Step 1: Theoretical Foundation of Heat Exchange

When two bodies of masses $m_1$ and $m_2$ at initial temperatures $T_1$ and $T_2$ (where $T_1 > T_2$) are brought into thermal contact, heat flows from the hotter body to the colder body until thermal equilibrium is reached at a common temperature $T_f$. [Per the Zeroth Law of Thermodynamics and the Principle of Calorimetry].

Step 2: Energy Conservation Equation

Assuming the system is isolated from the surroundings, the heat lost by the hotter body must equal the heat gained by the colder body:

$-Q_{lost} = Q_{gained}$

$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$

Where:

• $m_1, m_2$: Masses of the two bodies.
• $c_1, c_2$: Specific heat capacities of the materials.
• $T_f$: Final equilibrium temperature.

Step 3: Derivation of the Final Temperature

Expanding the equation to solve for $T_f$:

$m_1 c_1 T_1 - m_1 c_1 T_f = m_2 c_2 T_f - m_2 c_2 T_2$

$m_1 c_1 T_1 + m_2 c_2 T_2 = T_f (m_1 c_1 + m_2 c_2)$

$T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}$

Step 4: Analysis of the "Mean Temperature" Condition

The equilibrium temperature $T_f$ will equal the arithmetic mean $\frac{T_1 + T_2}{2}$ if and only if the denominator coefficients are identical, i.e., $m_1 c_1 = m_2 c_2$.

If the bodies have different masses ($m_1 \neq m_2$) or different materials (different specific heat capacities $c_1 \neq c_2$), then $m_1 c_1 \neq m_2 c_2$. Consequently, the final temperature is a weighted average rather than a simple arithmetic mean.

Step 5: Logical Conclusion

Since the thermal inertia (heat capacity, $C = mc$) of the two bodies is rarely equal, the temperature change per unit heat transfer differs for each body. Therefore, the equilibrium temperature is determined by the ratio of the products of mass and specific heat capacity, not by simple geometric midpoint logic.

Final Solution: The equilibrium temperature is a weighted average $T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}$. It only equals $(T_1 + T_2)/2$ if the heat capacities $m_1 c_1$ and $m_2 c_2$ are exactly equal, which is not a universal condition.

Step 1: Theoretical Foundation

The relationship between the change in internal energy ($\Delta U^{\ominus}$) and the change in enthalpy ($\Delta H^{\ominus}$) for a chemical reaction involving gases is defined by the first law of thermodynamics and the definition of enthalpy ($H = U + PV$):

$$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta (PV)$$

[Per the Ideal Gas Law $PV = nRT$], for a reaction at constant temperature ($T$):

$$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT$$

Where $\Delta n_g$ represents the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants.

Step 2: Defining the Combustion Reaction

The combustion of methane ($CH_4$) in oxygen is represented by the following balanced chemical equation:

$$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

Note: $H_2O$ is produced in the liquid phase ($l$) under standard conditions at $298 \, K$. Therefore, it is excluded from the gaseous mole count.

Step 3: Calculating Change in Gaseous Moles ($\Delta n_g$)

The change in the number of gaseous moles is calculated as follows:

$$\Delta n_g = \sum n_{g(\text{products})} - \sum n_{g(\text{reactants})}$$

$$\Delta n_g = (n_{CO_2}) - (n_{CH_4} + n_{O_2})$$

$$\Delta n_g = (1) - (1 + 2) = 1 - 3 = -2$$

Step 4: Deriving the Expression for $\Delta H^{\ominus}$

Substitute the given values into the enthalpy relationship formula:

Given: $\Delta U^{\ominus} = -X \, \text{kJ mol}^{-1}$

$$\Delta H^{\ominus} = (-X) + (-2)RT$$

$$\Delta H^{\ominus} = -X - 2RT$$

Visual Representation of Enthalpy vs Internal Energy

Step 5: Analysis of the Result

Since $\Delta n_g$ is negative ($-2$), the work done on the system by the atmosphere (or the work performed during the volume contraction) results in:

$$\Delta H^{\ominus} < \Delta U^{\ominus}$$

The enthalpy of combustion will be more negative than the internal energy change by a factor of $2RT$.

Final Solution: The value of $\Delta H^{\ominus}$ is $(-X - 2RT) \, \text{kJ mol}^{-1}$.

Theoretical Foundation: The Second Law of Thermodynamics

The behavior of an isolated system—a system that exchanges neither energy nor matter with its surroundings ($Q = 0$ and $W = 0$)—is governed by the Second Law of Thermodynamics. This law dictates the direction of spontaneous processes and the criterion for equilibrium.

