Question 7

   Question: Match List I with List II.Choose the correct answer from the options given below:

   Options: 

       A. A-IV, B-III, C-II, D-I 

       B. A-IV, B-II, C-III, D-I 

       C. A-I, B-II, C-III, D-IV 

       D. A-II, B-III, C-IV, D-I

   Correct Answer: D

   Year: NEET 2024

   Solution (Source): (A) Isothermal process $\Rightarrow$ Temperature is constant throughout the process. (B) Isochoric process $\Rightarrow$ Volume is constant throughout the process. (C) Isobaric process $\Rightarrow$ Pressure is constant throughout the process. (D) Adiabatic process $\Rightarrow$ No exchange of heat (q) between system and surrounding.

   Step Solution:

    1.  Identify "Isothermal": 'Iso' means same and 'thermal' refers to temperature; therefore, $T$ is constant (Matches II).

    2.  Identify "Isochoric": 'Choric' refers to volume; therefore, $V$ is constant (Matches III).

    3.  Identify "Isobaric": 'Baric' refers to pressure (measured in bars); therefore, $P$ is constant (Matches IV).

    4.  Identify "Adiabatic": Defined as a process where no heat ($q$) is exchanged with the surroundings (Matches I).

    5.  Match the pairs: A-II, B-III, C-IV, D-I.

   Difficulty Level: Easy

   Concept Name: Thermodynamic Processes

   Shortcut Solution: Remember the prefixes: Thermal = Temp, Choric = Volume, Baric = Pressure, Adiabatic = No Heat.

Question 14

   Question: Which one among the following is the correct option for right relationship between $C_P$ and $C_V$ for one mole of ideal gas?

   Options: 

       A. $C_P + C_V = R$

       B. $C_P - C_V = R$

       C. $C_P = R C_V$

       D. $C_V = R C_P$

   Correct Answer: B

   Year: NEET 2021

   Solution (Source): At constant volume, $q_V = C_V \Delta T = \Delta U$. At constant pressure, $q_P = C_P \Delta T = \Delta H$. For a mole of an ideal gas, $\Delta H = \Delta U + \Delta(PV)$. On putting the values, $C_P \Delta T = C_V \Delta T + R \Delta T \Rightarrow C_P = C_V + R \Rightarrow C_P - C_V = R$.

   Step Solution:

    1.  Start with the definition of enthalpy for an ideal gas: $H = U + PV$.

    2.  For 1 mole of an ideal gas, use the ideal gas law $PV = RT$ to substitute: $H = U + RT$.

    3.  Differentiate the equation with respect to temperature ($T$): $dH/dT = dU/dT + R$.

    4.  Substitute molar heat capacities: $C_P = dH/dT$ and $C_V = dU/dT$.

    5.  Rearrange the resulting equation $C_P = C_V + R$ to get $C_P - C_V = R$.

   Difficulty Level: Easy

   Concept Name: Mayer’s Relation

   Shortcut Solution: For ideal gases, $C_P$ is always greater than $C_V$ by exactly the gas constant $R$.

Question 35

   Question: Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats ($C_P/C_V$) of the mixture will be?

   Options: 

       A. 0.83

       B. 1.50

       C. 3.3

       D. 1.67

   Correct Answer: D

   Year: 2012 Mains

   Solution (Source): $C_P$ for monoatomic gas mixture of same volume $= \frac{5}{2} R$. $C_V = \frac{3}{2} R$. Therefore $\frac{C_P}{C_V} = \frac{5/2 R}{3/2 R} = 5/3 = 1.67$.

   Step Solution:

    1.  Determine $C_V$ for a monoatomic gas: $C_V = \frac{3}{2} R$.

    2.  Determine $C_P$ for a monoatomic gas using $C_P = C_V + R$: $C_P = \frac{5}{2} R$.

    3.  Mixing two monoatomic gases results in a mixture that is still monoatomic; thus, the mixture retains these molar heat capacities.

    4.  Calculate the ratio $\gamma = C_P / C_V = (\frac{5}{2} R) / (\frac{3}{2} R)$.

    5.  Simplify the fraction: $5/3 \approx 1.67$.

   Difficulty Level: Medium

   Concept Name: Poisson's Ratio for Monoatomic Gases

   Shortcut Solution: The ratio $\gamma$ for any monoatomic gas or mixture of monoatomic gases is always 1.67 ($5/3$).

Question 51

   Question: Which of the following are not state functions? (I) $q + w$, (II) $q$, (III) $w$, (IV) $H - TS$

   Options: 

       A. (I) (II) and (III)

       B. (II) and (III)

       C. (I) and (IV)

       D. (II) (III) and (IV)

   Correct Answer: B

   Year: 2008

   Solution (Source): State functions depend only on the state of the system and not on how it was reached. Path function depends on the path followed during a process as well as the end states. Work and heat are the path functions

   Step Solution:

    1.  Analyze (I): According to the First Law, $q + w = \Delta U$. Since internal energy ($U$) depends only on initial and final states, it is a state function.

