Masterclass: 25 Solved JEE Advanced Numericals on Thermodynamics
Conquer the laws of energy! This exhaustive guide features complex multi-step problems on Reversible/Irreversible Work, Entropy changes, Hess's Law, and Gibbs Free Energy. Click "View Solution" to reveal the step-by-step breakdown.
Thermodynamics is the absolute foundation of all physical sciences. It requires impeccable sign conventions and unit conversions. Before attempting these rigorous problems, ensure you clearly understand the difference between intensive and extensive properties, state functions vs. path functions, and the IUPAC sign convention for work ($w = -P_{ext}\Delta V$).
View Solution
Step 1: Calculate Reversible Isothermal Work
The formula for maximum work in a reversible isothermal expansion is:
$w_{\text{rev}} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$
Substitute the values:
$w_{\text{rev}} = -2.303 \times 2.0 \times 8.314 \times 300 \times \log_{10}\left(\frac{20.0}{2.0}\right)$
$w_{\text{rev}} = -2.303 \times 4988.4 \times \log_{10}(10)$
$w_{\text{rev}} = -11488.3 \text{ J} = -11.49 \text{ kJ}$.
Step 2: Calculate Irreversible Isothermal Work
For an irreversible process against constant external pressure, the formula is:
$w_{\text{irr}} = -P_{\text{ext}} (V_2 - V_1)$
$w_{\text{irr}} = -1.0 \text{ atm} \times (20.0 \text{ L} - 2.0 \text{ L}) = -18.0 \text{ L atm}$.
Step 3: Convert to Joules
$w_{\text{irr}} = -18.0 \times 101.3 \text{ J} = -1823.4 \text{ J} = -1.82 \text{ kJ}$.
Thermodynamic Insight: Notice that the magnitude of work done by the system is significantly greater in the reversible process ($11.49 \text{ kJ}$) compared to the irreversible one ($1.82 \text{ kJ}$). Reversible paths always yield maximum work.
View Solution
Step 1: Relate Constant Volume Heat to $\Delta U$
Heat measured at constant volume ($q_v$) is exactly equal to the change in internal energy ($\Delta U$).
Thus, $\Delta U = -3263.9 \text{ kJ mol}^{-1}$.
Step 2: Write the balanced combustion equation
You must write the equation with correct physical states at $298 \text{ K}$:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
Step 3: Calculate $\Delta n_g$ (Change in moles of gas)
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta n_g = 6 - \frac{15}{2} = 6 - 7.5 = -1.5 \text{ moles}$.
Step 4: Use the thermodynamic relation
$\Delta H = \Delta U + \Delta n_g RT$
First, calculate the $RT$ term in $\text{kJ}$ to match $\Delta U$:
$\Delta n_g RT = -1.5 \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298 \text{ K} = -3.716 \text{ kJ mol}^{-1}$.
$\Delta H = -3263.9 + (-3.716) = -3267.6 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Find Final Temperature ($T_2$)
For a reversible adiabatic process, the relationship between $T$ and $P$ is:
$T_1^{\gamma} P_1^{1-\gamma} = T_2^{\gamma} P_2^{1-\gamma}$ or $\frac{T_2}{T_1} = \left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}}$
Rearranging for a simpler form: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}$
$\frac{\gamma-1}{\gamma} = \frac{(5/3) - 1}{5/3} = \frac{2/3}{5/3} = \frac{2}{5} = 0.4$.
$T_2 = 300 \times \left(\frac{1}{10}\right)^{0.4} = 300 \times (0.1)^{0.4} = 300 \times 0.398 = 119.4 \text{ K}$.
Step 2: Calculate Work Done ($w$)
In an adiabatic process, $q = 0$. By the First Law of Thermodynamics, $\Delta U = q + w$, so $w = \Delta U$.
$\Delta U = n C_v \Delta T = n C_v (T_2 - T_1)$.
$w = 1 \times \left(\frac{3}{2} \times 8.314\right) \times (119.4 - 300)$
$w = 12.471 \times (-180.6) = -2252.3 \text{ J}$.