Step 1: Defining the Entropy Criterion

According to the Clausius inequality, for any process, the change in the entropy of the universe ($\Delta S_{univ}$) must satisfy the condition:
$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \geq 0$
[Per the Second Law of Thermodynamics]

Step 2: Analyzing the Isolated System

For an isolated system, by definition, the system does not interact with its surroundings. Therefore:

• $\Delta Q_{surr} = 0$
• $\Delta S_{surr} = \frac{\Delta Q_{surr}}{T} = 0$

Step 3: Deriving the Entropy Constraint

Since $\Delta S_{surr} = 0$, the change in the entropy of the universe reduces to the change in the entropy of the system itself:
$\Delta S_{univ} = \Delta S_{sys} + 0 \geq 0$
Thus, for any process within an isolated system, the entropy must either increase or remain constant:
$\Delta S \geq 0$

Step 4: Interpretation of States

Final Solution: For an isolated system with $\Delta U = 0$, the change in entropy ($\Delta S$) must be greater than or equal to zero ($\Delta S \geq 0$). It increases for spontaneous processes and remains zero for reversible processes at equilibrium.

Step 1: Evaluation of Thermodynamic Stability

Thermodynamic stability of a compound is determined by its enthalpy of formation ($\Delta_{f}H^{\ominus}$). A compound is considered thermodynamically stable relative to its constituent elements if its enthalpy of formation is negative (exothermic). If the enthalpy of formation is positive (endothermic), the compound is thermodynamically unstable relative to its constituent elements.

Step 2: Analysis of the Given Enthalpy Data

We are provided with the following thermochemical equations:

• Reaction (1): $\frac{1}{2} N_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow NO(g)$; $\Delta_{r}H^{\ominus} = +90 \text{ kJ mol}^{-1}$
• Reaction (2): $NO(g) + \frac{1}{2} O_{2}(g) \rightarrow NO_{2}(g)$; $\Delta_{r}H^{\ominus} = -74 \text{ kJ mol}^{-1}$

Step 3: Determining the Stability of NO(g)

The enthalpy of formation for $NO(g)$ is given directly by Reaction (1), as it represents the formation of one mole of $NO$ from its constituent elements in their standard states [Per the definition of standard enthalpy of formation, $\Delta_{f}H^{\ominus}$].

Since $\Delta_{r}H^{\ominus} = +90 \text{ kJ mol}^{-1} > 0$, the formation of $NO(g)$ is an endothermic process. This indicates that $NO(g)$ possesses higher energy than its constituent elements ($N_2$ and $O_2$). Therefore, $NO(g)$ is thermodynamically unstable with respect to its elements at standard conditions.

Step 4: Comparative Stability Context

While $NO(g)$ is unstable, we can observe its reactivity via Reaction (2). By applying Hess's Law of Constant Heat Summation [Per Hess's Law], we can determine the enthalpy of formation for $NO_2(g)$:

The calculation reveals that $NO(g)$ tends to react spontaneously with oxygen to form $NO_2(g)$ because the process releases energy ($\Delta H = -74 \text{ kJ mol}^{-1}$), despite $NO_2$ itself remaining endothermic relative to its elements.

Step 5: Visual Representation of Energy Levels

Final Solution: The positive enthalpy of formation ($\Delta_{f}H^{\ominus} = +90 \text{ kJ mol}^{-1}$) confirms that $NO(g)$ is thermodynamically unstable with respect to its constituent elements at standard conditions.

Step 1: Analyzing the Initial Conditions

The system consists of two identical cylinders, $A$ and $B$, each having a volume $V$. Cylinder $A$ is filled with an ideal gas at standard temperature and pressure ($STP$), defined as $P_i = 1.013 \times 10^5 \, \text{Pa}$ and $T_i = 273.15 \, \text{K}$. Cylinder $B$ is completely evacuated, meaning its initial pressure $P_B = 0$ and it contains no mass.

Initial state variables for the system:

• Total Volume ($V_{total}$) = $V_A + V_B = V + V = 2V$
• Initial Pressure ($P_i$) = $1.013 \times 10^5 \, \text{Pa}$
• Initial Temperature ($T_i$) = $273.15 \, \text{K}$

Step 2: Evaluating the Thermodynamic Process

When the stopcock is opened, the gas undergoes free expansion into the vacuum of cylinder $B$. Since the system is thermally insulated, there is no heat exchange with the surroundings ($Q = 0$). Furthermore, because the expansion occurs into a vacuum, the gas does no work on the surroundings ($W = 0$).

By the First Law of Thermodynamics, $\Delta U = Q - W$. Substituting $Q = 0$ and $W = 0$, we find that $\Delta U = 0$. [Per the First Law of Thermodynamics, for an ideal gas, internal energy $U$ is a function of temperature only. Therefore, $\Delta T = 0$, meaning the process is isothermal].

