacement to drive an analog pointer across a graduated dial.

• Fluid Statics and Barometer Examples:

  • Example 1: Building Height Determination:
  • Problem: Kim Atienza carries a barometer atop a building in Quezon City. At the top of the building, the barometer reads $754\,\text{mmHg}$; at the building base, it reads $760\,\text{mmHg}$. Air density is assumed constant at $\rho = 1.2\,\text{kg}_m/\text{m}^3$, and local gravity is $g = 9.785\,\text{m/s}^2$. Compute the building height.
  • Solution: $\Delta P = \rho_{Hg} g \Delta h_{Hg} = \rho_{air} g h_{building}$$h_{building} = \frac{\rho_{Hg} \Delta h_{Hg}}{\rho_{air}} = \frac{(13600\,\text{kg/m}^3)(0.760\,\text{m} - 0.754\,\text{m})}{1.2\,\text{kg/m}^3} = \frac{(13600)(0.006)}{1.2} = 68.0\,\text{m}$

• Example 2: Multi-Layer Stratified Storage Tank:

  • Problem: A closed vertical tank contains four distinct layers:
  • Layer 1 (top): Air at a gage pressure $P_g = 20\,\text{kPag}$ with thickness $h_1 = 1.0\,\text{m}$.
  • Layer 2: Fresh water ($H_2O$) with specific gravity $SG = 1.0$ and thickness $h_2 = 1.2\,\text{m}$.
  • Layer 3: Saltwater with specific gravity $SG = 1.1$ and thickness $h_3 = 1.3\,\text{m}$.
  • Layer 4 (bottom): Mercury ($Hg$) with specific gravity $SG = 13.6$ and thickness $h_4 = 1.5\,\text{m}$.
  • Compute the gage pressure at the middle of the fluid strata and the absolute pressure at the bottom in both $\text{kPag}$ and $\text{psia}$.

Manometry and Pressure Measurement

• Principles of Manometry:
  • Manometry is the science and methodology of measuring fluid pressure differences using columns of liquid.
  • A manometer is a device typically consisting of a U-shaped glass or plastic tube containing one or more fluids of known specific gravity ($SG$), whose liquid interfaces adjust position until hydrostatic equilibrium is established.
  • Open (Atmospheric) Manometer: One leg is open to the atmosphere, directly reading gage pressures relative to atmospheric reference (P_{atm}$).\n * **Closed (Differential) Manometer**: Both ends are connected across pressure ports, lines, or containers without atmospheric venting; used to determine the differential pressure (\Delta P = P_A - P_B).\n\n\n\n* **Worked Manometer Examples**:\n * **Example 1: Simple Open U-Tube Manometer**:\n * A U-tube manometer filled with mercury (\rho = 13600\,\text{kg/m}^3$) is connected to a gas line with the opposite leg open to the atmosphere. The mercury level in the gas leg is$50\,\text{mm} lower than in the open leg.\n\n\n\n* *Calculation*:\nP_{gas} = \rho g h = (13600\,\text{kg/m}^3)(9.8066\,\text{m/s}^2)(0.050\,\text{m}) = 6668.5\,\text{Pa} = 6.67\,\text{kPa (gauge)}\n * \text{Answer: } P_g = 6.67\,\text{kPa (gauge)}\n\n* **Example 2: Differential Manometer with Water**:\n * A differential water manometer (\rho = 1000\,\text{kg/m}^3$) is attached across points A and B in a pipeline. The water column in the right limb stands$0.30\,\text{m} higher than in the left limb.\n\n\n\n* *Calculation*:\nP_A - P_B = \rho g \Delta h = (1000\,\text{kg/m}^3)(9.8066\,\text{m/s}^2)(0.30\,\text{m}) = 2942\,\text{Pa} = 2.94\,\text{kPa}\n * \text{Answer: } P_A - P_B = 2.94\,\text{kPa}\n\n* **Example 3: Differential Manometer with Mercury**:\n * A differential mercury manometer (\rho = 13600\,\text{kg/m}^3$) measures the pressure differential between air ducts A and B. The mercury column on side B is$120\,\text{mm} higher than on side A.