Thermodynamic Terms and First Law
System, Surrounding, and Boundary
A system is any portion of the universe that is under study or observation. The rest of the universe outside the system is called the surroundings. In practice, the system consists of a specific amount of one or more substances — for example, one mole of oxygen confined in a cylinder fitted with a piston, or a cup of water on a table. Everything outside the defined system — the cylinder walls, the piston, the air, the table — belongs to the surroundings.
Gas in a cylinder: The gas is the system; the cylinder, piston, and everything outside are the surroundings
Cup of water: The water is the system; the surrounding air, the table it rests on, and the room are the surroundings
Chemical reaction: The reacting mixture (e.g., Zn and CuSO₄ solution in a flask) is the system; the flask, the air, and the laboratory are the surroundings
The boundary is the real or imaginary surface that separates the system from the surroundings. A boundary can be a physical wall — like the walls of a cylinder or the glass of a beaker — or it can be an imaginary dividing surface with no material existence. The boundary defines where the system ends and the surroundings begin, and it is across this boundary that heat and work are exchanged between the system and its surroundings.
State Functions
Every system exists in a particular condition called its state, defined by measurable properties such as temperature, pressure, and volume. When a process acts on the system, its state changes. The condition before the change is the initial state and the condition after the change is the final state. The difference between the final and initial values of any property is called the change in that property.
$$\Delta T = T_2 - T_1$$
Calculates the change in temperature between two states of a system
$\Delta T$=Change in temperature(K or °C)
$T_2$=Final temperature(K or °C)
$T_1$=Initial temperature(K or °C)
A state function is a macroscopic property of a system that has a definite value for any given state. The change in a state function depends only on the initial and final states — it is entirely independent of the path taken to get from one state to the other. By convention, state functions are represented by capital letters, such as P (pressure), T (temperature), V (volume), E (internal energy), and H (enthalpy).
$$\Delta V = V_2 - V_1$$
Calculates the change in volume between two states of a system
$\Delta V$=Change in volume(L or m³)
$V_2$=Final volume(L or m³)
$V_1$=Initial volume(L or m³)
$\Delta V > 0$
→System has expanded
$\Delta V < 0$
→System has been compressed
Common State Functions
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Pressure ($P$)
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Temperature ($T$)
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Volume ($V$)
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Internal energy ($E$)
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Enthalpy ($H$)
Internal Energy
The internal energy ($E$) of a system is the sum total of all possible kinds of energies possessed by the particles within it. It includes kinetic energy from the translational, rotational, and vibrational motions of molecules, and potential energy from all attractive forces between particles, including chemical bonds and van der Waals forces. Internal energy is itself a state function, meaning only its change ($\Delta E$) between states is measurable.
Kinetic energy contributions: Translational motion (movement through space), rotational motion (spinning about an axis), and vibrational motion (oscillation of atoms within bonds)
Potential energy contributions: All types of bonds (covalent, ionic, metallic, hydrogen) and intermolecular forces (van der Waals forces, dipole-dipole interactions)
Molecular Motions Contributing to Internal Energy
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All molecules: translational motion
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Diatomic and larger molecules: rotational motion
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Diatomic and larger molecules: vibrational motion
Although it is not possible to measure the absolute value of the internal energy of a system, it is possible to measure the change in internal energy ($\Delta E$) for any change in the state of the system. This is because $\Delta E$ is a state function, and its value depends only on the initial and final states. Energy can be transferred into or out of a system through two fundamental mechanisms: heat and work.
Absolute internal energy cannot be measured: There is no reference point for zero internal energy
Change in internal energy is measurable: $\Delta E = E_2 - E_1$ is determined by the initial and final states alone
Energy transfer occurs via heat and work: These are the only two ways a system exchanges energy with its surroundings
Heat and Work
Heat is the quantity of energy that flows across the boundary of a system during a change in its state, driven by a temperature difference between the system and the surroundings. Heat is denoted by the symbol $q$. Crucially, heat is not a property of a system and is therefore NOT a state function — the amount of heat transferred depends on the path taken.