Step 3: Calculating Final Pressure

Using the Ideal Gas Law $PV = nRT$, we observe that because $T$ remains constant and the amount of substance $n$ is conserved, the product of pressure and volume must be conserved [Boyle's Law]:

$P_i V_i = P_f V_f$

Substituting the known values:

$P_i V = P_f (2V)$

$P_f = \frac{P_i V}{2V} = \frac{P_i}{2}$

Step 4: Numerical Computation

$P_f = \frac{1.013 \times 10^5 \, \text{Pa}}{2}$

$P_f = 0.5065 \times 10^5 \, \text{Pa}$

$P_f \approx 5.065 \times 10^4 \, \text{Pa}$

Visual Representation of Expansion

Final Solution: The final pressure of the gas in both cylinders A and B is $0.5065 \times 10^5 \, \text{Pa}$ (or $50.65 \, \text{kPa}$).

Step 1: Analyzing the Thermodynamic Process

The formation of liquid water from its constituent elements in their standard states is represented by the chemical equation:

$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)$

For this reaction, the standard enthalpy of formation ($\Delta_{f}H^{\ominus}$) is given as $-286\text{ kJ mol}^{-1}$. This value represents the heat released by the system into the surroundings at constant pressure and standard temperature ($T = 298\text{ K}$). [Per the First Law of Thermodynamics, energy conserved implies $q_{surr} = -q_{sys}$].

Step 2: Calculating Heat Exchange with Surroundings

Under standard pressure ($1\text{ bar}$), the heat absorbed by the surroundings is equal in magnitude and opposite in sign to the enthalpy change of the system:

$q_{surr} = - \Delta_{f}H^{\ominus}_{sys}$

$q_{surr} = - (-286\text{ kJ mol}^{-1}) = +286\text{ kJ mol}^{-1}$

$q_{surr} = 286,000\text{ J mol}^{-1}$

Step 3: Determining Entropy Change of Surroundings

The entropy change of the surroundings ($\Delta S_{surr}$) is defined by the heat exchange divided by the absolute temperature at which the process occurs [Per the Second Law of Thermodynamics, for a process at constant temperature $T$]:

$\Delta S_{surr} = \frac{q_{surr}}{T}$

Given the standard condition $T = 298\text{ K}$:

$\Delta S_{surr} = \frac{286,000\text{ J mol}^{-1}}{298\text{ K}}$

Step 4: Final Arithmetic Computation

$\Delta S_{surr} \approx 959.73\text{ J K}^{-1}\text{ mol}^{-1}$

Visual Representation of Entropy Flow

Final Solution: The entropy change in the surroundings is $\Delta S_{surr} = +959.73\text{ J K}^{-1}\text{ mol}^{-1}$.

Theoretical Foundation: Enthalpy of Formation

In chemical thermodynamics, the enthalpy of formation ($\Delta H_f^\circ$) of a substance is defined as the change in enthalpy when one mole of a substance is formed from its constituent elements in their most stable forms at a specified temperature (usually $298.15 \, K$) and standard pressure ($1 \, bar$).

Step 1: The Reference State Definition

By international convention established by IUPAC (International Union of Pure and Applied Chemistry), the enthalpy of all elements in their standard reference states is defined as zero.

• Standard State: The pure form of the element at $1 \, bar$ pressure and a specified temperature (commonly $298 \, K$).
• Stable Form: The most thermodynamically stable allotrope of the element at the specified conditions (e.g., graphite for carbon, $O_2(g)$ for oxygen, $Fe(s)$ for iron).

Step 2: Mathematical Justification

The total enthalpy of a system is given by $H = U + PV$. Since absolute enthalpy cannot be measured, we measure the enthalpy relative to an arbitrary reference point. Setting the standard enthalpy of elements to zero simplifies the thermochemical calculations for compounds via Hess’s Law:

$\Delta_r H^\circ = \sum \nu_p \Delta_f H^\circ (\text{products}) - \sum \nu_r \Delta_f H^\circ (\text{reactants})$

[Per the Hess’s Law of Constant Heat Summation]

Step 3: Illustrative Representation

Step 4: Analytical Conclusion

Because there is no chemical reaction required to "form" an element from itself, the enthalpy change for the formation of an element in its standard state from itself is inherently zero. This convention serves as the thermodynamic "zero point" scale for all enthalpy measurements.

Final Solution: The enthalpies of all elements in their standard states are defined as zero.

Step 1: Establishing the Thermochemical Equations

To determine the enthalpy change for the atomization of $\text{CCl}_{4}(\text{g})$, we first define the standard enthalpy values provided:

• 1. $\text{C(s)} + 2\text{Cl}_{2}(\text{g}) \rightarrow \text{CCl}_{4}(\text{l}) \quad \Delta_{f}H^{\ominus} = -135.5 \text{ kJ mol}^{-1}$
• 2. $\text{CCl}_{4}(\text{l}) \rightarrow \text{CCl}_{4}(\text{g}) \quad \Delta_{\text{vap}}H^{\ominus} = 30.5 \text{ kJ mol}^{-1}$
• 3. $\text{C(s)} \rightarrow \text{C(g)} \quad \Delta_{a}H^{\ominus} = 715.0 \text{ kJ mol}^{-1}$
• 4. $\text{Cl}_{2}(\text{g}) \rightarrow 2\text{Cl(g)} \quad \Delta_{a}H^{\ominus} = 242.0 \text{ kJ mol}^{-1}$

