MCAT Physics Chapter 3: Thermodynamics

Previous Topics Recap

• Discussed Zeroth Law and First Law of Thermodynamics.
• Defined heat and examined three methods of heat energy transfer.
• Discussed specific heat and phase changes.

Thermodynamic Processes

• Review of work as a change of energy in a system:
  • As a function of force and displacement.
  • As a function of volume and pressure.
• Thermodynamic processes move systems from one equilibrium state to another:
  • Represented graphically with volume on x-axis and pressure/temperature on y-axis.

Key Thermodynamic Processes

1. Isothermal Process

   • Constant temperature, no change in internal energy.
   • Appears as a slight tilt on a pressure vs. volume graph.

2. Adiabatic Process

   • No heat exchange.
   • Appears distinct on a pressure vs. volume graph.

3. Isovolumetric (Isochoric) Process

   • No change in volume, only pressure changes.
   • No work is done.

4. Isobaric Process

   • Constant pressure.
   • Appears as a horizontal line on a pressure vs. volume graph.

Second Law of Thermodynamics

• Objects in thermal contact not in equilibrium will exchange heat to reach equilibrium.
• Energy spontaneously disperses unless hindered.
• Entropy
  • Misconception: Entropy as disorder; actually, it's about energy dispersal.
  • Measure of energy dispersal at a specific temperature.
  • Example: Melting ice increases microstates and entropy due to more freedom of movement.
• Entropy Equation
  • ( \Delta S = \frac{Q}{T} )
  • Units: Joules per mole per Kelvin.
  • Entropy increases with energy distribution into a system, decreases with energy distribution out.

Key Points

• Concentration of energy requires external work.
• Refrigerators work against spontaneous heat flow using energy.

Entropy and Process Types

• Second law correlates with increasing universe entropy.
• Reversible Process: Delta S = 0; reversible without affecting surroundings.
• Irreversible Process: Delta S > 0; irreversible with some effect on surroundings.

Example Problem: Change in Entropy

• Given:
  • 6.66 x 10^4 Joules of heat.
  • 200g ice at 273K melts to water.
  • Heat of fusion of ice: 333 J/K.
• Solution:
  • Use ( \Delta S = \frac{Q}{T} )
  • Calculate ( \Delta S ):
    • ( \Delta S = \frac{6.66 \times 10^4}{273} = 244 ) Joules per Kelvin.
• Result: Heat exactly enough to melt ice without changing water temperature.

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Ensure understanding of each thermodynamic process and the Second Law of Thermodynamics for MCAT success.

Good luck with your studies!