Positive $q$: Heat is absorbed by the system from the surroundings (energy enters the system)
Negative $q$: Heat is released by the system to the surroundings (energy leaves the system)
Work is a form of energy transfer defined as the product of force and distance ($W = F \times S$). In thermochemistry, the most common type is pressure-volume work, which occurs when a system expands or contracts against an external pressure. For example, when a gas is evolved during a chemical reaction (such as $CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$), the gas does work by pushing back the atmosphere. Like heat, work is not a state function.
$$W = -P\Delta V$$
Calculates the pressure-volume work done by a system expanding against (or being compressed by) an external pressure
$W$=Work done (negative when done by the system, positive when done on the system)(J or kJ)
$P$=External pressure acting on the system(Pa or atm)
$\Delta V$=Change in volume of the system ($V_f - V_i$)(L or m³)
$\Delta V > 0$ (expansion)
→$W$ is negative — the system does work on the surroundings
$\Delta V < 0$ (compression)
→$W$ is positive — the surroundings do work on the system
$\Delta V = 0$ (constant volume)
→$W = 0$ — no pressure-volume work is done
Positive $W$: Work is done on the system by the surroundings (energy enters the system, e.g., compression)
Negative $W$: Work is done by the system on the surroundings (energy leaves the system, e.g., expansion)
First Law of Thermodynamics
The first law of thermodynamics states that energy can neither be created nor destroyed, but can only be changed from one form to another. This is also known as the law of conservation of energy. A system cannot create or destroy energy — it can only exchange energy with its surroundings through heat and work. The total energy of the system and its surroundings always remains constant.
$$\Delta E = q + w$$
States that the change in a system's internal energy equals the net heat absorbed plus the net work done on the system
$\Delta E$=Change in internal energy of the system(J or kJ)
$q$=Heat absorbed by the system (positive if absorbed, negative if released)(J or kJ)
$w$=Work done on the system (positive if done on, negative if done by)(J or kJ)
$\Delta E > 0$
→System gains energy overall
$\Delta E < 0$
→System loses energy overall
When the pressure-volume work is the only form of work, the first law becomes $\Delta E = q - P\Delta V$. If the volume of the system is held constant (as in a sealed rigid container), then $\Delta V = 0$ and no pressure-volume work is done. Under these conditions, the entire energy change of the system is due to heat transfer alone.
$$\Delta E = q_v$$
At constant volume, the change in internal energy equals the heat absorbed by the system
$\Delta E$=Change in internal energy(J or kJ)
$q_v$=Heat absorbed at constant volume(J or kJ)
Constant volume condition: No expansion or compression occurs, so $P\Delta V = 0$
Practical significance: Reactions carried out in sealed containers (e.g., bomb calorimeters) occur at constant volume
Enthalpy
Enthalpy ($H$) is a thermodynamic property of a system defined as the sum of its internal energy and the product of its pressure and volume. Like internal energy, enthalpy is a state function and is measured in joules. While the absolute value of enthalpy cannot be determined, the change in enthalpy ($\Delta H$) for a process can be measured and is extremely useful in chemistry.
$$H = E + PV$$
Defines enthalpy as internal energy plus the pressure-volume product of the system
$H$=Enthalpy of the system(J or kJ)
$E$=Internal energy of the system(J or kJ)
$P$=Pressure of the system(Pa or atm)
$V$=Volume of the system(L or m³)
When a system absorbs heat at constant pressure, part of the energy increases the internal energy and the rest is used to do pressure-volume work as the system expands. The change in enthalpy ($\Delta H$) accounts for both effects. By combining the first law of thermodynamics with the enthalpy definition and applying the constant-pressure condition, it can be shown that the change in enthalpy equals the heat of reaction at constant pressure.
$$\Delta H = q_p$$
At constant pressure, the change in enthalpy equals the heat exchanged by the system
$\Delta H$=Change in enthalpy of the system(J or kJ)
$q_p$=Heat absorbed or released at constant pressure(J or kJ)
Most reactions occur at constant pressure: Open beakers and flasks are at atmospheric pressure, so $\Delta H = q_p$ is more commonly used than $\Delta E$
Derivation: From $\Delta E = q - P\Delta V$ and $\Delta H = \Delta E + P\Delta V$, substituting gives $\Delta H = q - P\Delta V + P\Delta V = q$ at constant pressure
Chemical processes are classified by the sign of $\Delta H$. In an exothermic process, the system releases heat to the surroundings and $\Delta H$ is negative. In an endothermic process, the system absorbs heat from the surroundings and $\Delta H$ is positive. The magnitude of $\Delta H$ tells us how much energy is transferred during the process.