Step 2: Calculating Enthalpy of Atomization for $\text{CCl}_{4}(\text{g})$

The target reaction is: $\text{CCl}_{4}(\text{g}) \rightarrow \text{C(g)} + 4\text{Cl(g)}$. We apply Hess's Law [Per Hess's Law of Constant Heat Summation] by combining the equations:

$\Delta_{a}H^{\ominus}(\text{CCl}_{4}) = [\Delta_{a}H^{\ominus}(\text{C}) + 2\Delta_{a}H^{\ominus}(\text{Cl}_{2})] - [\Delta_{f}H^{\ominus}(\text{CCl}_{4}) + \Delta_{\text{vap}}H^{\ominus}(\text{CCl}_{4})]$

Substituting the given values:

$\Delta_{a}H^{\ominus} = [715.0 + 2(242.0)] - [-135.5 + 30.5]$

$\Delta_{a}H^{\ominus} = [715.0 + 484.0] - [-105.0]$

$\Delta_{a}H^{\ominus} = 1199.0 + 105.0 = 1304.0 \text{ kJ mol}^{-1}$

Step 3: Calculating Bond Enthalpy of C–Cl

In $\text{CCl}_{4}$ gas, there are 4 identical C–Cl bonds. The enthalpy of atomization corresponds to the energy required to break all 4 bonds [By Definition of Bond Dissociation Enthalpy].

$\text{Bond Enthalpy (C-Cl)} = \frac{\Delta_{a}H^{\ominus}(\text{CCl}_{4})}{4}$

$\text{Bond Enthalpy (C-Cl)} = \frac{1304.0}{4} = 326.0 \text{ kJ mol}^{-1}$

Visual Representation of Bond Cleavage

Final Solution: The enthalpy change for the process is $1304.0 \text{ kJ mol}^{-1}$, and the average C–Cl bond enthalpy is $326.0 \text{ kJ mol}^{-1}$.

Given Variables & Initial Setup

To determine the change in internal energy ($\Delta U$), we utilize the First Law of Thermodynamics, which establishes the principle of conservation of energy for a thermodynamic system.

• Heat absorbed by the system ($Q$) = $+701\ J$ [By convention, heat added to the system is positive].
• Work done by the system ($W$) = $+394\ J$ [By convention, work done by the system is positive].

Step 1: Applying the First Law of Thermodynamics

The First Law of Thermodynamics is mathematically expressed as:

$\Delta U = Q - W$

Where:

• $\Delta U$ represents the change in the internal energy of the system.
• $Q$ is the net heat transferred to the system.
• $W$ is the net work done by the system on its surroundings.

Step 2: Substitution and Calculation

Substituting the given values into the equation:

$\Delta U = 701\ J - 394\ J$

Step 3: Analytical Conclusion

Performing the arithmetic subtraction:

$\Delta U = 307\ J$

Since the resulting value for $\Delta U$ is positive, it indicates an increase in the internal energy of the system. This confirms that the energy absorbed as heat was greater than the energy expended as work, leading to a net accumulation of internal energy within the system.

Final Solution: The change in internal energy for the process is $307\ J$.

Step 1: Establishing the Governing Thermodynamic Relation

The relationship between the standard Gibbs free energy change ($\Delta G^{\ominus}$) and the equilibrium constant ($K_{eq}$) for a chemical reaction is defined by the following fundamental thermodynamic equation:

$\Delta G^{\ominus} = -RT \ln(K_{eq})$

[Per the definition of the reaction quotient at equilibrium, where $\Delta G = \Delta G^{\ominus} + RT \ln(Q) = 0$ and $Q = K_{eq}$]

Step 2: Identification of Given Variables

• Gas Constant ($R$) = $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
• Temperature ($T$) = $300 \text{ K}$
• Equilibrium Constant ($K_{eq}$) = $10$

Step 3: Conversion to Natural Logarithm

To compute the value, we express the natural logarithm of the constant:

$\ln(10) \approx 2.303$

Step 4: Substitution and Calculation

Substitute the known values into the governing equation:

$\Delta G^{\ominus} = -(8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (300 \text{ K}) \times \ln(10)$

$\Delta G^{\ominus} = -(8.314) \times (300) \times (2.303)$

$\Delta G^{\ominus} = -2494.2 \times 2.303$

$\Delta G^{\ominus} = -5744.1426 \text{ J mol}^{-1}$

Step 5: Final Unit Conversion and Precision

Converting the result into kilojoules ($1 \text{ kJ} = 1000 \text{ J}$):

$\Delta G^{\ominus} = -5.744 \text{ kJ mol}^{-1}$

Final Solution: The standard Gibbs free energy change ($\Delta G^{\ominus}$) for the reaction is $-5744.14 \text{ J mol}^{-1}$ or $-5.744 \text{ kJ mol}^{-1}$.