1. $\Delta H_c^{\circ} (C_{(graphite)}) = -393 \text{ kJ mol}^{-1}$
2. $\Delta H_c^{\circ} (H_{2(g)}) = -286 \text{ kJ mol}^{-1}$
3. $\Delta H_c^{\circ} (CH_3OH_{(l)}) = -726 \text{ kJ mol}^{-1}$
View Solution
Step 1: Write the target equation
The standard enthalpy of formation corresponds to forming 1 mole of the substance from its constituent elements in their standard states:
$C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(l)} \quad (\Delta H_f^{\circ} = ?)$
Step 2: Write out the given combustion equations
Eq 1: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H_1 = -393 \text{ kJ}$
Eq 2: $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H_2 = -286 \text{ kJ}$
Eq 3: $CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H_3 = -726 \text{ kJ}$
Step 3: Manipulate equations to yield the target
- We need 1 mole of $C_{(s)}$ on the reactant side: Keep Eq 1 as is.
- We need 2 moles of $H_{2(g)}$ on the reactant side: Multiply Eq 2 by 2.
- We need 1 mole of $CH_3OH_{(l)}$ on the product side: Reverse Eq 3.
Target $\Delta H_f^{\circ} = (\Delta H_1) + 2(\Delta H_2) - (\Delta H_3)$
$\Delta H_f^{\circ} = (-393) + 2(-286) - (-726)$
$\Delta H_f^{\circ} = -393 - 572 + 726 = -965 + 726 = -239 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Understand Resonance Energy
Resonance Energy = Expected theoretical enthalpy of atomization (based on a single Kekulé structure) - Actual experimental enthalpy of atomization.
Step 2: Calculate Theoretical Enthalpy of Atomization
A single Kekulé structure of benzene contains:
- 3 single $C-C$ bonds
- 3 double $C=C$ bonds
- 6 single $C-H$ bonds
$\Delta H_{\text{theoretical}} = 3(\Delta H_{C-C}) + 3(\Delta H_{C=C}) + 6(\Delta H_{C-H})$
$\Delta H_{\text{theoretical}} = 3(347) + 3(615) + 6(414)$
$\Delta H_{\text{theoretical}} = 1041 + 1845 + 2484 = 5370 \text{ kJ mol}^{-1}$.
Step 3: Calculate Resonance Energy
Wait, the standard convention is that resonance stabilizes the molecule, meaning it requires more energy to atomize than predicted.
Resonance Energy (magnitude) = Actual $\Delta H_{\text{atomization}} - \text{Theoretical } \Delta H_{\text{atomization}}$
Resonance Energy = $5536 - 5370 = 166 \text{ kJ mol}^{-1}$.
Note: Resonance energy is technically a stabilization energy, so it is often reported with a negative sign ($-166 \text{ kJ mol}^{-1}$) indicating the molecule is lower in energy than the hypothetical Kekulé structure.
View Solution
Step 1: Set up the First Law for Irreversible Adiabatic Process
Since adiabatic, $q = 0$. Thus, $\Delta U = w$.
$\Delta U = n C_v (T_2 - T_1)$
$w = -P_{\text{ext}} (V_2 - V_1)$
Equating them: $n C_v (T_2 - T_1) = -P_{\text{ext}} \left( \frac{n R T_2}{P_2} - \frac{n R T_1}{P_1} \right)$
Step 2: Substitute values and solve for $T_2$
At equilibrium, the final pressure of the gas $P_2$ equals the external pressure $P_{\text{ext}} = 1 \text{ atm}$. The number of moles $n = 1$ cancels out.
$\frac{5}{2} R (T_2 - 300) = -1 \cdot \left( \frac{R T_2}{1} - \frac{R \cdot 300}{5} \right)$
Cancel $R$ from both sides:
$2.5 (T_2 - 300) = - (T_2 - 60)$
$2.5 T_2 - 750 = -T_2 + 60$
$3.5 T_2 = 810 \implies T_2 = \frac{810}{3.5} = 231.4 \text{ K}$.
Step 3: Calculate Work Done
$w = \Delta U = n C_v (T_2 - T_1)$
$w = 1 \times \left(\frac{5}{2} \times 8.314\right) \times (231.4 - 300)$
$w = 20.785 \times (-68.6) = -1425.8 \text{ J}$.
Given: $C_p(\text{ice}) = 37.7 \text{ J K}^{-1} \text{ mol}^{-1}$, $C_p(\text{water}) = 75.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\Delta H_{\text{fusion}} = 6.01 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Identify the Steps
Moles of water $n = 18 \text{ g} / 18 \text{ g mol}^{-1} = 1 \text{ mole}$.
