Work, Heat, Energy — Revision Notes

⚡ 30-Second Revision

Internal Energy ($U$) — Total energy of system. State function. $\Delta U = U_{final} - U_{initial}$.
First Law of Thermodynamics — $\Delta U = q + w$.
Heat ($q$) — Energy transfer due to $\Delta T$. Path function.

* $q > 0$: absorbed by system. * $q < 0$: released by system. * $q = mcDelta T$ or $q = nC_mDelta T$.

Work ($w$) — Energy transfer not due to $\Delta T$. Path function.

* $w > 0$: done on system (compression). * $w < 0$: done by system (expansion). * Irreversible P-V work: $w = -P_{ext}\Delta V$. * Reversible isothermal P-V work (ideal gas): $w = -nRT \ln \left( \frac{V_{final}}{V_{initial}} \right)$.

Isochoric ($ \Delta V = 0 $) — $w = 0 \implies \Delta U = q_v$.
Isothermal ($ \Delta T = 0 $, ideal gas) — $\Delta U = 0 \implies q = -w$.
Adiabatic ($q = 0$) — $\Delta U = w$.
Units — $1\,\text{L atm} = 101.3\,\text{J}$.

2-Minute Revision

The core of Work, Heat, and Energy in chemistry is the First Law of Thermodynamics: $\Delta U = q + w$. This law states that the change in a system's internal energy ($\Delta U$) is the sum of heat ($q$) absorbed by the system and work ($w$) done on the system. Internal energy ($U$) is a state function, meaning its value depends only on the system's current state, not how it got there. For an ideal gas, $U$ depends solely on temperature.

Heat ($q$) is energy transferred due to a temperature difference. It's a path function. Remember the sign convention: positive $q$ for heat absorbed by the system (endothermic), negative $q$ for heat released (exothermic).

Work ($w$) is energy transferred not due to temperature difference, primarily P-V work in chemistry. It's also a path function. Sign convention: positive $w$ for work done on the system (compression), negative $w$ for work done by the system (expansion). Key formulas are $w = -P_{ext}\Delta V$ for irreversible work and $w = -nRT \ln(V_{final}/V_{initial})$ for reversible isothermal work.

Understand how these terms simplify for different processes: isochoric ($\Delta V = 0 \implies w=0 \implies \Delta U = q_v$), isothermal (ideal gas, $\Delta T = 0 \implies \Delta U = 0 \implies q = -w$), and adiabatic ($q=0 \implies \Delta U = w$). Always pay close attention to units and sign conventions in numerical problems.

5-Minute Revision

A solid understanding of Work, Heat, and Energy begins with the First Law of Thermodynamics, $\Delta U = q + w$, which is essentially the law of conservation of energy. Let's break down each term:

**Internal Energy ($U$): This is the total energy stored within a system, including all forms of kinetic and potential energy of its molecules. It's a state function**, meaning $\Delta U$ depends only on the initial and final states, not the path. For an ideal gas, $U$ is solely dependent on temperature. If $\Delta T = 0$ for an ideal gas, then $\Delta U = 0$.

**Heat ($q$): Heat is the transfer of thermal energy due to a temperature difference. It's a path function**.

Sign Convention — $q > 0$ when the system absorbs heat (endothermic), $q < 0$ when the system releases heat (exothermic).
Calculation — $q = mcDelta T$ (where $c$ is specific heat capacity) or $q = nC_mDelta T$ (where $C_m$ is molar heat capacity).

**Work ($w$): Work is energy transfer not due to a temperature difference. In chemistry, we primarily deal with P-V work (expansion/compression). It's also a path function**.

Sign Convention — $w > 0$ when work is done on the system by surroundings (compression), $w < 0$ when work is done by the system on surroundings (expansion).
Irreversible Work (constant external pressure) — $w = -P_{ext}\Delta V$. Example: A gas expands from $1\,\text{L}$ to $3\,\text{L}$ against $1\,\text{atm}$ external pressure. $w = -(1\,\text{atm})(3-1\,\text{L}) = -2\,\text{L atm} = -202.6\,\text{J}$.
Reversible Isothermal Work (ideal gas) — $w = -nRT \ln \left( \frac{V_{final}}{V_{initial}} \right)$. Example: $1\,\text{mol}$ of ideal gas expands reversibly from $1\,\text{L}$ to $2\,\text{L}$ at $300\,\text{K}$. $w = -(1)(8.314)(300) \ln(2/1) \approx -1729\,\text{J}$.