Step 1: Identifying Given Parameters

Based on the provided information, we extract the variables for the thermodynamic system:

• Rate of heat supplied to the system ($dQ/dt$): $100\text{ W} = 100\text{ J/s}$
• Rate of work done by the system ($dW/dt$): $75\text{ J/s}$

Step 2: Theoretical Framework

The First Law of Thermodynamics, derived from the Law of Conservation of Energy, states that the heat supplied to a system is equal to the sum of the increase in internal energy and the work done by the system. Mathematically, this is expressed as:

$dQ = dU + dW$

[Per the First Law of Thermodynamics, where $dQ$ is heat supplied, $dU$ is the change in internal energy, and $dW$ is the work done by the system.]

Step 3: Calculating the Rate of Change

To find the rate of change of internal energy ($dU/dt$), we differentiate the expression with respect to time ($t$):

$\frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}$

Substituting the given values into the equation:

$\frac{dU}{dt} = 100\text{ J/s} - 75\text{ J/s}$

$\frac{dU}{dt} = 25\text{ J/s}$

Step 4: Visualizing the Energy Flux

Final Solution: The internal energy of the system is increasing at a rate of $25\text{ J/s}$ (or $25\text{ Watts}$).

Step 1: Given Variables & Initial Setup

The standard molar internal energy change ($\Delta U^{\ominus}$) and the standard entropy change ($\Delta S^{\ominus}$) for the reaction $2A(g) + B(g) \rightarrow 2D(g)$ are provided as follows:

• $\Delta U^{\ominus} = -10.5 \text{ kJ} = -10500 \text{ J}$
• $\Delta S^{\ominus} = -44.1 \text{ J K}^{-1}$
• Temperature ($T$) = $298 \text{ K}$ (Standard ambient temperature is assumed for $\Delta^{\ominus}$ calculations)
• $\Delta n_g$ (change in number of moles of gas) = $(\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (2 + 1) = -1 \text{ mol}$
• Gas constant ($R$) = $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

Step 2: Calculating Standard Enthalpy Change ($\Delta H^{\ominus}$ )

The relationship between $\Delta H^{\ominus}$ and $\Delta U^{\ominus}$ is defined by the enthalpy definition: $\Delta H = \Delta U + \Delta(PV)$. For ideal gases, this transitions to the equation [Per the First Law of Thermodynamics and the Equation of State for Ideal Gases]:

$$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT$$

Substituting the values:

$$\Delta H^{\ominus} = -10500 \text{ J} + (-1) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$$

$$\Delta H^{\ominus} = -10500 - 2477.57 \text{ J} = -12977.57 \text{ J} \approx -12.98 \text{ kJ}$$

Step 3: Calculating Standard Gibbs Free Energy Change ($\Delta G^{\ominus}$)

The standard Gibbs free energy is related to enthalpy and entropy by the Gibbs-Helmholtz equation [Per the Second Law of Thermodynamics]:

$$\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$$

Substituting the calculated values:

$$\Delta G^{\ominus} = -12977.57 \text{ J} - (298 \text{ K}) \times (-44.1 \text{ J K}^{-1})$$

$$\Delta G^{\ominus} = -12977.57 \text{ J} + 13141.8 \text{ J}$$

$$\Delta G^{\ominus} = +164.23 \text{ J} \approx +0.164 \text{ kJ}$$

Step 4: Spontaneity Prediction

According to the criteria for spontaneity at constant temperature and pressure:

• If $\Delta G < 0$, the reaction is spontaneous.
• If $\Delta G > 0$, the reaction is non-spontaneous.

Since the calculated value is $\Delta G^{\ominus} = +0.164 \text{ kJ}$, which is greater than zero, the reaction is predicted to be non-spontaneous under standard conditions at 298 K.

Final Solution: The standard Gibbs free energy change ($\Delta G^{\ominus}$) is approximately $+0.164 \text{ kJ}$, indicating that the reaction is non-spontaneous under standard conditions.

Step 1: Analyzing the Adiabatic Process (State A to B)

In an adiabatic process, the heat exchange $Q$ with the surroundings is zero by definition. [Per the First Law of Thermodynamics: $\Delta U = Q - W$, where $W$ is the work done by the system].

Given that work is done on the system, the work done by the system ($W_{ad}$) is negative:

$W_{ad} = -22.3 \, \text{J}$

Applying the First Law of Thermodynamics ($\Delta U = Q - W$):

$\Delta U = 0 - (-22.3 \, \text{J})$

$\Delta U = 22.3 \, \text{J}$

Since the internal energy is a state function, the change in internal energy ($\Delta U$) for the transition from state A to state B is path-independent. Thus, $\Delta U = 22.3 \, \text{J}$ for any process connecting state A to state B.