The process occurs in 3 distinct steps:
1. Heating ice from $263.15 \text{ K}$ to $273.15 \text{ K}$.
2. Melting ice at $273.15 \text{ K}$.
3. Heating water from $273.15 \text{ K}$ to $283.15 \text{ K}$.
Step 2: Calculate $\Delta S$ for each step
$\Delta S_1 = n C_p(\text{ice}) \ln\left(\frac{T_f}{T_i}\right) = 1 \times 37.7 \times \ln\left(\frac{273.15}{263.15}\right) = 37.7 \times 0.0373 = 1.41 \text{ J K}^{-1}$.
$\Delta S_2 (\text{Phase Change}) = \frac{\Delta H_{\text{fus}}}{T_{\text{melt}}} = \frac{6010 \text{ J}}{273.15 \text{ K}} = 22.00 \text{ J K}^{-1}$.
$\Delta S_3 = n C_p(\text{water}) \ln\left(\frac{T_f}{T_i}\right) = 1 \times 75.3 \times \ln\left(\frac{283.15}{273.15}\right) = 75.3 \times 0.0360 = 2.71 \text{ J K}^{-1}$.
Step 3: Sum the Entropy Changes
$\Delta S_{\text{total}} = \Delta S_1 + \Delta S_2 + \Delta S_3 = 1.41 + 22.00 + 2.71 = 26.12 \text{ J K}^{-1}$.
View Solution
Step 1: Understand Spontaneity Condition
A reaction is spontaneous when Gibbs Free Energy change ($\Delta G$) is negative ($\Delta G < 0$).
Using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$.
Step 2: Find the Equilibrium Temperature
At equilibrium, $\Delta G = 0$. This is the threshold temperature.
$0 = \Delta H - T_{\text{eq}}\Delta S \implies T_{\text{eq}} = \frac{\Delta H}{\Delta S}$.
Step 3: Calculate
Convert $\Delta H$ to Joules: $30 \text{ kJ} = 30000 \text{ J}$.
$T_{\text{eq}} = \frac{30000 \text{ J mol}^{-1}}{100 \text{ J K}^{-1} \text{ mol}^{-1}} = 300 \text{ K}$.
Step 4: Determine Spontaneity Direction
Since both $\Delta H$ and $\Delta S$ are positive, the $T\Delta S$ term must be larger than $\Delta H$ to make $\Delta G$ negative. Therefore, the temperature must be greater than the equilibrium temperature.
View Solution
Step 1: Work Done ($w$)
Expansion into a vacuum means the external opposing pressure is exactly zero ($P_{\text{ext}} = 0$).
$w = -P_{\text{ext}} \Delta V = -0 \times \Delta V = 0 \text{ Joules}$.
Step 2: Internal Energy ($\Delta U$) and Heat ($q$)
For an ideal gas, Internal Energy is a function of temperature only. Since the process is isothermal ($\Delta T = 0$), $\Delta U = 0$.
By the First Law: $\Delta U = q + w \implies 0 = q + 0 \implies q = 0$. (The process is inherently adiabatic!).
Step 3: Enthalpy ($\Delta H$)
$\Delta H = \Delta U + \Delta(PV)$. For an ideal gas at constant temperature, $PV = nRT = \text{constant}$. Thus $\Delta(PV) = 0$.
$\Delta H = 0 + 0 = 0$.
Step 4: Entropy of System ($\Delta S_{\text{sys}}$)
Entropy is a state function. We can calculate it using a reversible isothermal path between the same initial and final states.
$\Delta S = n R \ln\left(\frac{V_2}{V_1}\right)$.
We need $V_1$. $V_1 = \frac{nRT_1}{P_1} = \frac{5 \times 0.0821 \times 298}{2} = 61.16 \text{ L}$.
Wait, the gas expands into a $50 \text{ L}$ bulb. The final volume $V_2 = V_1 + 50 = 111.16 \text{ L}$.
$\Delta S = 5 \times 8.314 \times \ln\left(\frac{111.16}{61.16}\right) = 41.57 \times \ln(1.817) = 41.57 \times 0.597 = 24.8 \text{ J K}^{-1}$.
View Solution
Step 1: Formula for Entropy of Mixing
Mixing ideal gases leads to an increase in disorder. The formula is derived from statistical thermodynamics:
$\Delta S_{\text{mix}} = -R \sum n_i \ln(x_i) = -R [n_1 \ln(x_1) + n_2 \ln(x_2)]$
Step 2: Calculate Mole Fractions
Total moles $n_{\text{total}} = 2 + 3 = 5 \text{ moles}$.