\n\n\n\n* *Calculation*:\nP_A - P_B = \rho g \Delta h = (13600\,\text{kg/m}^3)(9.8066\,\text{m/s}^2)(0.120\,\text{m}) = 16004\,\text{Pa} = 16.0\,\text{kPa}\n * \text{Answer: } P_A - P_B = 16.0\,\text{kPa}\n\n* **Example 4: Multi-Fluid Open Manometer**:\n * A U-tube manometer containing mercury (\rho = 13600\,\text{kg/m}^3$) has its left limb coupled to an oil tank ($\rho_{oil} = 850\,\text{kg/m}^3$) while the right limb is open to the atmosphere. The oil layer above the mercury interface in the left limb is$0.25\,\text{m}$. The mercury level in the right limb stands$0.10\,\text{m} above the interface in the left limb.\n\n\n\n* *Calculation*:\nP_g + \rho_{oil} g h_{oil} = \rho_{Hg} g h_{Hg}\nP_g = (13600)(9.8066)(0.10) - (850)(9.8066)(0.25) = 13337 - 2084 = 11253\,\text{Pa} = 11.25\,\text{kPa}\n* Documented reported answer: \n\text{Answer: } P_g = 5.46\,\text{kPa}\n\n# The Ideal Gas Equation of State\n\n* **Definition of an Ideal Gas**:\n * An ideal gas is defined as a hypothetical gas possessing zero intermolecular forces of attraction and negligible molecular volume.\n * Gases that adhere strictly to the ideal gas laws across all ranges of pressure and temperature are termed ideal gases.\n\n* **Gas Constants and State Formulations**:\n * **Individual Specific Gas Constant (R$)**: Unique to each gaseous species, established from the universal gas constant ($\bar{R}$) and molar mass ($MW):\nR = \frac{\bar{R}}{MW}\n* For uniformity, reference values follow Item B1 from *Thermodynamics* by Faires and Simmang, 6th Edition.\n* Dimensional units of R:\nR \rightarrow \frac{(\text{pressure unit})\left(\frac{\text{volume unit}}{\text{unit mass}}\right)}{\text{absolute temperature}} = \frac{\text{energy unit}}{(\text{mass})(\text{absolute temperature})}\n* Specific constants for atmospheric air:\nR_{air} = 0.28708\,\frac{\text{kJ}}{\text{kg}m\cdot\text{K}} = 0.28708\,\frac{\text{kJ}}{\text{kg}_m\cdot^\circ\text{C}} = 53.342\,\frac{\text{ft}\cdot\text{lb}_f}{\text{lb}_m\cdot^\circ\text{R}} = 53.342\,\frac{\text{ft}\cdot\text{lb}_f}{\text{lb}_m\cdot^\circ\text{F}}\n* **Molar Mass (MW)**:\n * Mass is related to mole quantity via m = n(MW).\n * For atmospheric air: MW = 28.970\,\frac{\text{kg}}{\text{kmol}} = 28.970\,\frac{\text{kg}}{\text{kgmol}} = 28.970\,\frac{\text{lb}}{\text{pmole}} = 28.970\,\frac{\text{lb}}{\text{lbmol}}.\n * For molecular nitrogen (N_2$):$MW = 28\,\frac{\text{kg}}{\text{kmol}} = 28\,\frac{\text{lb}_m}{\text{pmole}}.\n* **Universal Gas Constant (\bar{R})**:\n * Invariant across all chemical substances:\n\bar{R} = 8.3143\,\frac{\text{kJ}}{\text{kmol}\cdot\text{K}} = 8.3143\,\frac{\text{kJ}}{\text{kmol}\cdot^\circ\text{C}} = 8.3143\,\frac{\text{kJ}}{\text{kgmol}\cdot\text{K}} = 8.3143\,\frac{\text{kJ}}{\text{kgmol}\cdot^\circ\text{C}}\n\bar{R} = 8.3143\,\frac{\text{J}}{\text{mol}\cdot\text{K}} = 8.3143\,\frac{\text{J}}{\text{mol}\cdot^\circ\text{C}} = 8.3143\,\frac{\text{J}}{\text{gmol}\cdot\text{K}} = 8.3143\,\frac{\text{J}}{\text{gmol}\cdot^\circ\text{C}}\n\bar{R} = 1545.32\,\frac{\text{ft}\cdot\text{lb}_f}{\text{pmole}\cdot^\circ\text{R}} = 1545.32\,\frac{\text{ft}\cdot\text{lb}_f}{\text{pmole}\cdot^\circ\text{F}} = 1545.32\,\frac{\text{ft}\cdot\text{lb}_f}{\text{lbmol}\cdot^\circ\text{R}} = 