Step 2: Analyzing the Second Process (Heat Exchange)

In the second process, the gas moves from state A to state B with a net heat absorption ($Q_{new}$):

$Q_{new} = 9.35 \, \text{cal}$

Converting calories to Joules using the given conversion factor ($1 \, \text{cal} = 4.19 \, \text{J}$):

$Q_{new} = 9.35 \times 4.19 \, \text{J}$

$Q_{new} = 39.1765 \, \text{J}$

Step 3: Calculating Work Done in the Second Process

Using the First Law of Thermodynamics for the second process:

$\Delta U = Q_{new} - W_{new}$

Substituting the known values of $\Delta U$ and $Q_{new}$:

$22.3 \, \text{J} = 39.1765 \, \text{J} - W_{new}$

$W_{new} = 39.1765 \, \text{J} - 22.3 \, \text{J}$

$W_{new} = 16.8765 \, \text{J}$

Final Solution: The net work done by the system in the second process is 16.88 J (rounded to two decimal places).

Theoretical Definition & Criterion

A thermodynamic state function is defined as a property of a system whose value depends exclusively on the current state of the system (characterized by variables such as pressure $P$, volume $V$, temperature $T$, and internal energy $U$), rather than the historical path taken to reach that state.

Logical Analysis of State Functions

To identify the nature of a state function, we evaluate the mathematical properties of the differentials involved [Per the Path Independence Axiom]:

• Differential Property: If $f(P, V, T)$ is a state function, its infinitesimal change $df$ is an exact differential. The integral of $df$ around any closed path is zero: $\oint df = 0$.
• Path Independence: The change in a state function $\Delta f$ between two states, $1$ and $2$, is given by $\int_{1}^{2} df = f_2 - f_1$. This result is invariant regardless of the path chosen.
• Path Functions (Contrast): Quantities such as heat ($Q$) and work ($W$) are path functions because their values depend on the specific sequence of thermodynamic processes (isobaric, isothermal, adiabatic, etc.) used to transition the system.

Comparison Table: State vs. Path Functions

Final Solution: A thermodynamic state function is a quantity whose value depends only on the state of the system and is independent of the path by which that state is attained.

Theoretical Foundation of Adiabatic Processes

An adiabatic process is defined as a thermodynamic transformation of a system in which there is no heat exchange between the system and its surroundings. This implies that the net flow of heat ($Q$) across the boundaries of the system is zero.

Step 1: Identifying the Mathematical Constraint

By the First Law of Thermodynamics, the change in internal energy ($\Delta U$) of a system is given by the relation:

$\Delta U = Q - W$

Where:

• $Q$ represents the heat exchanged.
• $W$ represents the work done by the system.

For a process to satisfy adiabatic conditions, we define the constraint as $Q = 0$ [Per the definition of adiabatic insulation]. Substituting this into the First Law yields:

$\Delta U = -W$

Step 2: Differential Representation

In infinitesimal form, the adiabatic condition is expressed as:

$dQ = 0$

This necessitates that for any infinitesimal change in the state of the gas, no differential heat energy is absorbed or rejected [Per the Second Law of Thermodynamics, implies no entropy change due to heat transfer, $dS = dQ_{rev}/T = 0$].

Step 3: State Variable Relationship

For an ideal gas undergoing a reversible adiabatic process, the state variables pressure ($P$) and volume ($V$) are governed by the Poisson equation:

$PV^{\gamma} = \text{constant}$

Where $\gamma = \frac{C_p}{C_v}$ is the adiabatic index [Derived from the kinetic theory of gases].

Visual Representation of Adiabatic vs Isothermal Processes

Summary of Conditions

Final Solution: The correct condition for an adiabatic process is $dQ = 0$, meaning there is no exchange of heat between the system and its environment during the process.

Step 1: Identification of the Target Reaction

The standard enthalpy of formation ($\Delta_{f}H^{\ominus}$) for methanol ($\text{CH}_{3}\text{OH(l)}$) corresponds to the enthalpy change of the reaction where one mole of the substance is formed from its constituent elements in their most stable states at standard conditions ($298\text{ K}$, $1\text{ bar}$).

Target Equation: $\text{C(graphite)} + 2\text{H}_{2}\text{(g)} + \frac{1}{2}\text{O}_{2}\text{(g)} \rightarrow \text{CH}_{3}\text{OH(l)}$

Step 2: Analysis of Given Thermochemical Data

We are provided with the following equations:

1. $\text{CH}_{3}\text{OH(l)} + \frac{3}{2}\text{O}_{2}\text{(g)} \rightarrow \text{CO}_{2}\text{(g)} + 2\text{H}_{2}\text{O(l)}$ ; $\Delta_{r}H_{1}^{\ominus} = -726 \text{ kJ mol}^{-1}$
2. $\text{C(graphite)} + \text{O}_{2}\text{(g)} \rightarrow \text{CO}_{2}\text{(g)}$ ; $\Delta_{r}H_{2}^{\ominus} = -393 \text{ kJ mol}^{-1}$
3. $\text{H}_{2}\text{(g)} + \frac{1}{2}\text{O}_{2}\text{(g)} \rightarrow \text{H}_{2}\text{O(l)}$ ; $\Delta_{r}H_{3}^{\ominus} = -286 \text{ kJ mol}^{-1}$

Step 3: Application of Hess's Law

[Per Hess's Law of Constant Heat Summation, the total enthalpy change of a reaction is independent of the pathway taken.]