$x_{N_2} = \frac{2}{5} = 0.4$
$x_{Ar} = \frac{3}{5} = 0.6$
Step 3: Calculate $\Delta S_{\text{mix}}$
$\Delta S_{\text{mix}} = -8.314 \times [2 \ln(0.4) + 3 \ln(0.6)]$
$\Delta S_{\text{mix}} = -8.314 \times [2(-0.916) + 3(-0.510)]$
$\Delta S_{\text{mix}} = -8.314 \times [-1.832 - 1.530] = -8.314 \times (-3.362) = +27.95 \text{ J K}^{-1}$.
View Solution
Step 1: Relate $\Delta G^{\circ}$ to $K_p$
The fundamental isotherm relates standard free energy to the equilibrium constant:
$\Delta G^{\circ} = -RT \ln K_p = -2.303 RT \log_{10} K_p$
Step 2: Calculate $\Delta G^{\circ}$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log_{10}(0.15)$
$\Delta G^{\circ} = -5705.8 \times (-0.8239) = +4701 \text{ J mol}^{-1} = +4.70 \text{ kJ mol}^{-1}$.
Since $\Delta G^{\circ}$ is positive, the reaction is non-spontaneous under standard conditions (1 atm partial pressures).
Step 3: Actual $\Delta G$ at Equilibrium
By pure definition, when any chemical system reaches dynamic equilibrium, it can do no more net work. Therefore, the actual free energy change $\Delta G$ must be exactly zero.
View Solution
Step 1: Write the formation reaction
$\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightarrow NH_{3(g)}$
Step 2: Calculate $\Delta C_p$ for the reaction
$\Delta C_p = \sum C_p(\text{products}) - \sum C_p(\text{reactants})$
$\Delta C_p = C_p(NH_3) - \left[ \frac{1}{2}C_p(N_2) + \frac{3}{2}C_p(H_2) \right]$
$\Delta C_p = 35.1 - \left[ 0.5(29.1) + 1.5(28.8) \right]$
$\Delta C_p = 35.1 - [14.55 + 43.2] = 35.1 - 57.75 = -22.65 \text{ J K}^{-1} \text{ mol}^{-1}$.
Step 3: Apply Kirchhoff's Equation
$\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p (T_2 - T_1)$
Ensure units match! Convert $\Delta C_p$ to $\text{kJ}$:
$\Delta C_p = -0.02265 \text{ kJ K}^{-1} \text{ mol}^{-1}$.
$\Delta H_{400} = -46.1 + (-0.02265)(400 - 298)$
$\Delta H_{400} = -46.1 + (-0.02265)(102) = -46.1 - 2.31 = -48.41 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Understand the Thermodynamics of Weak Acids
A weak acid must first absorb energy to completely dissociate (ionize) before it can neutralize. Thus, the observed heat evolved is the standard strong-strong heat minus the enthalpy of ionization.
$\Delta H_{\text{obs}} = \Delta H_{\text{neutralization(strong)}} + \Delta H_{\text{ionization}}$
Step 2: Scale standard values to actual moles used
Standard $\Delta H$ is for $1 \text{ equivalent}$. We used $0.5 \text{ equivalents}$.
Expected heat evolved for $0.5 \text{ eq}$ of strong acid/base $= 0.5 \times (-57.3) = -28.65 \text{ kJ}$.
Step 3: Calculate Enthalpy of Ionization
Observed heat evolved $= -26.1 \text{ kJ}$.
$-26.1 = -28.65 + \Delta H_{\text{ionization(for } 0.5 \text{ mol)}}$
$\Delta H_{\text{ionization(0.5 mol)}} = -26.1 + 28.65 = +2.55 \text{ kJ}$.
Step 4: Convert to per mole basis
$\Delta H_{\text{ionization}} \text{ per mole} = \frac{2.55 \text{ kJ}}{0.5 \text{ mol}} = +5.1 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Calculate Heat Absorbed by Calorimeter
$q_{\text{cal}} = C \times \Delta T = 20.7 \text{ kJ K}^{-1} \times (299.1 - 298.0) \text{ K}$
$q_{\text{cal}} = 20.7 \times 1.1 = 22.77 \text{ kJ}$.
