Thermodynamics Review

Heat Capacity (c)

• The amount of heat needed to change the temperature of an object.

Molar Heat Capacity

• Amount of substance in moles.

Specific Heat Capacity

• Amount of substance in grams.

Constants and Values

• $R = 0.08206 \frac{L \cdot atm}{mol \cdot K}$
• $1 \frac{L \cdot atm}{mol \cdot K} = 101.3 J$
• $R = 8.3145 \frac{J}{mol \cdot K}$

Heat Capacity at Constant Volume and Pressure

• C_M = \frac{#J}{mol \cdot K}
• C_V = \frac{#J}{g \cdot K}
• Constant Volume: $C_V$
• Constant Pressure: $C_P$
• For monatomic gases: $C_P = \frac{5}{2} R$
• Because $\Delta T$ (in Kelvin) = $\Delta T$ (in Celsius), K and °C can be interchanged when calculating temperature differences.

Heat Capacity at Constant Volume (Cᵥ)

• Ideal Gas: $(KE)_{avg} = \frac{3}{2} RT$
• No change in volume ($\Delta V$) means no work is done.
• Translational movement:
  $C_V = \frac{3}{2} R$

Heat Capacity at Constant Pressure (Cₚ)

• $C_P =$ translational + work
• $C<em>P = C</em>V + R = \frac{5}{2}R$

Heat Capacity Problem (Non-Monoatomic Ideal Gas)

• Water has a heat capacity of $4.18 \frac{J}{g \cdot °C}$.
• A $20.0 g$ block of iron heated to $234 °C$ is added to $100. g$ of water at $25.0 °C$.
• The water's final temperature is $29.4 °C$.
• Goal: Find the heat capacity of iron.
• Final temperature of iron & water is the same due to the 0th Law Consequence.
• $q<em>{H</em>2O} = -q_{Fe}$
• 1st Law Consequence:
  $m<em>{Fe} \cdot C</em>{S,Fe} \cdot \Delta T<em>{Fe} = - m</em>{H<em>2O} \cdot C</em>{S,H<em>2O} \cdot \Delta T</em>{H<em>2O}$$C</em>{S,Fe} = \frac{m<em>{H</em>2O} \cdot C<em>{S,H</em>2O} \cdot \Delta T<em>{H</em>2O}}{- m<em>{Fe} \cdot \Delta T</em>{Fe}}$
• This calculates the heat needed to change the temperature of a mass of a substance.
• $C_{S,Fe} = \frac{100.g \cdot 4.18 \frac{J}{g \cdot °C} \cdot (29.4 - 25.0)°C}{-20.0g \cdot (29.4 - 234)°C} = 0.450 \frac{J}{g \cdot °C}$.

State Functions 9.1

• A state function depends only on its present position, not how it got there.
• Also called pathway-independent functions.

Examples of State Functions

• P, V, T, E, H, G, S (Entropy)

Examples of Non-State Functions

• q and w

• If $\frac{a}{b} =$ Not state function, but $\frac{\Delta a}{\Delta b} =$ state function.

State Functions

• $\Delta E = q + w$
• If $w = 0 J$, $\Delta E = q$
• q and w are pathway dependent.
• Conclusion: $\Delta E$ is a state function, while q and w are not.

$\Delta E$, q, w, and Pathways

• $\Delta E = q + w$
• $w = -P \Delta V$
• The work is the area under the curve on a P-V diagram.
• Pathways 1 and 2:
  • $\Delta E<em>1 = \Delta E</em>2$ because $\Delta E$ is path independent
  • $q<em>1 \neq q</em>2$