To obtain the target equation, we manipulate the given equations:

• Keep Eq(2) as is: $\text{C(graphite)} + \text{O}_{2}\text{(g)} \rightarrow \text{CO}_{2}\text{(g)}$ ($\Delta H = -393 \text{ kJ mol}^{-1}$)
• Multiply Eq(3) by $2$: $2\text{H}_{2}\text{(g)} + \text{O}_{2}\text{(g)} \rightarrow 2\text{H}_{2}\text{O(l)}$ ($\Delta H = 2 \times -286 = -572 \text{ kJ mol}^{-1}$)
• Reverse Eq(1): $\text{CO}_{2}\text{(g)} + 2\text{H}_{2}\text{O(l)} \rightarrow \text{CH}_{3}\text{OH(l)} + \frac{3}{2}\text{O}_{2}\text{(g)}$ ($\Delta H = +726 \text{ kJ mol}^{-1}$)

Step 4: Algebraic Summation of Enthalpies

Adding the modified equations:

Canceling common species on both sides ($\text{CO}_{2}$, $2\text{H}_{2}\text{O}$, and $\frac{3}{2}\text{O}_{2}$):

$\Delta_{f}H^{\ominus} = -393 - 572 + 726$

$\Delta_{f}H^{\ominus} = -965 + 726$

$\Delta_{f}H^{\ominus} = -239 \text{ kJ mol}^{-1}$

Final Solution: The standard enthalpy of formation of $\text{CH}_{3}\text{OH(l)}$ is $-239 \text{ kJ mol}^{-1}$.

Step 1: Analyzing the Thermodynamic Process

The problem states that the cylinder walls and the piston are made of heat insulators. This implies that no heat exchange occurs between the system and the surroundings ($dQ = 0$). [Per the definition of an adiabatic process]. Consequently, the compression of the hydrogen gas follows the adiabatic relation between pressure ($P$) and volume ($V$).

Step 2: Defining Physical Constants for Hydrogen

Hydrogen ($H_2$) is a diatomic gas. For a diatomic gas, the degree of freedom ($f$) is 5 at room temperature. The adiabatic exponent ($\gamma$), defined as the ratio of molar heat capacity at constant pressure ($C_p$) to constant volume ($C_v$), is calculated as follows:

$\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1.4$

Step 3: Applying the Adiabatic Gas Law

The state equation for an adiabatic process is given by: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$ [Per the Adiabatic Equation of State]

Given the condition that the gas is compressed to half its original volume: $V_2 = \frac{1}{2} V_1$

Step 4: Algebraic Derivation of the Pressure Factor

Rearranging the adiabatic equation to find the ratio of final pressure ($P_2$) to initial pressure ($P_1$): $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}$

Substituting $V_2 = \frac{V_1}{2}$ into the equation: $\frac{P_2}{P_1} = \left( \frac{V_1}{V_1 / 2} \right)^{1.4}$

$\frac{P_2}{P_1} = (2)^{1.4}$

Step 5: Numerical Calculation

To evaluate $(2)^{1.4}$:

$(2)^{1.4} \approx 2.639$

Summary Table of Parameters

Final Solution: The pressure of the gas increases by a factor of approximately 2.64.

Step 1: Identifying Given Parameters and Physical Constants

To determine the rate of fuel consumption, we identify the following variables provided in the problem statement:

• Rate of water flow ($V_f$): $3.0\text{ litres/minute} = 3.0 \times 10^{-3}\text{ m}^3\text{/min}$
• Initial temperature ($T_1$): $27^\circ\text{C}$
• Final temperature ($T_2$): $77^\circ\text{C}$
• Density of water ($\rho$): $10^3\text{ kg/m}^3$ [Standard physical constant]
• Specific heat capacity of water ($c_w$): $4.2 \times 10^3\text{ J/(kg}\cdot\text{K)}$ [Standard physical constant]
• Heat of combustion of fuel ($H$): $4.0 \times 10^4\text{ J/g} = 4.0 \times 10^7\text{ J/kg}$

Step 2: Calculating Mass Flow Rate of Water

The mass flow rate ($\dot{m}$) is the product of volume flow rate and the density of the fluid.

