Q.Match the following :
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Showing the 12 most recent of 19 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.For an ideal gas undergoing an isothermal change, there is $\_\_\_\_$ (A) no change in Internal energy of the system and heat released by the system is equal to the work done by the system (B) an increase in Internal energy of the system and heat absorbed by the system is greater than the work done on the system (C) a decrease in Internal energy of the system and heat released by the system is equal to the work done by the system (D) no change in Internal energy of the system and heat absorbed by the system is equal to the work done by the system

   ›Reveal solutionSolution

   For an ideal gas undergoing an isothermal process, temperature is constant, so internal energy (which depends only on temperature) does not change. By the first law of thermodynamics, any heat absorbed equals the work done by the gas. The correct option is (D).

   The key concept here is the first law of thermodynamics combined with the properties of an ideal gas. The first law states:

   $\Delta U = Q - W$

   where $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system, and $W$ is the work done by the system. For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature is constant, so $\Delta U = 0$. This immediately tells us that $Q = W$. The sign convention matters: if the gas expands, it does positive work and absorbs heat; if compressed, work is done on it and heat is released. The question’s phrasing “heat absorbed by the system is equal to the work done by the system” matches the expansion case, which is the standard interpretation.

   Let’s walk through the reasoning step by step:

   1. Identify the process: “Isothermal change” means the temperature of the gas remains constant throughout the process. For an ideal gas, internal energy is a function of temperature only (since intermolecular forces are negligible). Therefore, $\Delta U = 0$.

   2. Apply the first law: With $\Delta U = 0$, the first law $\Delta U = Q - W$ becomes $0 = Q - W$, so $Q = W$. This means the heat added to the system equals the work done by the system. If the gas expands, $W > 0$ and $Q > 0$ (heat absorbed). If compressed, $W < 0$ and $Q < 0$ (heat released). The statement in option (D) says “heat absorbed by the system is equal to the work done by the system,” which is exactly the expansion scenario.

   3. Evaluate the options:

      • (A) says “no change in internal energy” (correct) but “heat released by the system is equal to the work done by the system.” This would imply $Q$ negative and $W$ negative, which is possible for compression, but the phrasing “heat released” and “work done by the system” (positive work) are contradictory in sign. Typically, “work done by the system” is taken as positive, so this option is inconsistent.
      • (B) says internal energy increases — false for isothermal.
      • (C) says internal energy decreases — false. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.1 kg of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature. Volume of 1 cc of water changes to $1671 \times 10^3 \mathrm{cc}$ on boiling. The change in internal energy of the system is (Latent Heat of vaporisation of water is $22.68 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1} ; 1 \mathrm{~atm}=1.0 \times 10^5 \mathrm{~Pa}$ ) (A) $21.01 \times 10^3 J$ (B) $19.01 \times 10^5 J$ (C) $20.1 \times 10^5 J$ (D) $21.01 \times 10^5 J$

   ›Reveal solutionSolution

   The change in internal energy is found from the first law: $\Delta U = Q - W$, where $Q$ is the latent heat absorbed and $W$ is the work done against the atmosphere during expansion. The result is $21.01 \times 10^5 \, \mathrm{J}$, which corresponds to option (D).

   Concept & Intuition

   When water boils at constant pressure (1 atm), it absorbs latent heat to break intermolecular bonds, but it also does work pushing the atmosphere back as it expands dramatically (from 1 cc to 1671 litres). The first law of thermodynamics tells us that the heat added ($Q$) goes partly into increasing internal energy ($\Delta U$) and partly into doing work ($W$). So we compute $Q$ from the latent heat, $W$ from $P\Delta V$, and subtract.

   Step-by-step solution

   1. Heat absorbed ($Q$) The latent heat of vaporisation is $L = 22.68 \times 10^5 \, \mathrm{J/kg}$. For 1 kg of water:

   $Q = m L = 1 \times 22.68 \times 10^5 = 22.68 \times 10^5 \, \mathrm{J}.$

   2. Work done ($W$) At constant pressure, $W = P \Delta V$.
      • Initial volume of water: $V_i = 1 \, \mathrm{cc} = 1 \times 10^{-6} \, \mathrm{m}^3$.
      • Final volume of steam: $V_f = 1671 \times 10^3 \, \mathrm{cc} = 1671 \times 10^{-3} \, \mathrm{m}^3 = 1.671 \, \mathrm{m}^3$.
      • Change in volume: $\Delta V = V_f - V_i \approx 1.671 \, \mathrm{m}^3$ (the tiny initial volume is negligible).
      • Pressure: $P = 1 \, \mathrm{atm} = 1.0 \times 10^5 \, \mathrm{Pa}$. Hence, $W = P \Delta V = (1.0 \times 10^5) \times 1.671 = 1.671 \times 10^5 \, \mathrm{J}.$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.The Enthalpy of combustion of 1.0 mole of a reactive metal X at $27^{\circ} \mathrm{C}$ and 1.0 bar pressure to form a solid oxide ( XO ) is $-601.83 \mathrm{~kJ} / \mathrm{mol}$. The Internal energy change for this reaction is ___________ kJ. $\left(\mathrm{R}=8.314 \mathrm{j} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ (A) 701.33 (B) 754.58 (C) $-$614.30 (D) $-$600.58