1545.32\,\frac{\text{ft}\cdot\text{lb}_f}{\text{lbmol}\cdot^\circ\text{F}}\n\bar{R} = \frac{1545.32}{778.16}\,\frac{\text{BTU}}{\text{pmole}\cdot^\circ\text{R}} = \frac{1545.32}{778.16}\,\frac{\text{BTU}}{\text{pmole}\cdot^\circ\text{F}} = \frac{1545.32}{778.16}\,\frac{\text{BTU}}{\text{lbmol}\cdot^\circ\text{R}} = \frac{1545.32}{778.16}\,\frac{\text{BTU}}{\text{lbmol}\cdot^\circ\text{F}}\n\bar{R} = 0.08205\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\n\n* **Ideal Gas State Equations**:\n * Mass form:\nPV = mRT\n* Molar form:\nPV = [n(MW)]RT = [n(MW)]\left[\frac{\bar{R}}{MW}\right]T \implies PV = n\bar{R}T\n* Molar volume form (\bar{v} = \frac{V}{n}):\nP\bar{v} = \bar{R}T\n\n* **Specific Heat Capacities and Mayer's Relation**:\n * Specific heat capacity at constant volume (c_v$): The energy needed to increase the temperature of a unit mass of gas by$1\,^\circ\text{unit} at constant volume:\nc_v = \left(\frac{\partial u}{\partial T}\right)_v = \frac{du}{dT}\n* Specific heat capacity at constant pressure (c_p$): The energy needed to increase the temperature of a unit mass of gas by$1\,^\circ\text{unit} at constant pressure:\nc_p = \left(\frac{\partial h}{\partial T}\right)_p = \frac{dh}{dT}\n* **Mayer's Formula (Mayer's Relation)**: Discovered by German physicist Julius von Mayer:\nh = u + Pv = u + RT\n\frac{dh}{dT} = \frac{du}{dT} + R\left(\frac{dT}{dT}\right) = \frac{du}{dT} + R\nR = \frac{dh}{dT} - \frac{du}{dT} = c_p - c_v\n* On a molar basis:\n\bar{R} = \bar{c}_p - \bar{c}_v\n* Ratio of specific heats (k):\nk = \frac{c_p}{c_v}\n\n* **Worked Ideal Gas Problems**:\n * **Example 1: Air in a Rigid Tank**:\n * *Problem*: Two kilograms of air (m = 2\,\text{kg}$) at$T = 280\,\text{K}$are contained inside a$V = 0.2\,\text{m}^3 rigid tank. Treating air as an ideal gas, find the absolute pressure, mole quantity, and specific volume on both a mass and molal basis.\n * *Calculations*:\nP = \frac{mRT}{V} = \frac{(2\,\text{kg})(0.28708\,\text{kJ}/(\text{kg}\cdot\text{K}))(280\,\text{K})}{0.2\,\text{m}^3} = 803.824\,\text{kPa}\nn = \frac{m}{MW} = \frac{2\,\text{kg}}{28.970\,\text{kg/kmol}} = 0.069037\,\text{kmol}\nv = \frac{V}{m} = \frac{0.2\,\text{m}^3}{2\,\text{kg}} = 0.10\,\text{m}^3/\text{kg}\n\bar{v} = \frac{V}{n} = \frac{0.2\,\text{m}^3}{0.069037\,\text{kmol}} = 2.897\,\text{m}^3/\text{kmol}\n\n* **Example 2: Acetylene Drum Depletion**:\n * *Problem*: A cylindrical storage drum of diameter D = 6\,\text{in}$and length$L = 40\,\text{in}$contains acetylene gas at$p_1 = 250\,\text{psia}$and$T_1 = 90\,^\circ\text{F}$. Following usage, the pressure drops to$p_2 = 200\,\text{psia}$at$T_2 = 85\,^\circ\text{F}$. The gas constant for acetylene is$R = 59.35\,\frac{\text{ft}\cdot\text{lb}_f}{\text{lb}_m\cdot^\circ\text{R}}.\n * (a) What portion of the acetylene was used?\n * (b) What volume would the used acetylene occupy at 14.7\,\text{psia}$and$80\,^\circ\text{F}?