$\dot{m} = V_f \times \rho$

Converting minutes to seconds for SI consistency ($1\text{ min} = 60\text{ s}$):

$\dot{m} = \left(\frac{3.0 \times 10^{-3}\text{ m}^3}{60\text{ s}}\right) \times 10^3\text{ kg/m}^3 = 0.05\text{ kg/s}$

Step 3: Determining Heat Required per Second (Power Output)

The rate of heat energy ($Q/t$) required to raise the water temperature is given by the calorimeter equation [Per the First Law of Thermodynamics, $Q = mc\Delta T$]:

$\dot{Q} = \dot{m} \cdot c_w \cdot (T_2 - T_1)$

$\Delta T = 77^\circ\text{C} - 27^\circ\text{C} = 50\text{ K}$

$\dot{Q} = 0.05\text{ kg/s} \times 4.2 \times 10^3\text{ J/(kg}\cdot\text{K)} \times 50\text{ K}$

$\dot{Q} = 0.05 \times 210,000 = 10,500\text{ J/s (or Watts)}$

Step 4: Calculating Rate of Fuel Consumption

The rate of fuel consumption ($\dot{m}_{fuel}$) is obtained by dividing the required heat rate by the energy density (heat of combustion) of the fuel.

$\dot{m}_{fuel} = \frac{\dot{Q}}{H}$

$\dot{m}_{fuel} = \frac{10,500\text{ J/s}}{4.0 \times 10^4\text{ J/g}}$

$\dot{m}_{fuel} = \frac{10,500}{40,000}\text{ g/s} = 0.2625\text{ g/s}$

Step 5: Visual Representation of the Energy Flux

To convert the result into a more standard unit ($g/min$):

$0.2625\text{ g/s} \times 60\text{ s/min} = 15.75\text{ g/min}$

Final Solution: The rate of consumption of the fuel is 15.75 g/min.

Step 1: Theoretical Foundation of Spontaneity

The spontaneity of a reaction at constant temperature ($T$) and pressure ($P$) is governed by the change in Gibbs Free Energy ($\Delta G$), defined by the Gibbs-Helmholtz equation:

$$\Delta G = \Delta H - T\Delta S$$

[Per the Second Law of Thermodynamics, a process is spontaneous if $\Delta G < 0$].

Step 2: Analysis of the Given Reaction Variables

The reaction is given as: $A + B \rightarrow C + D + q$.

• Enthalpy Change ($\Delta H$): The presence of $+ q$ in the product side indicates that energy is released into the surroundings. Therefore, the reaction is exothermic, implying $\Delta H < 0$.
• Entropy Change ($\Delta S$): The problem explicitly states that the reaction has a positive entropy change, implying $\Delta S > 0$.

Step 3: Evaluating the Gibbs Free Energy Equation

Substitute the signs of the variables into the Gibbs-Helmholtz equation:

$$\Delta G = (\text{negative}) - T(\text{positive})$$

Since the temperature $T$ (in Kelvin) is always positive, the term $-T\Delta S$ will be negative.

$$\Delta G = (\text{negative}) - (\text{positive}) = \text{negative}$$

[By the algebraic summation of negative values, the resultant $\Delta G$ must be negative regardless of the magnitude of $T$].

Step 4: Graphical Representation of Thermodynamic Spontaneity

Step 5: Conclusion on Reaction Nature

Because $\Delta H < 0$ and $\Delta S > 0$, the condition $\Delta G < 0$ is satisfied at all temperatures. A reaction where $\Delta G$ is always negative is defined as a spontaneous reaction.

Final Solution: The reaction will be spontaneous at all temperatures.

Step 1: Chemical Equation and Standard Enthalpy

The combustion of carbon to form carbon dioxide is represented by the thermochemical equation:

$C(s) + O_2(g) \rightarrow CO_2(g)$

The standard enthalpy of combustion for this reaction is given as:

$\Delta_c H = -393.5 \text{ kJ mol}^{-1}$ [Given]

Step 2: Calculating Molar Mass of $CO_2$

To determine the number of moles in a given mass, we first calculate the molar mass ($M$) of $CO_2$ using atomic masses of Carbon ($12.01 \text{ g mol}^{-1}$) and Oxygen ($16.00 \text{ g mol}^{-1}$):

$M(CO_2) = (1 \times 12.01) + (2 \times 16.00) = 44.01 \text{ g mol}^{-1}$

Step 3: Calculating Number of Moles ($n$)

Given the mass of $CO_2$ ($m = 35.2 \text{ g}$), we calculate the amount in moles ($n$) using the formula $n = \frac{m}{M}$:

$n = \frac{35.2 \text{ g}}{44.01 \text{ g mol}^{-1}} \approx 0.80 \text{ mol}$ [By Stoichiometric Definition]

Step 4: Calculating Heat Released

The heat released ($q$) upon the formation of $n$ moles of product is calculated by multiplying the molar enthalpy of combustion by the number of moles:

$q = n \times \Delta_c H$

$q = 0.80 \text{ mol} \times (-393.5 \text{ kJ mol}^{-1})$

$q = -314.8 \text{ kJ}$

Final Solution: The total heat released upon the formation of 35.2 g of $CO_2$ is $314.8 \text{ kJ}$.