    2.  Analyze (II): Heat ($q$) depends on the specific process path (e.g., adiabatic vs. isothermal), making it a path function.

    3.  Analyze (III): Work ($w$) depends on the process path (e.g., reversible vs. irreversible), making it a path function.

    4.  Analyze (IV): $H - TS$ is the formula for Gibbs Free Energy ($G$), which is a fundamental thermodynamic potential and a state function.

    5.  Conclusion: Path functions (not state functions) are (II) and (III).

   Difficulty Level: Easy

   Concept Name: State vs. Path Functions

   Shortcut Solution: $q$ and $w$ are the only common path functions; their sum ($\Delta U$) or other thermodynamic properties are state functions.

Question 55

   Question: Consider the following reactions: (i)$\ \mathrm{H^+{(aq)} + OH^-{(aq)} \rightarrow H_2O_{(l)}},\ \Delta H = -X_1\ \mathrm{kJ\ mol^{-1}}$

$(ii)\ \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O_{(l)}},\ \Delta H = -X_2\ \mathrm{kJ\ mol^{-1}}$

$(iii)\ \mathrm{CO_2(g) + H_2(g) \rightarrow CO(g) + H_2O_{(l)}},\ \Delta H = X_3\ \mathrm{kJ\ mol^{-1}}$

$(iv)\ \mathrm{C_2H_2(g) + \tfrac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O_{(l)}},\ \Delta H = +X_4\ \mathrm{kJ\ mol^{-1}}$

Enthalpy of formation of $ \mathrm{H_2O_{(l)}} \text{ is }$

   Options: 

       A. $+X kJ/l$ $mol^{{-1}}$

       B. $-X kJ/$ $mol^{{-1}}$

       C. $+X kJ/$$mol^{{-1}}$

       D. $-X kJ/$$mol^{{-1}}$

   Correct Answer: D 

   Year: 2007

   Solution (Source): The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent elements is called the heat of formation. Equation (i) represents a neutralization reaction. The formation of $H_2O(l)$ must be from $H_2(g)$ and $O_2(g)$.

   Step Solution:

    1.  Define Enthalpy of Formation ($\Delta H_f^\circ$): The change in enthalpy when 1 mole of a compound is formed from its elements in their most stable states.

    2.  Identify the stable elements for $H_2O$: These are Hydrogen gas ($H_2$) and Oxygen gas ($O_2$).

    3.  Write the formation reaction: $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l)$.

    4.  Evaluate Reaction (i): $H^+(aq) + OH^-(aq) \to H_2O(l)$ is a reaction between ions, which is Enthalpy of Neutralization, not formation.

    5.  The correct enthalpy of formation is associated with the reaction of the pure gaseous elements.

   Difficulty Level: Medium

   Concept Name: Enthalpy of Formation

   Shortcut Solution: Formation always starts from pure elements (like $H_2, O_2$), not ions or other compounds.

Question 71

   Question: For which one of the following equations is $\Delta H_{reaction}^\circ$ equal to $\Delta H_f^\circ$ for the product?

   Options: 

       A. $N_{2(g)} + O_{3(g)} \to N_2O_{3(g)}$

       B. $CH_{4(g)} + 2Cl_{2(g)} \to CH_2Cl_{2(l)} + 2HCl_{(g)}$

       C. $Xe_{(g)} + 2F_{2(g)} \to XeF_{4(g)}$

       D. $2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$

   Correct Answer: C

   Year: 2003

   Solution (Source): For (c), $\Delta H_{reaction}^\circ = \Delta H_f^\circ(XeF_4) - [\Delta H_f^\circ(Xe) + 2\Delta H_f^\circ(F_2)]$. Enthalpies of formation of elementary substances Xe and $F_2$ are taken as zero. Thus, $\Delta H_{reaction}^\circ = \Delta H_f^\circ(XeF_4)$.

   Step Solution:

    1.  Recall the condition for $\Delta H_{reaction}^\circ = \Delta H_f^\circ$: All reactants must be elements in their most stable standard states.

    2.  Analyze (A): Uses $O_3$ (Ozone); the stable standard state of oxygen is $O_2$.

    3.  Analyze (B): Uses $CH_4$, which is a compound, not an element.

    4.  Analyze (D): Uses $CO$, which is a compound, not an element.

    5.  Analyze (C): Uses $Xe(g)$ and $F_2(g)$, both of which are elements in their stable standard states. Thus, the heat of reaction equals the heat of formation of the product.