   ›Reveal solutionSolution

   The key idea is to relate enthalpy change ($\Delta H$) to internal energy change ($\Delta U$) via $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in moles of gas. For the reaction $\text{X(s)} + \frac12 \text{O}_2\text{(g)} \to \text{XO(s)}$, $\Delta n_g = -0.5$, giving $\Delta U = -601.83 - (-0.5)(8.314 \times 10^{-3})(300) = -600.58$ kJ/mol. The correct option is (D).

   The relationship between enthalpy and internal energy is fundamental in thermochemistry. Enthalpy ($H$) is defined as $H = U + PV$, so for a constant-pressure process, the change in enthalpy equals the heat exchanged. The difference between $\Delta H$ and $\Delta U$ comes from the $PV$ work done when gases expand or contract. For reactions involving gases, the key formula is:

   $\Delta H = \Delta U + \Delta n_g \, RT$

   where $\Delta n_g$ = (moles of gaseous products) – (moles of gaseous reactants).

   Here, the reaction is:

   $\text{X(s)} + \frac12 \text{O}_2\text{(g)} \to \text{XO(s)}$

   We are given $\Delta H = -601.83$ kJ/mol at $27^\circ$C (300 K) and 1 bar. We need $\Delta U$.

   Step-by-step:

   1. Determine $\Delta n_g$

      Reactants: 1 solid + $\frac12$ mole of gas. Products: 1 solid (no gas).

      So $\Delta n_g = 0 - \frac12 = -\frac12$ mol.

   2. Convert units consistently

      $R = 8.314$ J K$^{-1}$ mol$^{-1}$ = $8.314 \times 10^{-3}$ kJ K$^{-1}$ mol$^{-1}$.

      Temperature $T = 27 + 273 = 300$ K.

   3. Compute the $PV$ work term

      $\Delta n_g RT = (-0.5) \times (8.314 \times 10^{-3}) \times 300$

      $= -0.5 \times 2.4942 = -1.2471$ kJ/mol.

   4. Solve for $\Delta U$ …

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4. COMEDK 2024Set 2024-A1 markMCQ
   Q.At Constant volume, the heat required to raise the temperature of $4.48 \mathrm{~L}$ of an ideal gas at STP by $15^{\circ} \mathrm{C}$ is 12.0 calories. The $\mathrm{C_p}$ of the gas is _____________ $(\mathrm{R}=2 \mathrm{~Cal} \mathrm{~kg}^{-1} \mathrm{~mol}^{-1})$ (A) 6 cal (B) 9 cal (C) 3 cal (D) 12 cal

   ›Reveal solutionSolution

   Using the first law at constant volume, we find $C_v$ from the given heat and moles, then use $C_p = C_v + R$ to get $C_p = 6\ \text{cal mol}^{-1} \text{K}^{-1}$, which corresponds to option (A).

   Concept & Intuition

   At constant volume, no work is done ($W=0$), so the heat added equals the change in internal energy: $Q = n C_v \Delta T$. Once we know $C_v$, we can find $C_p$ using the relation $C_p - C_v = R$ for an ideal gas. The trick is to first find the number of moles from the given volume at STP.

   Step-by-step solution

   1. Find the number of moles of gas At STP (Standard Temperature and Pressure: $0^\circ\text{C}$, 1 atm), 1 mole of an ideal gas occupies 22.4 L. Given volume = 4.48 L, so

   $n = \frac{4.48}{22.4} = 0.2\ \text{moles}.$

   2. Use the constant-volume heat equation At constant volume, $Q = n C_v \Delta T$. Given $Q = 12.0$ cal, $\Delta T = 15^\circ\text{C} = 15\ \text{K}$, and $n = 0.2$:

   $12 = 0.2 \times C_v \times 15$

   $12 = 3 C_v \quad \Rightarrow \quad C_v = 4\ \text{cal mol}^{-1} \text{K}^{-1}.$

   3. Apply the relation between $C_p$ and $C_v$ For an ideal gas: $C_p - C_v = R$. Here $R = 2\ \text{cal mol}^{-1} \text{K}^{-1}$ (note: the problem gives $R$ in these units, so we use them directly).

