Q.The reaction of cyanamide, $NH_2CN(s)$, with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be –742.7 kJ mol$^{-1}$ at 298 K. Calculate the enthalpy change for the reaction at 298 K. $NH_2CN(g) + \tfrac{3}{2} O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)$

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Which of the following statements are not correct? I. The process of adding an electron to a gaseous atom is always exothermic II. The process of removing an electron from a gaseous atom is always endothermic III. The first ionization enthalpy of oxygen is greater than nitrogen IV. In group I elements lithium is the least electropositive (A) I, III, IV only (B) I & II only (C) I & III only (D) II & IV only

   ›Reveal solutionSolution

   The question tests your understanding of electron gain enthalpy, ionization enthalpy, and periodic trends. Statements I and III are false; II and IV are true. So the incorrect statements are I and III only, which corresponds to option (C).

   Concept and Intuition

   We need to evaluate each statement based on fundamental chemistry principles:

   • Electron gain enthalpy can be exothermic or endothermic depending on the element (e.g., noble gases have positive values).
   • Ionization enthalpy is always endothermic because energy is required to remove an electron.
   • Periodic trends: Ionization enthalpy generally increases across a period, but nitrogen has a higher first ionization enthalpy than oxygen due to its half-filled p-subshell stability.
   • Electropositivity increases down a group; lithium is the least electropositive in Group 1.

   Let’s examine each statement step by step.

   1. Statement I: “The process of adding an electron to a gaseous atom is always exothermic”

      • Adding an electron releases energy (exothermic) for most atoms, but not all. For example, noble gases (like neon) have a completely filled valence shell, so adding an electron requires energy input (endothermic).
      • Also, elements like beryllium and magnesium have stable s² configurations, making electron gain endothermic.
      • Therefore, the statement is false because it says “always.”

   2. Statement II: “The process of removing an electron from a gaseous atom is always endothermic”

      • Removing an electron always requires energy to overcome the electrostatic attraction between the electron and the nucleus.
      • Even for highly electropositive metals, energy is needed (though it may be small).
      • This statement is true.

   3. Statement III: “The first ionization enthalpy of oxygen is greater than nitrogen”

      • In the periodic table, across period 2, ionization enthalpy generally increases from left to right. However, nitrogen (atomic number 7) has a half-filled 2p³ configuration, which is extra stable. Oxygen (atomic number 8) has one electron in a doubly occupied 2p orbital, causing electron-electron repulsion that makes it easier to remove an electron.
      • Thus, nitrogen’s first ionization enthalpy (1402 kJ/mol) is actually higher than oxygen’s (1314 kJ/mol).
      • So the statement “oxygen is greater than nitrogen” is false. …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Identify the sets which contain either only extensive properties or only intensive properties I. Density, specific heat, surface tension II. Mole fraction, Internal energy, viscosity III. Molality, pressure, Entropy IV. Number of moles, Enthalpy, Heat capacity (A) I, IV only (B) I, II only (C) II, III only (D) II, IV only

   ›Reveal solutionSolution

   The key is to classify each property as extensive (depends on amount) or intensive (independent of amount). Only sets where all entries are of the same type qualify. The correct sets are I (all intensive) and IV (all extensive), so the answer is (A).

   ---

   Concept & Intuition

   In thermodynamics, properties are split into two fundamental categories:

   • Extensive properties scale with the size or amount of the system (e.g., mass, volume, energy). If you double the system, these double.
   • Intensive properties are independent of the system’s size (e.g., temperature, pressure, density). They remain the same whether you have a drop or a bucket.

   The trick is that some properties can be derived from extensive ones (like density = mass/volume) and become intensive. Also, some properties (like heat capacity) are extensive, but their specific versions (like specific heat) are intensive. We must check each item carefully.

   ---

   Step-by-step analysis

   1. Set I: Density, specific heat, surface tension

      • Density = mass/volume → intensive (ratio of two extensive properties).
      • Specific heat = heat capacity per unit mass → intensive.
      • Surface tension = force per unit length → does not depend on how much liquid you have → intensive. → All three are intensive. ✓

   2. Set II: Mole fraction, Internal energy, viscosity

      • Mole fraction = ratio of moles of one component to total moles → intensive.
      • Internal energy = total energy of the system → doubles if you double the amount → extensive.
      • Viscosity = resistance to flow → a material property, independent of amount → intensive. → Mixed: one extensive, two intensive. ✗