Applying the First Law to Different Processes:

Isochoric Process ($ \Delta V = 0 $) — Volume is constant, so $w = -P_{ext}\Delta V = 0$. Thus, $\Delta U = q_v$ (heat at constant volume).
Isothermal Process ($ \Delta T = 0 $) — Temperature is constant. For an ideal gas, $\Delta U = 0$. Thus, $q = -w$.
Adiabatic Process ($q = 0$) — No heat exchange. Thus, $\Delta U = w$.
Isobaric Process ($ \Delta P = 0 $) — Pressure is constant. $\Delta U = q_p + w = q_p - P_{ext}\Delta V$. Here, $q_p$ is heat at constant pressure, which is equal to $\Delta H$ (enthalpy change).

Key Takeaway: Always identify the type of process, correctly assign signs to $q$ and $w$, and use the appropriate formulas. Pay attention to units, especially converting L atm to Joules.

Prelims Revision Notes

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First Law of Thermodynamics — $\Delta U = q + w$. This is the fundamental equation. $\Delta U$ is change in internal energy, $q$ is heat, $w$ is work.
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Internal Energy ($U$)

Total energy of a system (kinetic + potential of molecules). State function: Depends only on initial and final states ($\Delta U = U_{final} - U_{initial}$). * For ideal gases, $U$ depends only on temperature ($\Delta T = 0 \implies \Delta U = 0$ for ideal gas). * Absolute value of $U$ cannot be determined, only $\Delta U$.

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Heat ($q$)

Energy transfer due to temperature difference. Path function: Depends on the process. * Sign Convention: * $q > 0$: heat absorbed by system (endothermic). * $q < 0$: heat released by system (exothermic). * Formulas: $q = mcDelta T$ (specific heat capacity, $m$ in grams) or $q = nC_mDelta T$ (molar heat capacity, $n$ in moles).

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Work ($w$)

Energy transfer not due to temperature difference (e.g., P-V work). Path function: Depends on the process. * Sign Convention: * $w > 0$: work done on the system by surroundings (compression).

* $w < 0$: work done by the system on surroundings (expansion). * P-V Work Formulas: * Irreversible (constant external pressure): $w = -P_{ext}\Delta V$. * Reversible Isothermal (ideal gas): $w = -nRT \ln \left( \frac{V_{final}}{V_{initial}} \right) = -nRT \ln \left( \frac{P_{initial}}{P_{final}} \right)$.

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Thermodynamic Processes and First Law Simplifications

* **Isochoric Process ($\Delta V = 0$)**: Volume constant. $w = 0$. So, $\Delta U = q_v$. * **Isothermal Process ($\Delta T = 0$)**: Temperature constant. For ideal gas, $\Delta U = 0$. So, $q = -w$. * **Adiabatic Process ($q = 0$)**: No heat exchange. So, $\Delta U = w$. * **Isobaric Process ($\Delta P = 0$)**: Pressure constant. $\Delta U = q_p - P_{ext}\Delta V$.

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Units — Energy is typically in Joules (J). $1\,\text{L atm} = 101.3\,\text{J}$. $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$ or $0.0821\,\text{L atm mol}^{-1}\text{K}^{-1}$.
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Common Traps — Incorrect sign conventions for $q$ and $w$, confusing state vs. path functions, and unit conversion errors.

Vyyuha Quick Recall

To remember the First Law sign conventions: 'Q-W' for 'Quit Work'

Q — (Heat): Quickly Qain (positive) or Quit (negative) heat.
W — (Work): Work Won (positive, on system) or Work Wasted (negative, by system).

So, $\Delta U = Q_{gain} + W_{on}$ or $\Delta U = Q_{absorbed} + W_{compression}$.

For the formula $\Delta U = q + w$:

Q — is positive when Quickly Qaining heat (system absorbs).
W — is positive when Work is done Within (on the system).