   $C_p = C_v + R = 4 + 2 = 6\ \text{cal mol}^{-1} \text{K}^{-1}.$

   Watch out

   A common mistake is to forget that $R$ is given in $\text{cal mol}^{-1} \text{K}^{-1}$ and to try converting units. The problem explicitly states $R=2\ \text{Cal kg}^{-1} \text{mol}^{-1}$ — but note the capital C in "Cal" is a typographical quirk; it means calories per mole per Kelvin. So just use $R=2$ directly.

   4. Check the options …

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5. COMEDK 2024Set 2024-A1 markMCQ
   Q.The correct option for free expansion of an ideal gas under adiabatic condition is: (A) $\mathrm{q}=0, \quad \Delta \mathrm{T} \neq 0, \quad \mathrm{w}=0$ (B) $\mathrm{q}=0, \quad \Delta \mathrm{T}<0, \quad \mathrm{w} \neq 0$ (C) $\mathrm{q} \neq 0, \quad \Delta \mathrm{T}=0, \quad \mathrm{w}=0$ (D) $\mathrm{q}=0, \quad \Delta \mathrm{T}=0, \quad \mathrm{w}=0$

   ›Reveal solutionSolution

   For the free expansion of an ideal gas under adiabatic conditions, no heat is exchanged (q = 0), no work is done (w = 0), and because the internal energy of an ideal gas depends only on temperature, the temperature remains constant (ΔT = 0). The correct option is (D).

   The key concept here is free expansion combined with adiabatic conditions. Free expansion means the gas expands into a vacuum — there is no opposing pressure, so the gas does no work. Adiabatic means no heat enters or leaves the system. For an ideal gas, internal energy depends only on temperature. Since no work is done and no heat is transferred, internal energy doesn’t change, so temperature stays constant.

   Let’s walk through it step by step:

   1. Adiabatic condition means the system is thermally insulated: no heat exchange with the surroundings. Therefore, $q = 0$. This eliminates option (C), which claims $q \neq 0$.

   2. Free expansion means the gas expands into a vacuum. There is no external pressure to push against, so the gas does no work: $w = 0$. This eliminates option (B), which claims $w \neq 0$.

   3. First Law of Thermodynamics states:

   $\Delta U = q + w$

   With $q = 0$ and $w = 0$, we get $\Delta U = 0$. …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.An ideal gas changes its state from $\mathrm{A}$ to $\mathrm{C}$ in two different paths $\mathrm{ABC}$ and $\mathrm{AC}$. The internal energy of the gas at state $\mathrm{C}$ is $20 \mathrm{~J}$ and at state $\mathrm{B}$ is $10 \mathrm{~J}$. Heat supplied to the gas to go from $\mathrm{B} \rightarrow \mathrm{C}$ is (A) 90 J (B) zero (C) 70 J (D) 60 J

   ›Reveal solutionSolution

   Internal energy is a state function, so $\Delta U_{B\to C} = 10$ J; adding the isobaric work $W_{B\to C} = 60$ J from the P–V diagram gives $Q_{B\to C} = 70$ J — option (C).

   Because internal energy depends only on state,

   $\Delta U_{B\to C} = U_C - U_B = 20 - 10 = 10\text{ J}.$

   The step $B\to C$ is an isobaric expansion; the work done by the gas (area under $B\to C$ on the P–V diagram) is $W_{B\to C} = 60$ J. Applying the first law of thermodynamics, with $Q$ the heat supplied to the gas and $W$ the work done by the gas: …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.The latent heat of vaporisation of water is $2240 \mathrm{~J}$. If the work done in the process of vaporisation of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, the increase in internal energy is (A) 1408 J (B) 2072 J (C) 2208 J (D) 2408 J

   ›Reveal solutionSolution

   $\Delta U=Q-W=2240-168=2072$ J.

   Heat supplied (latent heat for 1 g) $Q=2240$ J; work done by the vapour $W=168$ J.

   First law of thermodynamics: …

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8. KCET 2023Set A-31 markMCQ
   Q.The P-V diagram of a Carnot's engine is shown in the graph below. The engine uses 1 mole of an ideal gas as working substance. From the graph, the area enclosed by the P-V diagram is [The heat supplied to the gas is $8000\,\mathrm{J}$] (A) $2000\,\mathrm{J}$ (B) $3000\,\mathrm{J}$ (C) $1000\,\mathrm{J}$ (D) $1200\,\mathrm{J}$

   ›Reveal solutionSolution

   The area enclosed by any closed $P$–$V$ loop is the net work done; for a Carnot engine $W = \eta Q_H$, and $\eta$ follows from the two isotherm temperatures read off the graph via $PV=nRT$.

   1. What the enclosed area means

   For a cyclic process, $\Delta U = 0$, so the first law gives $W_{\text{net}} = Q_{\text{net}}$. Graphically, the area enclosed by the closed loop is exactly the net work done by the gas (positive for a clockwise loop, which this is). So the question is really: what is the work output of this Carnot engine?