Exothermic ($\Delta H < 0$): Products have lower enthalpy than reactants — energy is released to the surroundings
Endothermic ($\Delta H > 0$): Products have higher enthalpy than reactants — energy is absorbed from the surroundings
Exothermic vs Endothermic
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Exothermic: $H_{\text{products}} < H_{\text{reactants}}$, $\Delta H$ is negative, surroundings get warmer
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Endothermic: $H_{\text{products}} > H_{\text{reactants}}$, $\Delta H$ is positive, surroundings get cooler
The change in enthalpy and the change in internal energy are related by $\Delta H = \Delta E + P\Delta V$. For gaseous reactions where the number of moles changes, the $P\Delta V$ term is significant. Using the ideal gas equation, $P\Delta V$ can be expressed in terms of the change in moles of gas. For reactions involving only liquids and solids, volume changes are negligible, so $\Delta H \approx \Delta E$.
$$P\Delta V = \Delta nRT$$
Expresses pressure-volume work in terms of the change in moles of gas during a reaction at constant temperature
$P\Delta V$=Pressure-volume work done by the system(J)
$\Delta n$=Change in moles of gas (moles of gaseous products minus moles of gaseous reactants)(mol)
$R$=Universal gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)(J K⁻¹ mol⁻¹)
$T$=Absolute temperature(K)
$\Delta n = 0$
→$P\Delta V = 0$, so $\Delta H = \Delta E$
Only solids and liquids involved
→$\Delta V \approx 0$, so $\Delta H \approx \Delta E$
For gases: Calculate $\Delta n = n_{\text{gaseous products}} - n_{\text{gaseous reactants}}$ and use $\Delta H = \Delta E + \Delta nRT$
For solids and liquids: Volume changes are negligible, so $\Delta H$ and $\Delta E$ are approximately equal
Heat Capacity and Calorimetry
The heat capacity ($C$) of a body or system is the quantity of heat required to change its temperature by 1 kelvin. When the mass of the substance is specified, the specific heat ($s$) is used — the heat required to raise the temperature of 1 gram of a substance by 1 kelvin. The product of mass and specific heat gives the total heat capacity of a system.
$$q = ms\Delta T$$
Calculates the heat absorbed or released by a substance using its mass, specific heat, and temperature change
$q$=Heat energy transferred (positive if absorbed, negative if released)(J or kJ)
$m$=Mass of the substance(g)
$s$=Specific heat capacity(J g⁻¹ K⁻¹)
$\Delta T$=Change in temperature ($T_f - T_i$)(K)
$C = ms$
→Total heat capacity can replace the product $ms$, giving $q = C\Delta T$
Key Heat Capacity Relationships
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$q = ms\Delta T$ (using mass and specific heat)
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$q = C\Delta T$ (using total heat capacity, where $C = ms$)
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Specific heat of water = $4.2 \text{ J g}^{-1} \text{ K}^{-1}$
Calorimeter Types
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Glass calorimeter: Insulated container with thermometer and stirrer — used for reactions in solution (neutralization, dissolution)
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Bomb calorimeter: Strong steel vessel with electrical ignition — used for combustion reactions at constant volume
Calorimetry is the experimental technique for measuring the heat absorbed or released during a chemical reaction. A calorimeter is an insulated vessel that prevents heat exchange with the surroundings. Reactants are mixed inside the calorimeter in stoichiometric amounts, and the temperature change is recorded before and after the reaction. Using the temperature change, the mass of the solution, and the specific heat, the quantity of heat transferred can be calculated using $q = ms\Delta T$.