\n * *Calculations*:\nV = \frac{\pi}{4}\left(\frac{6}{12}\right)^2\left(\frac{40}{12}\right) = \frac{\pi}{4}(0.25)(3.3333) = 0.6545\,\text{ft}^3\nT_1 = 90 + 459.67 = 549.67\,^\circ\text{R}\nT_2 = 85 + 459.67 = 544.67\,^\circ\text{R}\nm_1 = \frac{P_1 V}{R T_1} = \frac{(250 \times 144)(0.6545)}{(59.35)(549.67)} = \frac{23562}{32623} = 0.7222\,\text{lb}_m\nm_2 = \frac{P_2 V}{R T_2} = \frac{(200 \times 144)(0.6545)}{(59.35)(544.67)} = \frac{18849.6}{32326} = 0.5831\,\text{lb}_m\n\Delta m = m_1 - m_2 = 0.7222 - 0.5831 = 0.1391\,\text{lb}_m\n\text{Fraction used} = \frac{\Delta m}{m_1} = \frac{0.1391}{0.7222} = 0.1926 = 19.26\%\n* *Expanded Volume at Standard State* (P_3 = 14.7\,\text{psia}$,$T_3 = 80 + 459.67 = 539.67\,^\circ\text{R}):\nV_3 = \frac{\Delta m R T_3}{P_3} = \frac{(0.1391\,\text{lb}_m)(59.35)(539.67)}{(14.7 \times 144)} = \frac{4455.3}{2116.8} = 2.105\,\text{ft}^3\n\n* **Example 3: Constant Pressure Heating Process**:\n * *Problem*: A gas has gas constant R = 320\,\frac{\text{J}}{\text{kg}\cdot\text{K}} = 0.320\,\frac{\text{kJ}}{\text{kg}\cdot\text{K}}$and$c_v = 0.84\,\frac{\text{kJ}}{\text{kg}\cdot\text{K}}.\n * (a) Find c_p$and$k.\n * (b) If 5\,\text{kg}$of this gas undergoes a reversible nonflow constant pressure process from$V_1 = 1.333\,\text{m}^3$and$P_1 = 690\,\text{kPa}$to a state where$t_2 = 555\,^\circ\text{C}$, determine the changes in internal energy ($\Delta U$) and enthalpy ($\Delta H).\n * *Calculations*:\nc_p = c_v + R = 0.84 + 0.320 = 1.16\,\frac{\text{kJ}}{\text{kg}\cdot\text{K}}\nk = \frac{c_p}{c_v} = \frac{1.16}{0.84} = 1.381\n* *State Temperatures*:\nT_1 = \frac{P_1 V_1}{m R} = \frac{(690\,\text{kPa})(1.333\,\text{m}^3)}{(5\,\text{kg})(0.320\,\text{kJ}/(\text{kg}\cdot\text{K}))} = \frac{919.77}{1.60} = 574.86\,\text{K}\nT_2 = 555 + 273.15 = 828.15\,\text{K}\n\Delta T = T_2 - T_1 = 828.15 - 574.86 = 253.29\,\text{K}\n* *Property Changes*:\n\Delta U = m c_v \Delta T = (5\,\text{kg})(0.84\,\text{kJ}/(\text{kg}\cdot\text{K}))(253.29\,\text{K}) = 1063.8\,\text{kJ}\n\Delta H = m c_p \Delta T = (5\,\text{kg})(1.16\,\text{kJ}/(\text{kg}\cdot\text{K}))(253.29\,\text{K}) = 1469.1\,\text{kJ}\n\n# Classical Empirical Gas Laws and Gas Mixtures\n\n* **Boyle's Law**:\n * Discovered by Robert Boyle.\n * States that the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies when temperature and mass remain constant within a closed system:\nPV = C = \text{constant}\nP_1 V_1 = P_2 V_2\n* *Worked Example*: A gas occupies 12.3\,\text{L}$at a pressure of$40.0\,\text{mmHg}$gauge. Determine the volume when pressure increases to$60.0\,\text{mmHg}$gauge ($P{atm} = 760\,\text{mmHg}):\nP_1 = 760 + 40.0 = 800\,\text{mmHg}\nP_2 = 760 + 60.0 = 820\,\text{mmHg}\nV_2 = V_1 \left(\frac{P_1}{P_2}\right) = (12.3\,\text{L})\left(\frac{800}{820}\right) = 12.0\,\text{L}\n\n* **Charles' Law**:\n * Discovered by Jacques Charles.\n * States that the volume of a given mass of an ideal gas is directly proportional to its absolute temperature when absolute pressure and mass are held constant within a closed system:\n\frac{V}{T} = C = \text{constant}\n\frac{V_1}{T_1} = \frac{V_2}{T_2}\n* *Worked Example*: An ideal gas occupies 1.00\,\text{L}$at standard temperature ($0\,^\circ\text{C} = 273.15\,\text{K}$). Find its volume at$333.0\,^\circ\text{C}$($606.15\,\text{K}):\nV_2 = V_1 \left(\frac{T_2}{T_1}\right) = (1.00\,\text{L})\left(\frac{333.0 + 273.15}{273.15}\right) = (1.00)\left(\frac{606.15}{273.15}\right) = 2.22\,\text{L}\n\n* **Gay-Lussac's Law**:\n * Formulated by Joseph Louis Gay-Lussac.\n * States that the absolute pressure of a given mass of an ideal gas is directly proportional to its absolute temperature when volume and mass remain constant within a closed system:\n\frac{P}{T} = C = \text{constant}\n\frac{P_1}{T_1} = \frac{P_2}{T_2}\n* *Worked Example*: A 30.0\,\text{L}$rigid metal container of nitrogen at$20.0\,^\circ\text{C}$and$3.00\,\text{atm}$is heated inside an oven to$50.0\,^\circ\text{C}. Determine the final pressure:\nT_1 = 20.0 + 273.15 = 293.15\,\text{K}\nT_2 = 50.0 + 273.15 = 323.15\,\text{K}\nP_2 = P_1 \left(\frac{T_2}{T_1}\right) = (3.00\,\text{atm})\left(\frac{323.15}{293.15}\right) = 3.31\,\text{atm}\n\n* **Combined Gas Law**:\n * Synthesizes Boyle's, Charles', and Gay-Lussac's laws into a unified relationship for constant-mass systems:\n\frac{PV}{T} = C = \text{constant}\n\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\n* *Worked Example*: The pressure of 8.40\,\text{L}$of nitrogen gas in a flexible container is halved ($P_2 = 0.5 P_1$), and its absolute temperature is doubled ($T_2 = 2 T_1). Calculate the new volume:\nV_2 = V_1 \left(\frac{P_1}{P_2}\right)\left(\frac{T_2}{T_1}\right) = (8.40\,\text{L})\left(\frac{1}{0.5}\right)(2) = (8.40)(2)(2) = 33.6\,\text{L}\n\n* **Avogadro's Law and Microscopic Constants**:\n * Formulated by Lorenzo Romano Amedeo Carlo Avogadro.\n * States that the volume of a gas is directly proportional to the number of moles (or representative particles) at constant temperature and pressure:\n\frac{V}{n} = C = \text{constant}\n\frac{V_1}{n_1} = \frac{V_2}{n_2}\n* Mass and constant proportions:\n\frac{m_1}{m_2} = \frac{MW_1}{MW_2}\n\frac{R_1}{R_2} = \frac{MW_1}{MW_2}\n* **Avogadro's Number (N_A)**:\nN_A = 6.02214179 \times 10^{23}\,\frac{\text{molecules}}{\text{mole}} \approx 6.022 \times 10^{23}\,\frac{\text{particles}}{\text{mole}}\n* Total particle count (N):\nN = n N_A\n* **Standard Molar Volume at STP** (273\,\text{K}$,$1\,\text{atm}):\n * Exactly one mole (6.022 \times 10^{23}$particles) of any ideal gas occupies$22.4\,\text{L}.\n* *Worked Example*: A balloon has a volume of 1.90\,\text{L}$with$0.0920\,\text{mol}$of helium. If$0.0210\,\text{mol} of helium is added at constant temperature and pressure, determine the final volume:\nn_2 = 0.0920 + 0.0210 = 0.1130\,\text{mol}\nV_2 = V_1 \left(\frac{n_2}{n_1}\right) = (1.90\,\text{L})\left(\frac{0.1130}{0.0920}\right) = 2.334\,\text{L}\n\n* **Dalton's Law of Partial Pressures**:\n * Observed empirically by John Dalton in 1801 and published in 1802.\n * States that the total pressure exerted by a mixture of non-reacting ideal gases equals the sum of the partial pressures of each individual constituent gas when each occupies the entire mixture volume (V_m$) alone at the mixture temperature ($T_m):\nP_{total} = P_m = \sum_{i=1}^n P_i = P_1 + P_2 + \dots + P_n\n* Subject to:\nT_m = T_1 = T_2 = \dots = T_n\nV_m = V_1 = V_2 = \dots = V_n\n* *Worked Example*: A four-component gaseous mixture containing CO$,$CH_4$,$N_2$, and$C_3H_8$at$33\,^\circ\text{C}$has partial pressures of$7\,\text{kPaa}$,$12\,\text{kPaa}$,$14\,\text{kPaa}$, and$28\,\text{kPaa}. Find the total mixture pressure:\nP_m = \sum P_i = 7 + 12 + 14 + 28 = 61\,\text{kPaa}\n\n* **Amagat's Law of Partial Volumes**:\n * Also designated as Amagat's Law of Additive Volumes or Leduc's Law; named after French physicist Émile Hilaire Amagat.\n * States that the total volume of a non-reacting mixture of ideal gases is the sum of the individual component volumes each evaluated at the pressure and temperature of the mixture:\nV_{total} = V_m = \sum_{i=1}^n V_i = V_1 + V_2 + \dots + V_n\n* Subject to:\nT_m = T_1 = T_2 = \dots = T_n\nP_m = P_1 = P_2 = \dots = P_n$$