   3. Set III: Molality, pressure, Entropy

      • Molality = moles of solute per kg of solvent → intensive (ratio).
      • Pressure = force/area → intensive.
      • Entropy = total disorder measure → doubles with system size → extensive. → Mixed: two intensive, one extensive. ✗

   4. Set IV: Number of moles, Enthalpy, Heat capacity …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.At 298 K for reaction X ⇌ Y if $\Delta H^\circ = 28.4$ kJ and equilibrium constant is $10^{-7}$, then the standard entropy change for the reaction (in J K$^{-1}$) is (R = 8.3 JK$^{-1}$mol$^{-1}$) (A) +17.5 (B) +38.5 (C) -17.5 (D) -38.5

   ›Reveal solutionSolution

   Use $\Delta G^\circ = -RT\ln K$ to find $\Delta G^\circ$, then $\Delta S^\circ = (\Delta H^\circ - \Delta G^\circ)/T$ — the answer is –38.5 J K$^{-1}$.

   The key here is the thermodynamic relationship between standard Gibbs free energy, enthalpy, and entropy. At a given temperature, the equilibrium constant $K$ tells us $\Delta G^\circ$ directly, and once we have $\Delta G^\circ$ and $\Delta H^\circ$, the standard entropy change $\Delta S^\circ$ follows from $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

   Let’s work it through.

   1. Find $\Delta G^\circ$ from $K$ The standard Gibbs free energy change is related to the equilibrium constant by

   $\Delta G^\circ = -RT \ln K$

   Here $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$, $T = 298\ \text{K}$, and $K = 10^{-7}$.

   So

   $\Delta G^\circ = - (8.3)(298) \ln(10^{-7})$

   Since $\ln(10^{-7}) = -7 \ln 10$ and $\ln 10 \approx 2.3026$,

   $\ln(10^{-7}) = -7 \times 2.3026 = -16.1182$

   Therefore

   $\Delta G^\circ = - (8.3)(298)(-16.1182) = + (8.3)(298)(16.1182)$

   Compute stepwise: $8.3 \times 298 = 2473.4$, then $2473.4 \times 16.1182 \approx 39860\ \text{J mol}^{-1}$ (or $39.86\ \text{kJ mol}^{-1}$).

   So $\Delta G^\circ \approx +39.86\ \text{kJ mol}^{-1}$.

   Watch out

   A common mistake is forgetting the sign: $\ln(10^{-7})$ is negative, so the two negatives cancel, giving a positive $\Delta G^\circ$. A positive $\Delta G^\circ$ is consistent with a very small $K$ — the reaction hardly proceeds forward at standard conditions.

   2. Use $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ Rearranging for $\Delta S^\circ$: …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.In a Carnot's cycle, if $V_1$ and $V_2$ are respectively the initial and final volumes of the working substance during isothermal expansion process, $V_3$ and $V_4$ are respectively the initial and final volumes of the working substance during isothermal compression process, then the relation among $V_1$, $V_2$, $V_3$ and $V_4$ is (A) $V_1 + V_3 = V_2 + V_4$ (B) $V_1 V_2 = V_3 V_4$ (C) $V_1 V_3 = V_2 V_4$ (D) $V_1 + V_2 = V_3 + V_4$

   ›Reveal solutionSolution

   In a Carnot cycle, the isothermal steps are connected by the adiabatic steps, and the relation between the four volumes is $V_1 V_3 = V_2 V_4$, which corresponds to option (C).

   The Carnot cycle is the most efficient heat engine cycle possible, and it consists of two reversible isothermal processes and two reversible adiabatic processes. The question asks for the relationship between the four volumes at the start and end of each isothermal step.

   The key insight is that the two adiabatic processes connect the isothermal expansion to the isothermal compression. On a $P$-$V$ diagram, the cycle goes: isothermal expansion from $V_1$ to $V_2$ at temperature $T_1$, then adiabatic expansion from $V_2$ to $V_3$ (temperature drops to $T_2$), then isothermal compression from $V_3$ to $V_4$ at $T_2$, and finally adiabatic compression from $V_4$ back to $V_1$ (temperature rises back to $T_1$).

   The relation among the volumes comes from applying the adiabatic condition to both the expansion and compression legs.

   1. Adiabatic expansion (from state 2 to state 3): For a reversible adiabatic process of an ideal gas, $TV^{\gamma-1} = \text{constant}$, where $\gamma = C_P/C_V$.

      So $T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$.

   2. Adiabatic compression (from state 4 to state 1): Similarly,

      $T_2 V_4^{\gamma-1} = T_1 V_1^{\gamma-1}$.

   3. Combine the two equations: From the first, $\frac{T_1}{T_2} = \left(\frac{V_3}{V_2}\right)^{\gamma-1}$. From the second, $\frac{T_1}{T_2} = \left(\frac{V_4}{V_1}\right)^{\gamma-1}$.

      Equating the two expressions for $T_1/T_2$:

   $\left(\frac{V_3}{V_2}\right)^{\gamma-1} = \left(\frac{V_4}{V_1}\right)^{\gamma-1}$

   4. Since $\gamma-1 \neq 0$ (for any real gas), we can take the $(\gamma-1)$-th root of both sides, giving: …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.One mole of an ideal gas at 300 K and 20 atm expands to 2 atm under isothermal and reversible conditions. The work done by the gas is $-x$ kJ mol$^{-1}$. The value of $x$ is (R = 8.3 J K$^{-1}$ mol$^{-1}$) (A) 5.73 (B) 7.37 (C) 3.75 (D) 4.57

   ›Reveal solutionSolution

   For an isothermal reversible expansion of an ideal gas, the work done is given by $W = -nRT \ln(V_f/V_i) = -nRT \ln(P_i/P_f)$. Plugging in the values gives $W \approx -5.73$ kJ/mol, so $x = 5.73$.

   The key concept here is the work done in an isothermal reversible process for an ideal gas. Because the temperature is constant, the internal energy of an ideal gas doesn't change ($\Delta U = 0$). From the first law, $\Delta U = Q + W$, so $Q = -W$. But more importantly, the work is not simply $-P\Delta V$ (which only applies for constant pressure). Instead, we must integrate the pressure-volume work along the reversible path, using the ideal gas law to relate pressure and volume at every step.

   Why this approach works:

   For a reversible process, the external pressure is always infinitesimally close to the internal pressure of the gas. So we can write $W = -\int_{V_i}^{V_f} P_{\text{ext}} \, dV = -\int_{V_i}^{V_f} P_{\text{gas}} \, dV$. Using $PV = nRT$ and constant $T$, we get $P = nRT/V$, leading to a simple logarithmic integral.

   1. Identify the formula For an isothermal reversible expansion of $n$ moles of an ideal gas:

   $W = -nRT \ln\left(\frac{V_f}{V_i}\right)$

   Since pressure and volume are inversely related at constant temperature ($P_i V_i = P_f V_f$), we can also write:

   $W = -nRT \ln\left(\frac{P_i}{P_f}\right)$

   2. Plug in the given values

      • $n = 1$ mol
      • $R = 8.3$ J K$^{-1}$ mol$^{-1}$
      • $T = 300$ K
      • $P_i = 20$ atm, $P_f = 2$ atm

      So:

   $W = - (1)(8.3)(300) \ln\left(\frac{20}{2}\right)$

   $W = -2490 \ln(10)$

   3. Compute the natural logarithm $\ln(10) \approx 2.302585$ …

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6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The isobars of one mole of an ideal gas were obtained at three different pressures ($p_1$, $p_2$ and $p_3$). The slopes of these isobars are $m_1$, $m_2$ and $m_3$ respectively. If $p_1 < p_2 < p_3$, then the correct relation of the slopes is (A) $m_1 > m_2 > m_3$ (B) $m_1 < m_2 < m_3$ (C) $m_1 > m_3 > m_2$ (D) $m_1 = m_2 = m_3$

   ›Reveal solutionSolution

   For an ideal gas, the slope of an isobar on a V–T diagram is $\frac{nR}{p}$; since pressure is inversely proportional to slope, the smallest pressure gives the steepest slope, so $m_1 > m_2 > m_3$.

   The key concept is the ideal gas law $pV = nRT$. An isobar is a curve at constant pressure. If we plot volume $V$ against temperature $T$, the equation becomes $V = \frac{nR}{p} T$. This is a straight line through the origin with slope $\frac{nR}{p}$. The slope is inversely proportional to pressure: lower pressure → steeper slope.

   1. Identify the relationship

      For one mole ($n = 1$), the ideal gas law gives $V = \frac{R}{p} T$. So on a $V$ vs. $T$ graph, the slope $m = \frac{R}{p}$.

   2. Compare slopes for different pressures

      Since $R$ is constant, slope $m$ is inversely proportional to pressure $p$. That is, $m \propto \frac{1}{p}$.

   3. Apply the given ordering

      We have $p_1 < p_2 < p_3$. The smallest pressure gives the largest slope, and the largest pressure gives the smallest slope. Therefore:

   $m_1 > m_2 > m_3.$

   4. Match with options …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The $C_p$ of an ideal gas is $10.314 \ \mathrm{J \ mol^{-1} \ K^{-1}}$. One mole of this gas is expanded against a constant pressure of p atm. The change in temperature during expansion is $1.0 \ \mathrm{K}$. The values of $q$ (in J) and $\Delta H$ (in $\mathrm{J \ mol^{-1}}$) are respectively (A) $10.314, \ 10.314$ (B) $2.000, \ 10.314$ (C) $10.314, \ 2.000$ (D) $2.000, \ 2.000$

   ›Reveal solutionSolution

   At constant pressure the heat absorbed equals the enthalpy change, $q=\Delta H=nC_p\Delta T=10.314\ \text{J}$. Correct option: (A).

   Enthalpy change. For an ideal gas the enthalpy change depends only on temperature:

   $\Delta H=nC_p\,\Delta T=(1\ \text{mol})(10.314\ \text{J mol}^{-1}\text{K}^{-1})(1.0\ \text{K})=10.314\ \text{J mol}^{-1}.$

   Heat at constant pressure. The gas is taken through a constant-pressure change, and for a constant-pressure process the heat exchanged equals the enthalpy change:

   $q_p=\Delta H=10.314\ \text{J}.$

   So $q=10.314\ \text{J}$ and $\Delta H=10.314\ \text{J mol}^{-1}$. …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The isobars of one mole of an ideal gas were obtained at three different pressures ($p_1$, $p_2$ and $p_3$). The slopes of these isobars are $m_1$, $m_2$ and $m_3$ respectively. If $p_1 < p_2 < p_3$, then the correct relation of the slopes is (A) $m_1 > m_3 > m_2$ (B) $m_1 > m_2 > m_3$ (C) $m_1 < m_2 < m_3$ (D) $m_1 = m_2 = m_3$

   ›Reveal solutionSolution

   An isobar for an ideal gas is a graph of volume versus temperature at constant pressure. The slope of such a graph is inversely proportional to the pressure. Since $p_1 < p_2 < p_3$, the corresponding slopes will be in the order $m_1 > m_2 > m_3$.

   When we talk about an isobar for an ideal gas, we are referring to a process where the pressure of the gas remains constant. The ideal gas law, $PV = nRT$, describes the relationship between pressure ($P$), volume ($V$), number of moles ($n$), gas constant ($R$), and absolute temperature ($T$).

   To understand the slope of an isobar, we need to visualize it on a graph. Typically, an isobar is plotted on a Volume-Temperature ($V-T$) graph. In such a graph, the volume $V$ is usually on the y-axis and the temperature $T$ on the x-axis. The slope of this graph will tell us how volume changes with temperature at a constant pressure.

   Let's derive the relationship:

   1. Start with the Ideal Gas Law: The fundamental equation governing ideal gases is:

   $PV = nRT$

   Here, $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the absolute temperature.
   

   2. Express Volume as a function of Temperature for an Isobar:

   For an isobaric process, the pressure $P$ is constant. We are also given that we have one mole of gas, so $n=1$. The gas constant $R$ is, by definition, a constant.

   Rearranging the ideal gas law to express $V$ in terms of $T$:

   $V = \left(\frac{nR}{P}\right)T$

   Since $n=1$, this becomes:
   

   $V = \left(\frac{R}{P}\right)T$

   3. Identify the Slope: This equation, $V = \left(\frac{R}{P}\right)T$, is in the form of a straight line equation, $y = mx + c$, where $y=V$, $x=T$, and the y-intercept $c=0$. Therefore, the slope $m$ of the $V-T$ graph (the isobar) is given by:

   $m = \frac{R}{P}$

   4. Relate Slope to Pressure: From the expression $m = \frac{R}{P}$, we can see that the slope $m$ is inversely proportional to the pressure $P$, because $R$ is a constant.

   $m \propto \frac{1}{P}$

   This means that if the pressure $P$ increases, the slope $m$ decreases, and if the pressure $P$ decreases, the slope $m$ increases.
   

   5. Apply the Given Pressure Condition:

   We are given three different pressures $p_1$, $p_2$, and $p_3$, with the relation:

   $p_1 < p_2 < p_3$ …

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9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The $C_p$ of an ideal gas is $10.314~\mathrm{J~mol^{-1}~K^{-1}}$. One mole of this gas is expanded against a constant pressure of $p$ atm. The change in temperature during expansion is $1.0~\mathrm{K}$. The values of $q$ (in J) and $\Delta H$ (in J mol$^{-1}$) are respectively (A) 10.314, 10.314 (B) 2.000, 10.314 (C) 10.314, 2.000 (D) 2.000, 2.000

   ›Reveal solutionSolution

   The expansion is at constant pressure, so $q_p = \Delta H = nC_p\Delta T = 10.314\ \text{J}$, and $\Delta H = 10.314\ \text{J mol}^{-1}$. Both equal $10.314$.

   Enthalpy change

   For an ideal gas $\Delta H$ depends only on temperature:

   $\Delta H = nC_p\Delta T = (1)(10.314)(1.0) = 10.314\ \text{J mol}^{-1}.$

   Heat exchanged

   The gas is expanded against a constant pressure — a constant-pressure process — and the heat exchanged at constant pressure is exactly the enthalpy change:

   $q_p = \Delta H = 10.314\ \text{J}.$

   First-law check. With $C_v = C_p - R = 10.314 - 8.314 = 2.000\ \text{J mol}^{-1}\text{K}^{-1}$: …

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10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.The electron gain enthalpy (in kJ/mol) of oxygen, sulphur, selenium respectively are (A) $-195, -200, -141$ (B) $-195, -141, -200$ (C) $-141, -200, -195$ (D) $-141, -195, -200$

    ›Reveal solutionSolution

    Electron gain enthalpy generally becomes less negative down a group, but oxygen shows an anomaly due to its small size and high electron density. Sulfur has the most negative electron gain enthalpy, followed by selenium, and then oxygen. The correct values for oxygen, sulfur, and selenium respectively are $\boxed{-141, -200, -195}$.

    Concept and Intuition

    Electron gain enthalpy ($\Delta_{eg}H$) is the energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion. A negative value indicates an exothermic process (energy is released), meaning the atom has an affinity for electrons. A more negative value signifies a greater tendency to accept an electron.

    General Trend Down a Group:

    As we move down a group in the periodic table, the atomic size increases. The outermost electrons are further away from the nucleus, and the shielding effect from inner electrons also increases. This reduces the effective nuclear charge experienced by an incoming electron. Consequently, the attraction for an additional electron decreases, making the electron gain enthalpy less negative (or more positive). So, generally, for elements in the same group, the electron gain enthalpy becomes less negative as you go down.

    Anomaly for Second Period Elements:

    Elements of the second period (like oxygen, fluorine, nitrogen) are exceptionally small. When an electron is added to such a small atom, the existing electrons are confined to a very small volume. This leads to a high electron density and significant inter-electronic repulsion between the incoming electron and the already present electrons. This repulsion counteracts the nuclear attraction, making the electron gain process less exothermic than expected, or even endothermic. Therefore, the electron gain enthalpy of a second-period element is often less negative than that of its third-period counterpart.

    Step-by-Step Solution

    1. Understanding Electron Gain Enthalpy:

       Electron gain enthalpy is a measure of an atom's ability to accept an electron. A negative value means energy is released when an electron is added, indicating a stable anion is formed. The more negative the value, the greater the electron affinity.

    2. Trend for Sulfur (S) and Selenium (Se):

       Sulfur and selenium are in Group 16. Sulfur is above selenium in the group. According to the general trend, as we move down the group from sulfur to selenium, the atomic size increases. This means the attraction for an incoming electron decreases. Therefore, sulfur should have a more negative (more exothermic) electron gain enthalpy than selenium.

       So, $\Delta_{eg}H(\text{S}) < \Delta_{eg}H(\text{Se})$.

    3. Anomaly for Oxygen (O) compared to Sulfur (S):

       Oxygen is the first element in Group 16, and it is a second-period element. Due to its extremely small size, when an electron is added to an oxygen atom, the incoming electron experiences strong repulsion from the already existing electrons in the compact $2p$ subshell. This inter-electronic repulsion makes the electron gain process less favorable (less exothermic) than it would otherwise be. As a result, oxygen's electron gain enthalpy is less negative than that of sulfur, its larger congener in the same group.

       So, $\Delta_{eg}H(\text{O}) > \Delta_{eg}H(\text{S})$.

       Watch out

       A common mistake is to assume that because oxygen is smaller and has a higher effective nuclear charge, it should have a more negative electron gain enthalpy than sulfur. However, the dominant factor for second-period elements like oxygen is the strong inter-electronic repulsion due to their very small size.

    4. Combining the Trends and Ordering the Values:

       From step 2, we know $\Delta_{eg}H(\text{S})$ is more negative than $\Delta_{eg}H(\text{Se})$.

       From step 3, we know $\Delta_{eg}H(\text{O})$ is less negative than $\Delta_{eg}H(\text{S})$.

       This implies that sulfur has the most negative electron gain enthalpy among the three. …

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11. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Two statements are given below Statement I: Liquid A and liquid B form a non-ideal solution with positive deviation. The interactions between A and B are weaker than A–A and B–B interactions Statement II: For an ideal solution, $\Delta_{\text{mix}} H = 2$ kJ mol$^{-1}$; $\Delta_{\text{mix}} V = 0$ (A) Both statement I and II are correct (B) Both statement I and II are not correct (C) Statement I is correct but statement II is not correct (D) Statement I is not correct but statement II is correct

    ›Reveal solutionSolution

    Statement I is correct: positive deviation arises when A–B interactions are weaker than A–A and B–B interactions. Statement II is false: for an ideal solution, both $\Delta_{\text{mix}} H$ and $\Delta_{\text{mix}} V$ are zero. Therefore, the correct option is (C).

    Concept and intuition

    The key idea is understanding deviation from ideality in liquid mixtures. In an ideal solution, the intermolecular forces between unlike molecules (A–B) are equal in strength to those between like molecules (A–A and B–B). When A–B interactions are weaker, the molecules tend to escape more easily, leading to a higher vapor pressure than predicted by Raoult’s law — this is positive deviation. Also, for an ideal solution, mixing involves no change in enthalpy or volume; both $\Delta_{\text{mix}} H$ and $\Delta_{\text{mix}} V$ are exactly zero. Statement II incorrectly claims a nonzero $\Delta_{\text{mix}} H$.

    Step-by-step reasoning

    1. Analyze Statement I

       Positive deviation from Raoult’s law occurs when the vapor pressure of the mixture is greater than expected. This happens when A–B interactions are weaker than A–A and B–B interactions, because molecules find it easier to leave the liquid phase. The statement exactly matches this definition, so Statement I is correct.

    2. Analyze Statement II …

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12. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.In a process, the work done by the system is equal to the decrease in its internal energy. The process that the system undergoes is (A) isothermal process (B) adiabatic process (C) isobaric process (D) isochoric process

    ›Reveal solutionSolution

    The key idea is the first law of thermodynamics: $\Delta U = Q - W$. If work done by the system equals the decrease in internal energy ($W = -\Delta U$), then $Q = 0$, which defines an adiabatic process. The correct option is (B).

    The problem states: “In a process, the work done by the system is equal to the decrease in its internal energy.” That is, $W = -\Delta U$ (since a decrease means $\Delta U$ is negative, so $-\Delta U$ is positive). We need to identify which thermodynamic process satisfies this condition.

    Concept and intuition:

    The first law of thermodynamics is the energy conservation statement for a system: the change in internal energy equals heat added minus work done by the system. If the work done by the system exactly matches the drop in internal energy, then no heat is exchanged — the process is thermally insulated. That’s the hallmark of an adiabatic process.

    Let’s verify step by step.

    1. Write the first law The standard form is:

    $\Delta U = Q - W$

    where $Q$ is heat added to the system, $W$ is work done by the system, and $\Delta U$ is the change in internal energy.

    2. Translate the given condition “Work done by the system is equal to the decrease in its internal energy” means:

    $W = -\Delta U$

    (If internal energy decreases, $\Delta U < 0$, so $-\Delta U > 0$, matching positive work done by the system.)

    3. Substitute into the first law Replace $W$ with $-\Delta U$:

    $\Delta U = Q - (-\Delta U) = Q + \Delta U$

    Subtract $\Delta U$ from both sides:

    $0 = Q$

    So $Q = 0$ — no heat exchange.

    4. Identify the process A process with $Q = 0$ is called an adiabatic process.
       • Isothermal ($\Delta T = 0$) would have $\Delta U = 0$ for an ideal gas, so $W = Q$, not $W = -\Delta U$. …

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