Glass calorimeter procedure: Place reactants in the insulated container, stir continuously, record initial and final temperatures, then calculate $q = ms\Delta T$
Bomb calorimeter procedure: Place a known mass of sample in the steel bomb, pressurize with oxygen (~20 atm), immerse in water, ignite electrically, and record the temperature rise of the surrounding water. Use $q = C\Delta T$ where $C$ is the known heat capacity of the entire calorimeter assembly
Application Strategy
Worked Examples
Problem Statement
When 2.00 moles of $H_2$ and 1.00 mole of $O_2$ react at 100°C and 1 atm pressure to produce 2.00 moles of gaseous water, 484.5 kJ of energy are evolved. Calculate (a) $\Delta H$ and (b) $\Delta E$ for the production of one mole of $H_2O_{(g)}$.
enthalpyfirst_law
Strategic Toolkit
$$\Delta H = \Delta E + P\Delta V$$
Step 1
Calculate $\Delta H$ per mole of water
The reaction occurs at constant pressure, so the energy evolved equals the enthalpy change. The total enthalpy change for forming 2 moles of water vapour is -484.5 kJ. Divide by 2 to get the value per mole.
$$\Delta H = \frac{-484.5\text{ kJ}}{2\text{ mol}} = -242.2\text{ kJ mol}^{-1}$$
Step 2
Calculate the change in moles of gas
From the balanced equation, count the moles of gaseous products and gaseous reactants. The change in moles of gas determines how much pressure-volume work is involved in the reaction.
$$\Delta n = n_{\text{products}} - n_{\text{reactants}} = 2 - 3 = -1\text{ mol}$$
Step 3
Calculate $P\Delta V$ using the ideal gas equation
Apply the relationship $P\Delta V = \Delta nRT$ using the universal gas constant and the absolute temperature of 373 K. This gives the total pressure-volume work for the formation of 2 moles of water.
$$P\Delta V = \Delta nRT = (-1)(8.314\text{ J K}^{-1}\text{ mol}^{-1})(373\text{ K}) = -3100\text{ J} = -3.10\text{ kJ}$$
Step 4
Scale $P\Delta V$ to per mole of water
The calculated $P\Delta V$ corresponds to the formation of 2 moles of water vapour. Divide by 2 to find the work per mole of water produced.
$$P\Delta V \text{ per mole} = \frac{-3.10\text{ kJ}}{2\text{ mol}} = -1.55\text{ kJ mol}^{-1}$$
Step 5
Calculate $\Delta E$ from $\Delta H$ and $P\Delta V$
Rearrange the equation $\Delta H = \Delta E + P\Delta V$ to solve for $\Delta E$. Substitute the values and note that subtracting a negative $P\Delta V$ adds to the result.
$$\Delta E = \Delta H - P\Delta V = -242.2 - (-1.55) = -242.2 + 1.55 = -240.6\text{ kJ mol}^{-1}$$
Problem Statement
Neutralization of 100 cm³ of 0.5 M NaOH at 25°C with 100 cm³ of 0.5 M HCl at 25°C raised the temperature of the reaction mixture to 28.5°C. Calculate the enthalpy of neutralization. (Specific heat of water = 4.2 J K⁻¹ g⁻¹)
heat_capacityenthalpy
Strategic Toolkit
$$q = ms\Delta T$$
Step 1
Determine total mass and temperature change
The total volume of the solution is 200 cm³. With the density of water approximately 1 g cm⁻³, the total mass is 200 g. The temperature rises from 25.0°C to 28.5°C.
$$m = 200\text{ g}, \quad \Delta T = 28.5 - 25.0 = 3.5\text{ K}$$
Step 2
Calculate moles of acid and base
A 0.5 M solution contains 0.5 moles of solute per 1000 cm³. For 100 cm³ of each solution, calculate the number of moles present. These react in a 1:1 ratio.
$$n = \frac{0.5 \times 100}{1000} = 0.05\text{ mol of each}$$
Step 3
Calculate total heat evolved
Use the calorimetry formula to find the total heat released. Since the reaction releases heat (exothermic), the value is negative.