$\Delta E$, q, w, and Pathways

• System: 1.0 mole of monatomic ideal gas in a 5.0 L cylinder at 5.0 atm.
• Final state: 1.0 L and 7.0 atm.
• Calculate q, w, and $\Delta E$ (in J) for two paths:
  1. Compression in volume, then increase in pressure.
  2. Increase in pressure, then compression in volume.
• Important Equations:
  • $C_V = \frac{3}{2} R$
  • $C_P = \frac{5}{2} R$
  • $R = 8.314 \frac{J}{mol \cdot K}$
  • $\Delta E = q + w$
  • $w = -P \Delta V$
  • $q = nC \Delta T$
  • $\Delta T = \frac{\Delta (PV)}{nR}$
  • $R = 0.08204 \frac{L \cdot atm}{mol \cdot K}$

Path 1

• Path 1: A -> B -> D

  • A -> B: (5.0 L, 5.0 atm) -> (1.0 L, 5.0 atm) *constant P
  • B -> D: (1.0 L, 5.0 atm) -> (1.0 L, 7.0 atm) *constant V

• Step IA: A → B

  • $w_1 = -P \Delta V = -5.0 atm \cdot (-4.0 L) = +20.0 L \cdot atm$
  • $q<em>1 = nC</em>P \Delta T = n (\frac{5}{2}R) (\frac{\Delta (PV)}{nR}) = \frac{5}{2} (-20.0 L \cdot atm) = -50.0 L \cdot atm$

• Step IB: B → D

  • $w_2 = -P \Delta V = 0 L \cdot atm$
  • $q<em>2 = nC</em>V \Delta T = n (\frac{3}{2}R) (\frac{\Delta (PV)}{nR}) = \frac{3}{2} (2.0 L \cdot atm) = 3.0 L \cdot atm$

• Total:

  • $q = q<em>1 + q</em>2 = (-50.0 + 3.0) L \cdot atm = -47 L \cdot atm = -4800 J$
  • $w = w<em>1 + w</em>2 = 20 + 0 = 20 L \cdot atm = 2000 J$
  • $\Delta E = q + w = -4800 J + 2000 J = -2800 J$

Path 2

• Path 2: A -> C -> D

• $C<em>V = (\frac{3}{2})R; C</em>P = (\frac{5}{2})R; w = -P\Delta V; q = nC\Delta T$

• a) 5.0 L, 5.0 atm to 5.0 L, 7.0 atm

• b) 5.0 L, 7.0 atm to 1.0 L, 7.0 atm

• Step 2A: A→C

  • (5.0L, 5.0 atm) -> (5.0L, 7.0 atm) *const V
  • $w_1 = -PDV = 0 L \cdot atm$
  • $q_1 = \frac{3}{2} (\Delta PV) = \frac{3}{2} (10 L \cdot atm) = 15 L \cdot atm$

• Step 2B: C→D

  • (5.0L, 7.0 atm) -> (1.0L, 7.0 atm) *const P
  • $w_2 = -PAV = -(7.0 atm) (-4.0L) = 28 L \cdot atm$
  • $q_2 = \frac{5}{2} (\Delta PV) = \frac{5}{2} (-27 L \cdot atm) = -70. L \cdot atm$

• Total

  • $q = q<em>1 + q</em>2 = -55 L \cdot atm = -5400 J$
  • $w = w<em>1 + w</em>2 = 28 L \cdot atm = 2800 J$
  • $\Delta E = q + w = -2800J$

Enthalpy (H) and Its Relationship to Internal Energy (E) 9.2

Enthalpy

• By definition: $H = E + PV$. Thus, $\Delta H = \Delta E + \Delta (PV)$. E, P, and V are all state functions, so H is also a state function.

Work

• $w = -P \Delta V$
• \Delta E = q + w => q_P = \Delta E + P \Delta V
• $\Delta H = q_P$ This is why enthalpy is sometimes called a heat.

Hess's Law 9.5

• Hess's Law states that because $\Delta H$ is a state function, we can calculate an unknown $\Delta H_r$ by generating a thermocycle from a series of known $\Delta H$ values.
• $\Delta H<em>r = \Delta H</em>1 + \Delta H<em>2 + \Delta H</em>3$