   2. Carnot efficiency

   $\eta = \frac{W}{Q_H} = 1 - \frac{T_C}{T_H}$

   We are told $Q_H = 8000\ \mathrm{J}$, so we only need $T_C/T_H$.

   3. Get the temperatures from the graph

   For $n = 1$ mole of an ideal gas, $T = \dfrac{PV}{R}$, so temperature is proportional to the product $PV$. On a Carnot cycle, $A\to B$ is the hot isotherm ($T_H$) and $C\to D$ the cold isotherm ($T_C$), so we may take the $PV$ product at any point on each isotherm. Use the two points whose coordinates fall exactly on printed gridlines:

   • Point $A$ (on the hot isotherm): $P = 1600\ \mathrm{kPa},\ V = 2.5\ \mathrm{cm^3}$

   $(PV)_A = 1600 \times 2.5 = 4000 \ \ \mathrm{kPa\cdot cm^3}$

   • Point $C$ (on the cold isotherm): $P = 400\ \mathrm{kPa},\ V = 6.25\ \mathrm{cm^3}$ $(PV)_C = 400 \times 6.25 = 2500 \ \ \mathrm{kPa\cdot cm^3}$ …

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9. COMEDK 2023Set 2023-E1 markMCQ
   Q.Gaseous Nitrous oxide decomposes at $298 \mathrm{~K}$ to form Nitrogen gas and Oxygen gas. The $\Delta \mathbf{H}$ for the reaction at $1.0 \mathrm{~atm}$ pressure and $298 \mathrm{~K}$ is $\mathbf{- 1 6 3 . 1 5} \mathbf{~ k J}$. Calculate Internal energy change for the decomposition of $100 \mathrm{~g}$ of Nitrous oxide gas under the same conditions of temperature and pressure. (A) $-166$ kJ (B) $-188.2$ kJ (C) $-230.3$ kJ (D) $-376.43$ kJ

   ›Reveal solutionSolution

   Internal energy change: dU = dH - dn_g R T = -185.4 - (1.136)(8.314 x 10^-3 kJ/mol K)(298 K) = -185.4 - 2.81 = -188.2 kJ

   Concept: dU = dH - (dn_g) R T, applied to the amount actually decomposed.

   Reaction: 2 N2O(g) -> 2 N2(g) + O2(g); dH = -163.15 kJ (for 2 mol N2O)

   dn_g = (2 + 1) - 2 = +1 (per 2 mol N2O)

   Moles of N2O in 100 g (M = 44 g/mol):

   n = 100/44 = 2.273 mol

   Scale the reaction: factor = 2.273/2 = 1.136

   dH = -163.15 x 1.136 = -185.4 kJ …

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10. COMEDK 2023Set 2023-M1 markMCQ
    Q.For an adiabatic change in a system, the condition which is applicable is (A) $q=0$ (B) $w=0$ (C) $q=-w$ (D) $q=w$

    ›Reveal solutionSolution

    Adiabatic means “no heat transfer,” so $q=0$.

    By definition, in an adiabatic change the system is thermally insulated and exchanges no heat with the surroundings:

    $q = 0.$ …

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11. COMEDK 2022Set 20221 markMCQ
    Q.The difference between heat capacity at constant pressure and heat capacity at constant volume is (A) $R$ (B) $\frac{R}{T}$ (C) $\frac{R}{V}$ (D) $1$

    ›Reveal solutionSolution

    Derivation: at constant volume, qv = Cv dT = dU. At constant pressure, qp = Cp dT = dH = dU + d(pV) = Cv dT + R dT for one mole of an ideal gas. Hence Cp = Cv + R, i.e. Cp - Cv = R.

    Concept: Relation between molar heat capacities of an ideal gas (Mayer's relation).

    Cp - Cv = R …

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12. COMEDK 2022Set 20221 markMCQ
    Q.If heat engine is filled at temperature 27$^\circ$C and heat of 100 k cal is taken from source at temperature 677$^\circ$C. Work done (in J) is (A) $0.28\times10^6$ (B) $2.8\times10^6$ (C) $28\times10^6$ (D) $0.028\times10^6$

    ›Reveal solutionSolution

    W = ηQ = 0.684 × 4.2 × 10⁵ ≈ 2.87 × 10⁵ J = 0.28 × 10⁶ J (≈0.29 × 10⁶).

    Concept: Maximum (Carnot) efficiency η = 1 − T_sink/T_source, and W = ηQ.

    T_sink = 27 °C = 300 K; T_source = 677 °C = 950 K.

    η = 1 − 300/950 = 650/950 = 0.684.

    Q = 100 kcal = 10⁵ cal = 10⁵ × 4.2 J = 4.2 × 10⁵ J. …

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