$$q = ms\Delta T = (200\text{ g})(4.2\text{ J g}^{-1}\text{ K}^{-1})(3.5\text{ K}) = 2940\text{ J} = 2.94\text{ kJ}$$
Step 4
Calculate enthalpy of neutralization per mole
Divide the total heat by the number of moles (0.05 mol) to get the enthalpy change per mole of water formed. The negative sign indicates an exothermic process.
$$\Delta H_n^\circ = \frac{-2.94\text{ kJ}}{0.05\text{ mol}} = -58.8\text{ kJ mol}^{-1}$$
Problem Statement
10.16 g of graphite is burnt in a bomb calorimeter and the temperature rise recorded is 3.87 K. Calculate the enthalpy of combustion of graphite, if the heat capacity of the calorimeter (bomb, water, etc.) is 86.02 kJ K⁻¹.
heat_capacityenthalpy
Strategic Toolkit
$$q = C\Delta T$$
Step 1
Calculate total heat absorbed by the calorimeter
The calorimeter absorbs all the heat released by the combustion. Use the heat capacity formula with the calorimeter's known heat capacity and the measured temperature rise.
$$q = C\Delta T = (86.02\text{ kJ K}^{-1})(3.87\text{ K}) = 332.9\text{ kJ}$$
Step 2
Calculate moles of graphite burnt
Convert the mass of graphite to moles using the molar mass of carbon (12 g mol⁻¹).
$$n = \frac{10.16\text{ g}}{12\text{ g mol}^{-1}} = 0.847\text{ mol}$$
Step 3
Calculate enthalpy of combustion per mole
Divide the total heat evolved by the number of moles to get the molar enthalpy of combustion. The negative sign indicates that heat is released (exothermic process).
$$\Delta H_c^\circ = \frac{-332.9\text{ kJ}}{0.847\text{ mol}} = -393\text{ kJ mol}^{-1}$$
Memorize formulas
First Law of Thermodynamics
Relates change in internal energy to heat absorbed and work done on the system. q is positive if heat enters, w is positive if work is done on the system. Units: J or kJ.
Formula
$$\Delta E = q + w$$
Pressure-Volume Work
Work done by expansion against external pressure. Negative sign means expansion (W < 0) and compression (W > 0). Units: J or kJ, with P in Pa and V in m³.
Formula
$$W = -P\Delta V$$
Internal Energy at Constant Volume
When volume is held constant, no PV work is done, so all heat transfer changes internal energy. Used in bomb calorimetry.
Formula
$$\Delta E = q_v$$
Enthalpy Definition
Enthalpy combines internal energy with the PV term. It is a state function measured in J or kJ. Absolute H is not measurable, only $\Delta H$.
Formula
$$H = E + PV$$
Enthalpy at Constant Pressure
At constant pressure, the change in enthalpy equals the heat of reaction. Most chemical reactions occur at constant (atmospheric) pressure.
Formula
$$\Delta H = q_p$$
PV Work via Ideal Gas Equation
Converts PV work to moles using the ideal gas law. $\Delta n$ is moles of gaseous products minus moles of gaseous reactants. R = 8.314 J K⁻¹ mol⁻¹.
Formula
$$P\Delta V = \Delta nRT$$
Heat from Mass and Specific Heat
Calculates heat transfer using mass, specific heat, and temperature change. For water, s = 4.2 J g⁻¹ K⁻¹. Also written as q = C$\Delta T$ where C = ms.
Formula
$$q = ms\Delta T$$
Review
1
In thermodynamics, the portion of the universe that is under investigation is called the:
Ans
System
Review
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Identify thermodynamic definitions
A system is the portion of matter or space chosen for study, separated from the rest of the universe by a real or imaginary boundary.
System = portion under study
2
A sealed thermos flask containing hot coffee, which does not allow heat or matter to be exchanged with the surroundings, is an example of:
Ans
A closed system
Review
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Classify system types
Open system: exchanges both matter and energy. Closed system: exchanges energy only. Isolated system: exchanges neither.
Sealed flask allows heat flow but no matter flow → closed system
3
Which of the following is a state function?
Ans
Internal energy
Review
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Recall the definition of a state function
A state function has a value that depends only on the initial and final states, independent of the pathway between them.
