Chapter 8.7 – 8.11 Universal Units

8.7 Gas Pressure

• Ideal Gas Law: Describes the relationship between pressure, volume, amount, and temperature of an ideal gas.

  • An ideal gas occupies a volume of 22.4 L under standard conditions.
  • Standard conditions:
    • Pressure: 1 atm
    • Temperature: 273 K
    • Amount: 1 mole
  • R = ideal gas constant
  • Temperature must be in absolute units (Kelvin).
  • The ideal gas law equation is given by: PV = nRT
    • R = 0.08206 {atm L / (mol K)}
    • R = 8.314 {Pa L / (mol K)}

• Relationships between variables in the Ideal Gas Law:

  • n = amount of substance (moles)
  • T = Temperature
  • P = Pressure
  • V = Volume
  • R = Ideal gas constant

• Intuition Check: Relationship between P and V

  • Rewriting the Ideal Gas Law: P = (nRT) * (1/V)
  • If n and T are constant, P = C/V, where C = nRT
  • As volume increases, the particles have more room to move around, leading to a decrease in pressure.

• Intuition Check: Relationship between P and n

  • Rewriting the Ideal Gas Law: P = (RT/V) * n
  • If V and T are constant, P = Cn, where C = RT/V
  • As the amount (n) increases, the particles have less room to move around, leading to an increase in pressure.

• Intuition Check: Relationship between P and T

  • Rewriting the Ideal Gas Law: P = (nR/V) * T
  • If V and n are constant, P = CT, where C = nR/V
  • As the temperature increases, the particles move around more, leading to an increase in pressure.

• Example 8.7, Ex 1:

  • Problem: 21 g of nitrogen (N2) is placed in an 8-L container. The pressure gauge reads 2.25 atm. What is the temperature in K?
  • Solution:
    • Using the Ideal Gas Law: PV = nRT or T = PV / (nR)
    • n = (21 {g}) * (1 {mol} / 28 {g}) = 0.75 {mol}
    • T = (2.25 {atm} * 8 {L}) / (0.75 {mol} * 0.08206 {atm L / (mol K)}) = 292 {K}

• Example 8.7, Ex 2:

  • Problem: Humans require 85 mol/day of oxygen. An emergency backup oxygen supply is needed for six people for one week. The container must withstand a pressure of 150 psi and the oxygen is maintained at -10 °C. What is the volume of the container in m3?

• Example 8.7, Ex 3:

  • Problem: An ideal gas in a 5-L container at 300 K exhibits a pressure of 2 atm. The volume is decreased to 2.5 L, but the temperature remains the same. What is the pressure in the second container compared to the first?

8.8 Energy

• Energy: Abstract, conserved quantity that can shift forms.

• Types of Energy:

  • Work: Energy of exerting a force over a distance
  • Potential: Energy of position
  • Kinetic: Energy of motion
  • Thermal: Energy of change in temperature

• Equations for Energy:

  • Work: W = F * x
  • Kinetic Energy: KE = (1/2) * m * v^2
  • Potential Energy: PE = m * g * H
  • Thermal Energy: Q = m * Cp * ΔT

• Relationships

  • m = Mass
  • H = Height
  • PE = Potential Energy
  • g = gravitational constant
  • F = Force ( {N})
  • x = displacement ( {m})
  • W = Work ( {J})
  • v = Velocity ( {m/s})
  • KE = Kinetic Energy ( {J})

• Intuition Check: Potential Energy

  • The higher an object is off the ground, the more potential energy it has
  • The more an object weighs the more potential energy it has

• Intuition Check: Kinetic Energy

  • The faster an object is moving, the more kinetic energy it has.
  • The more an object weighs, the more kinetic energy it has

• Energy is Conserved

  • If an object is raised it has Potential Energy (PE)
  • If the object is released the Potential Energy (PE) is changed to Kinetic Energy (KE).

• Example 8.8, Ex 1

  • Problem: If it requires 5,650 J of energy to lift an object 15 ft into the air, what is the weight of the object in unites of N?

• Example Solution: 8.8, Ex 1

  • PE = m * g * H = W * H
  • H = 15 {ft} * (1 {m} / 3.28 {ft}) = 4.57 {m}
  • m = 5650 {kg m^2 / s^2} / (9.8 {m/s^2} * 4.57 {m}) = 126.16 {kg}
  • W = m * g = 126.16 {kg} * 9.8 {m/s^2} * (1 {N} / (1 {kg m/s^2})) = 1,236 {N}

• Intuition Check: Thermal Energy

  • The hotter the temperature needed, the more energy it takes.
  • The more mass an object has, the more energy it takes to heat it up.

• Example 8.8, Ex 2

  • Problem: A cup of coffee with a mass of 58 g is heated from 70°F to 125°F. Coffee has a specific heat of 4 J/(g K). How much energy in J is required to heat the coffee?

• Energy as…

  • Work, kinetic, potential. SI units [=] joule [J]

  • Heat British Thermal Unit [BTU] Heat to raise temp of 1 lbm water by 1°F

  • calorie [cal] Heat to raise temp of 1 g water by 1°C

  • 1 J ≡ (1 N)(1 m)

• Concept Check 8.8, Check 1

  • Which of the following quantities has the highest energy?
  • A. 1 J
  • B. 1 BTU
  • C. 1 cal
  • D. 1 ft lbf

8.9 Power

• Defintion: Power is the rate at which energy is transferred or converted.

• Relationship between Power, Time and Energy

  • P = E/t;
    • P = Power
    • t = time
    • E = Energy

• Example 8.9, Ex 1

  • Problem: A small nuclear reactor with an electric output power of 5 MW is to be designed to power an initial settlement on Mars. Express this power in joules per Earth day {J/d}. Use an appropiate metric prefix with joules so that there are either 1, 2, or 3 digits to the left of the decimal point.

8.10 Efficiency

• Efficiency:

  • Efficiency is represented by the Greek letter eta, {()eta} {(()eta)}.
  • The value of efficiency is between 0 and 1: 0 < (\eta) < 1
  • Input power is always greater than output power
  • Pout = (\eta) Pin

• Relationships

  • Pin: Required by device to operate
  • Pout: Supplied by device to process
  • η: A measure of "loss" during process

• Intutiion Check

  • If {Pin = 100 W} and {Pout = 75 W}. Then {Loss heat = 25 W}
  • (\eta) = (\frac{75 W}{100 W}) = 0.75
  • (\eta) = 0.75 or {75 %}

• Concept Check: 8.10, Check 1

  • The Input power _ the output power for a real world mechnical device.
    • A. is always greater than
    • B. is always less than
    • C. is always equal to
    • D. depends on the situation when compared to

• Example: 8.10, Ex 1

  • Problem: A 75-kg load was raised as high as possible in 30 s by a 150-W, 75% efficient motor. How high, in ft, was the load raised?

• Example: 8.10, Ex 2

  • Problem: A 150-W, 75% efficient motor is used to raise a 75 kg load 15 ft in 30 s. If we double the distance we lift the load, how long will it take the same motor to accomplish the task?

• Example Solution: 8.10, Ex 2

  • If all other parameters remain the same, distance and time are in a direct relationship:

  • \frac{E1}{t1} = \frac{E2}{t2}

  • \frac{(m)(g)(H1)}{t1} = \frac{(m)(g)(H2)}{t2}

  • \frac{H1}{t1} = \frac{H2}{t2}

  • \frac{H1}{t1} = \frac{2*H1}{t2}

  • t2 = 2*t1

8.11 Electrical Concepts

• Table 8-14: Summary of Electrical Properties

• Key electrical components: Resistors ( {R}), Capacitors ( {C}), Inductors ( {L})

• Capacitance