Therefore, heat released by the reaction at constant volume ($q_v$) is $-22.77 \text{ kJ}$.
Step 2: Relate to Molar Internal Energy ($\Delta U$)
This heat was released by burning $1.0 \text{ g}$ of graphite (Carbon, $M = 12 \text{ g/mol}$).
Moles of C burned = $1.0 / 12 = 0.0833 \text{ mol}$.
$\Delta U = \frac{q_v}{n} = \frac{-22.77 \text{ kJ}}{0.0833 \text{ mol}} = -273.3 \text{ kJ mol}^{-1}$.
Step 3: Relate $\Delta U$ to $\Delta H$
Combustion reaction: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$
$\Delta n_g = 1 - 1 = 0$.
Since $\Delta n_g = 0$, the $P\Delta V$ work is zero. Therefore, $\Delta H = \Delta U$.
View Solution
Step 1: Calculate Engine Efficiency ($\eta$)
$\eta = \frac{\text{Work Done}}{\text{Heat Absorbed}} = \frac{W}{Q_H} = \frac{400}{1000} = 0.40$ (or $40\%$).
Step 2: Find Cold Sink Temperature ($T_C$)
Thermodynamically, efficiency is also given by: $\eta = 1 - \frac{T_C}{T_H}$.
$0.40 = 1 - \frac{T_C}{600}$
$\frac{T_C}{600} = 0.60 \implies T_C = 600 \times 0.60 = 360 \text{ K}$.
Step 3: Entropy Change of the Universe
By definition, a Carnot cycle is comprised entirely of reversible processes. According to the Second Law of Thermodynamics, the total entropy change of the universe for any perfectly reversible cycle is exactly zero.
View Solution
Step 1: Write the Van der Waals equation
$\left( P + \frac{an^2}{V^2} \right)(V - nb) = nRT$
Step 2: Isolate Pressure ($P$)
$P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$
Step 3: Integrate to find Work ($w$)
For a reversible process, $w = - \int_{V_1}^{V_2} P \, dV$.
$w = - \int_{V_1}^{V_2} \left( \frac{nRT}{V - nb} - \frac{an^2}{V^2} \right) dV$
$w = - \left[ nRT \ln(V - nb) - \left( -\frac{an^2}{V} \right) \right]_{V_1}^{V_2}$
$w = -nRT \ln\left(\frac{V_2 - nb}{V_1 - nb}\right) - an^2\left(\frac{1}{V_2} - \frac{1}{V_1}\right)$
Insight: The first term represents the work modified by molecular volume (repulsion), and the second term accounts for the energy lost pulling attractive molecules apart.
View Solution
Step 1: Apply Trouton's Rule
Trouton's rule states that the entropy of vaporization for many non-polar liquids is approximately constant: $\Delta S_{\text{vap}} \approx 85 \text{ J K}^{-1} \text{ mol}^{-1}$.
Since $\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_b}$, we get:
$\Delta H_{\text{vap}} \approx 85 \times 350 = 29750 \text{ J mol}^{-1} = 29.75 \text{ kJ mol}^{-1}$.
Step 2: Entropy Change of the System
For 2 moles, $\Delta S_{\text{sys}} = 2 \times 85 = +170 \text{ J K}^{-1}$.
Step 3: Entropy Change of Surroundings
Reversible phase changes occur at equilibrium, meaning $\Delta S_{\text{universe}} = 0$.
Since $\Delta S_{\text{universe}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0$,
$\Delta S_{\text{surr}} = -\Delta S_{\text{sys}} = -170 \text{ J K}^{-1}$.
View Solution
Step 1: Set up Heat Balance Equation
According to the First Law, in an isolated system: Heat gained by ice = Heat lost by warm water.
Step 2: Define the Heat Terms
Let the final temperature be $T_f$ (in $^{\circ}\text{C}$).
Heat gained by ice = Heat to melt + Heat to warm the resulting water to $T_f$
$q_{\text{gain}} = m_{\text{ice}}L_f + m_{\text{ice}}C(T_f - 0) = (10 \times 334) + 10 \times 4.18 \times T_f = 3340 + 41.8 T_f$
Heat lost by warm water = $m_{\text{water}}C(50 - T_f)$
$q_{\text{loss}} = 100 \times 4.18 \times (50 - T_f) = 418 \times (50 - T_f) = 20900 - 418 T_f$
Step 3: Equate and Solve
$3340 + 41.8 T_f = 20900 - 418 T_f$
$459.8 T_f = 17560$
$T_f = \frac{17560}{459.8} \approx 38.19^{\circ}\text{C}$.
Enthalpy of Sublimation of $Na_{(s)} = +108$
Ionization Energy of $Na_{(g)} = +496$
Bond Dissociation Energy of $Cl_{2(g)} = +242$
Electron Affinity of $Cl_{(g)} = -349$
Enthalpy of Formation of $NaCl_{(s)} = -411$
View Solution
Step 1: Write the Born-Haber Cycle Equation
According to Hess's Law, the enthalpy of formation is the sum of all individual step enthalpies leading to the solid lattice.
$\Delta H_f^{\circ} = \Delta H_{\text{sub}}(Na) + IE(Na) + \frac{1}{2}\Delta H_{\text{diss}}(Cl_2) + EA(Cl) + \Delta H_{\text{lattice}}$
Step 2: Be careful with stoichiometry
We only need 1 mole of $Cl$ atoms, so we take half the bond dissociation energy of $Cl_2$.
$\frac{1}{2}(242) = +121 \text{ kJ mol}^{-1}$.
Step 3: Substitute and Solve for Lattice Energy
$-411 = 108 + 496 + 121 + (-349) + \Delta H_{\text{lattice}}$
$-411 = 725 - 349 + \Delta H_{\text{lattice}}$
$-411 = 376 + \Delta H_{\text{lattice}}$
$\Delta H_{\text{lattice}} = -411 - 376 = -787 \text{ kJ mol}^{-1}$.
Note: Lattice energy is defined as the energy released when gaseous ions form the solid, so it is highly negative. (Some textbooks define it as the energy required to break the lattice, which would be $+787 \text{ kJ}$).
View Solution
Step 1: Apply First Law
For an ideal rubber band, Internal Energy depends mainly on temperature. Since stretching is isothermal, $\Delta U \approx 0$.
$\Delta U = q + w \implies 0 = q + w \implies w = -q$.
Since heat is given off ($q$ is negative), $w$ must be positive. This means work is done on the system (stretching it).
Step 2: Apply Second Law (Entropy)
For a reversible isothermal process, $\Delta S_{\text{sys}} = \frac{q_{\text{rev}}}{T}$.
Since $q_{\text{rev}}$ is negative (exothermic), $\Delta S_{\text{sys}}$ must be negative.
Step 3: Physical Explanation
In a relaxed rubber band, polymer chains are curled up in a highly random, disordered state (high entropy). When stretched, these chains are forced to align parallel to each other, creating a highly ordered, crystalline-like state (low entropy). The decrease in entropy releases heat to the surroundings!
View Solution
Step 1: Identify the Path
The cycle A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D $\rightarrow$ A forms a rectangle on the P-V diagram. It proceeds in a counter-clockwise direction.
Step 2: Calculate Area
The magnitude of the work done is exactly equal to the area enclosed by the cycle on the P-V graph.
Area = $\Delta P \times \Delta V = (3 - 1) \text{ atm} \times (3 - 1) \text{ L} = 2 \times 2 = 4 \text{ L atm}$.
Step 3: Determine Sign Convention
For a clockwise cycle on a P-V diagram, the gas expands at high pressure and compresses at low pressure, meaning net work is done by the gas ($w < 0$).
For a counter-clockwise cycle, the gas compresses at high pressure and expands at low pressure. Net work is done on the gas ($w > 0$).
Thus, $w = +4 \text{ L atm}$.
Step 4: Convert to Joules
$w = +4 \times 101.3 \text{ J} = +405.2 \text{ J}$.
View Solution
Step 1: Understand the Third Law
The Third Law of Thermodynamics states that the entropy of a perfect crystal at $0 \text{ K}$ is exactly zero. Thus, we integrate from $0 \text{ K}$ to $10 \text{ K}$.
Step 2: Setup the Entropy Integral
$S(T) = \int_0^T \frac{C_p}{T} dT$
Substitute $C_p = aT^3$:
$S(T) = \int_0^{10} \frac{aT^3}{T} dT = \int_0^{10} aT^2 dT$
Step 3: Integrate and Solve
$S(T) = a \left[ \frac{T^3}{3} \right]_0^{10} = a \frac{10^3}{3} = \frac{1000a}{3}$
Substitute $a = 1.0 \times 10^{-4}$:
$S(10 \text{ K}) = \frac{1000 \times 1.0 \times 10^{-4}}{3} = \frac{0.1}{3} = 0.0333 \text{ J K}^{-1} \text{ mol}^{-1}$.
View Solution
Step 1: Formula for $\Delta G$ using Fugacity
$\Delta G = RT \ln\left(\frac{f_2}{f_1}\right)$
At $1 \text{ atm}$, the gas is essentially ideal, so fugacity equals pressure: $f_1 = P_1 = 1 \text{ atm}$.
Step 2: Find Final Fugacity
At $100 \text{ atm}$, the gas is non-ideal. $\gamma = f/P \implies f_2 = \gamma P_2$.
$f_2 = 0.8 \times 100 = 80 \text{ atm}$.
Step 3: Calculate $\Delta G$
$\Delta G = 8.314 \times 300 \times \ln\left(\frac{80}{1}\right)$
$\Delta G = 2494.2 \times \ln(80) = 2494.2 \times 4.382 = 10929 \text{ J mol}^{-1} = 10.93 \text{ kJ mol}^{-1}$.
Insight: If the gas were ideal, we would have used $\ln(100)$, yielding $\Delta G = 11.49 \text{ kJ}$. The attractive intermolecular forces making $\gamma < 1$ make the compression slightly "easier" thermodynamically.
View Solution
Step 1: Write the Van't Hoff Equation
$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^{\circ}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Step 2: Substitute Values
$\ln\left(\frac{100}{10}\right) = \frac{\Delta H^{\circ}}{8.314} \left( \frac{1}{300} - \frac{1}{400} \right)$
$\ln(10) = \frac{\Delta H^{\circ}}{8.314} \left( \frac{400 - 300}{120000} \right)$
$2.303 = \frac{\Delta H^{\circ}}{8.314} \left( \frac{100}{120000} \right)$
Step 3: Solve for $\Delta H^{\circ}$
$2.303 = \frac{\Delta H^{\circ}}{8.314} \left( \frac{1}{1200} \right)$
$\Delta H^{\circ} = 2.303 \times 8.314 \times 1200 = 22976 \text{ J mol}^{-1} = 22.98 \text{ kJ mol}^{-1}$.
View Solution
Step 1: Identify Knowns and Constants
$T = 373.15 \text{ K}$, $P = 1 \text{ atm}$, $n = 1 \text{ mole}$.
Step 2: Enthalpy ($\Delta H$) and Heat ($q$)
Since it's at constant pressure, heat exchanged equals enthalpy change.
$q = \Delta H = +40.6 \text{ kJ} = 40600 \text{ J}$.
Step 3: Work Done ($w$)
$w = -P\Delta V = -P(V_{\text{gas}} - V_{\text{liquid}})$. Ignoring $V_{\text{liquid}}$, $w \approx -PV_{\text{gas}}$.
For an ideal gas, $PV = nRT$. Thus, $w = -nRT$.
$w = -1 \times 8.314 \times 373.15 = -3102 \text{ J} = -3.1 \text{ kJ}$.
Step 4: Internal Energy ($\Delta U$)
By First Law: $\Delta U = q + w$.
$\Delta U = 40.6 \text{ kJ} - 3.1 \text{ kJ} = +37.5 \text{ kJ}$.
Step 5: Entropy ($\Delta S_{\text{sys}}$)
For a reversible phase change, $\Delta S = \frac{q_{\text{rev}}}{T} = \frac{\Delta H_{\text{vap}}}{T}$.
$\Delta S = \frac{40600}{373.15} = +108.8 \text{ J K}^{-1}$.
Step 6: Gibbs Free Energy ($\Delta G$)
Since liquid and vapor are in exact equilibrium, they have identical chemical potentials.
$\Delta G = 0$. (Can verify: $\Delta G = \Delta H - T\Delta S = 40600 - (373.15 \times 108.8) \approx 0$).
Mastering the Energy of Chemistry
Congratulations on completing these 25 highly advanced numericals on Chemical Thermodynamics. By mastering how to integrate heat capacities (Kirchhoff's law), balance energies across Born-Haber cycles, and apply exact sign conventions to the First and Second Laws, you have built the ultimate foundation for physical chemistry. Keep a close eye on your units—especially distinguishing between Joules and kiloJoules when bridging $\Delta U$ and $\Delta n_g RT$!
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