State functions: internal energy, enthalpy, pressure, volume, temperature
Identify path functions
Path functions depend on the specific process or pathway taken between states.
Path functions: heat (q) and work (w)
4
Which of the following is NOT a state function?
Ans
Heat
Review
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Check each option against state function criteria
Enthalpy, temperature, and volume all have fixed values for a given state. Heat transferred depends on the pathway.
Heat is the only path function listed
5
A system absorbs 500 J of heat and performs 200 J of work on its surroundings. What is the change in internal energy?
Ans
300 J
Review
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Assign signs to q and w
Heat absorbed by system: $q = +500$ J. Work done by system on surroundings: $w = -200$ J.
$$q = +500 \text{ J}, \quad w = -200 \text{ J}$$
Apply the first law of thermodynamics
Substitute values into the first law equation.
$$\Delta E = q + w = 500 + (-200) = 300 \text{ J}$$
6
A gas expands against a constant external pressure of 2 atm from 5 L to 10 L. The work done by the surroundings on the gas is: (1 L·atm = 101.3 J)
Ans
-1013 J
Review
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Calculate the volume change
$$\Delta V = V_f - V_i = 10 - 5 = 5 \text{ L}$$
$$\Delta V = 5 \text{ L}$$
Apply the PV work formula
$$w = -P\Delta V = -2 \times 5 = -10 \text{ L·atm}$$
$$w = -10 \text{ L·atm}$$
Convert to joules
$$w = -10 \times 101.3 = -1013 \text{ J}$$
$$w = -1013 \text{ J}$$
7
For a process carried out at constant volume, the change in internal energy equals:
Ans
The heat absorbed or released
Review
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Evaluate PV work at constant volume
Since $\Delta V = 0$, the work term vanishes.
$$w = -P\Delta V = -P(0) = 0$$
Simplify the first law
Substituting $w = 0$ into $\Delta E = q + w$.
$$\Delta E = q_v$$
8
The enthalpy ($H$) of a system is mathematically defined as:
Ans
$$H = E + PV$$
Review
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Write the enthalpy definition
Enthalpy is a useful state function for processes at constant pressure.
$$H = E + PV$$
9
For a reaction in which the total number of moles of gas increases ($\Delta n > 0$) carried out at constant temperature and pressure, which relationship between $\Delta H$ and $\Delta E$ is correct?
Ans
$$\Delta H > \Delta E$$
Review
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Start with the enthalpy change relation
$$\Delta H = \Delta E + P\Delta V$$
$$\Delta H = \Delta E + \Delta nRT$$
Evaluate the sign of $\Delta nRT$
Given $\Delta n > 0$ and $RT > 0$, the product $\Delta nRT$ is positive.
$$\Delta nRT > 0 \implies \Delta H > \Delta E$$
10
A process in which heat flows from the system to the surroundings is classified as:
Ans
Exothermic
Review
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Identify the direction of heat flow
Heat flows from system to surroundings means the system loses thermal energy.
$$q < 0 \text{ (system)} \implies \text{exothermic}$$
11
How much heat is required to raise the temperature of 200 g of water from 25°C to 75°C? (Specific heat of water = 4.18 J/g·°C)
Ans
41,800 J
Review
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Calculate the temperature change
$$\Delta T = T_f - T_i = 75 - 25 = 50°C$$
$$\Delta T = 50°C$$
Apply the heat formula
$$q = ms\Delta T = 200 \times 4.18 \times 50$$
$$q = 41{,}800 \text{ J}$$
12
In a bomb calorimeter experiment, the quantity measured directly is:
Ans
Internal energy change ($\Delta E$)
Review
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Recognize the calorimeter condition
A bomb calorimeter has a fixed-volume chamber, so $\Delta V = 0$.