   Difficulty Level: Easy

   Concept Name: Standard Enthalpy of Formation

   Shortcut Solution: Check the reactant side; if it contains only elements in their natural form (like $O_2, H_2, Xe$), it is a formation reaction.

Question 72

   Question: The molar heat capacity of water at constant pressure, C, is $75 \text{ J K}^{-1} \text{ mol}^{-1}$. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

   Options: 

       A. 1.2 K 

       B. 2.4 K 

       C. 4.8 K 

       D. 6.6 K

   Correct Answer: B

   Year: 2003

   Solution (Source): Molar heat capacity $= 75 \text{ J K}^{-1} \text{ mol}^{-1}$. 18 g of water $= 1 \text{ mole} = 75 \text{ J K}^{-1} \text{ mol}^{-1}$. 1 g of water $= \frac{75}{18} \text{ J K}^{-1}$. $Q = m \cdot C \cdot \Delta t$ or $1000 = 100 \times \frac{75}{18} \times \Delta t \Rightarrow \Delta t = \frac{10 \times 18}{75} = 2.4 \text{ K}$

   Step Solution:

    1.  Convert molar heat capacity ($C_m$) to specific heat capacity ($c$): $c = \frac{C_m}{\text{Molar Mass}} = \frac{75 \text{ J/mol K}}{18 \text{ g/mol}} = \frac{75}{18} \text{ J/g K}$.

    2.  Identify the heat supplied ($Q$) in Joules: $1.0 \text{ kJ} = 1000 \text{ J}$.

    3.  Use the calorimetry formula: $Q = m \cdot c \cdot \Delta T$.

    4.  Substitute the values: $1000 = 100 \cdot \left(\frac{75}{18}\right) \cdot \Delta T$.

    5.  Solve for $\Delta T$: $\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \text{ K}$.

   Difficulty Level: Medium

   Concept Name: Molar Heat Capacity vs. Specific Heat Capacity

   Short cut solution: Use the formula $\Delta T = \frac{Q}{n \cdot C_p}$. Number of moles $n = \frac{100}{18}$. So, $\Delta T = \frac{1000}{\left(\frac{100}{18}\right) \times 75} = \frac{10 \times 18}{75} = 2.4 \text{ K}$.

Question 73

   Question: Unit of entropy is

   Options: 

       A. $\text{J K}^{-1} \text{ mol}^{-1}$

       B. $\text{J mol}^{-1}$

       C. $\text{J}^{-1} \text{ K}^{-1} \text{ mol}^{-1}$

       D. $\text{J K mol}^{-1}$

   Correct Answer: A

   Year: 2002

   Solution (Source): Entropy change ($\Delta S$) is given by $\Delta S = q_{rev} / T$. ∴ Unit of entropy $= \text{J} / \text{K}$ mol (entropy per unit mole $= \text{J K}^{-1} \text{ mol}^{-1}$).

   Step Solution:

    1.  Recall the thermodynamic definition of entropy change: $\Delta S = \frac{q_{rev}}{T}$.

    2.  Identify the unit of heat ($q$): Joules (J).

    3.  Identify the unit of absolute temperature ($T$): Kelvin (K).

    4.  Combine them for molar entropy: $\frac{\text{Joules}}{\text{Kelvin} \times \text{moles}}$.

    5.  Resulting unit: $\text{J K}^{-1} \text{ mol}^{-1}$.

   Difficulty Level: Easy

   Concept Name: Units of Thermodynamic Variables

   Short cut solution: Entropy is "heat over temperature," so the unit must be Energy/Temperature (J/K).

Question 85

   Question: In an endothermic reaction, the value of $\Delta H$ is

   Options: 

       A. negative 

       B. positive 

       C. zero 

       D. constant

   Correct Answer: B

   Year: 1999

   Solution (Source): In endothermic reactions, energy of reactants is less than energy of products. Thus, $E_R < E_P$. $\Delta H = E_P - E_R = +ve$.

   Step Solution:

    1.  Define an endothermic reaction: A reaction where the system absorbs heat from the surroundings.

    2.  Compare energies: The total enthalpy of the products ($H_P$) is higher than the total enthalpy of the reactants ($H_R$).

    3.  Apply the formula for enthalpy change: $\Delta H = H_{\text{products}} - H_{\text{reactants}}$.

    4.  Since $H_P > H_R$, the result of the subtraction is a positive value.

   Difficulty Level: Easy

   Concept Name: Enthalpy of Reaction

   Short cut solution: Endo = Enter (Heat goes in, $\Delta H$ is positive); Exo = Exit (Heat goes out, $\Delta H$ is negative).