$$\text{Constant volume} \implies w = 0$$
Apply the first law
$$\Delta E = q_v + 0 = q_v$$
Bomb calorimeter measures $\Delta E$ directly
13
A gas in a cylinder absorbs 800 J of heat and expands against a constant external pressure of 3 atm from 2 L to 6 L. What is the change in internal energy? (1 L·atm = 101.3 J)
Ans
-415.6 J
Review
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Calculate PV work
$$w = -P\Delta V = -3 \times (6 - 2) = -12 \text{ L·atm}$$
$$w = -12 \times 101.3 = -1215.6 \text{ J}$$
Apply the first law of thermodynamics
$$\Delta E = q + w = 800 + (-1215.6)$$
$$\Delta E = -415.6 \text{ J}$$
14
Everything in the universe outside the defined system that can interact with it is called the:
Ans
Surroundings
Review
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Define surroundings thermodynamically
The surroundings are the portion of the universe external to the system with which the system can interact.
Surroundings = everything outside the system
15
For the reaction $$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$ carried out at constant temperature and pressure, the relationship between $\Delta H$ and $\Delta E$ is:
Ans
$$\Delta H = \Delta E - 2RT$$
Review
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Calculate the change in moles of gas
$$\Delta n = 2 - (1 + 3) = 2 - 4 = -2$$
$$\Delta n = -2$$
Apply the enthalpy-internal energy relationship
$$\Delta H = \Delta E + \Delta nRT = \Delta E + (-2)RT$$
$$\Delta H = \Delta E - 2RT$$
16
The real or imaginary surface that separates the system from its surroundings is known as the:
Ans
Boundary
Review
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Define the boundary in thermodynamics
The boundary delineates what is inside the system from what is outside. It can be rigid, flexible, permeable, or impermeable.
Boundary = separating surface between system and surroundings
17
At constant pressure, the heat exchanged by the system equals the change in:
Ans
Enthalpy
Review
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State the constant-pressure relation
From the definition $H = E + PV$, at constant pressure: $\Delta H = \Delta E + P\Delta V$.
$$\Delta H = \Delta E + P\Delta V$$
Substitute the first law
Since $\Delta E = q_p + w = q_p - P\Delta V$, substituting gives $\Delta H = q_p$.
$$\Delta H = q_p$$
18
A reaction at constant pressure releases 150 kJ of heat and the system's volume decreases. Which statement is correct?
Ans
$$\Delta H = -150 \text{ kJ} \text{ and } \Delta E > -150 \text{ kJ}$$
Review
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Determine $\Delta H$ from heat at constant pressure
Heat released means $q_p < 0$, so $\Delta H = -150$ kJ.
$$\Delta H = -150 \text{ kJ}$$
Evaluate $\Delta E$ using $\Delta H = \Delta E + P\Delta V$
Volume decreases → $\Delta V < 0$ → $P\Delta V < 0$. So $\Delta E = \Delta H - P\Delta V = -150 - (\text{negative}) > -150$ kJ.
$$\Delta E > -150 \text{ kJ}$$
19
The specific heat of aluminium is 0.90 J/g·°C and that of water is 4.18 J/g·°C. If equal masses of both absorb the same quantity of heat, which statement correctly describes the temperature change?
Ans
Water's temperature rises less than aluminium's
Review
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Rearrange the heat equation for $\Delta T$
$$\Delta T = \frac{q}{ms}$$
$$\Delta T \propto \frac{1}{s}$$
Compare specific heats
Water has a higher specific heat (4.18) than aluminium (0.90), so for the same $q$ and $m$, water's $\Delta T$ is smaller.
$$\Delta T_{\text{water}} < \Delta T_{\text{aluminium}}$$
20
In a thermodynamic process, a system has $w = +350$ J and $q = -120$ J. Which description of the process is correct?
Ans
Heat is released and work is done on the system
Review
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Interpret the sign of work
$w = +350$ J is positive, meaning work is done on the system.
$$w > 0 \implies \text{work done on system}$$
Interpret the sign of heat
$q = -120$ J is negative, meaning heat is released by the system.
$$q < 0 \implies \text{heat released}$$
Verify with the first law
$$\Delta E = q + w = -120 + 350 = +230 \text{ J}$$
$$\Delta E = +230 \text{ J} \text{ (net energy gain)}$$
Question 1 of 20
0:00
1
In thermodynamics, the portion of the universe that is under investigation is called the: