Q.In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The temperatures of source and sink of a Carnot heat engine are $127\,^\circ\mathrm{C}$ and $27\,^\circ\mathrm{C}$ respectively. If the working substance of the engine is 2 moles of a rigid diatomic gas, then the decrease in the internal energy of the substance during adiabatic expansion process is (Universal gas constant $= 8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$) (A) 2493 J (B) 4986 J (C) 3324 J (D) 4155 J

   ›Reveal solutionSolution

   During the adiabatic expansion of a Carnot cycle the gas cools from the source temperature to the sink temperature. For a rigid diatomic gas $\Delta U = nC_V\,\Delta T$ with $C_V=\tfrac{5}{2}R$, giving a decrease of $4155\text{ J}$ — option (D).

   Setup

   In a Carnot cycle the adiabatic expansion carries the gas from the high-temperature isotherm $T_1$ (source) down to the low-temperature isotherm $T_2$ (sink). During this step the gas does work entirely at the expense of its internal energy, so its temperature falls from $T_1$ to $T_2$.

   Step 1 — Temperatures in kelvin

   $T_1 = 127 + 273 = 400\text{ K}, \qquad T_2 = 27 + 273 = 300\text{ K}$

   so $\Delta T = T_2 - T_1 = -100\text{ K}$.

   Step 2 — Heat capacity of a rigid diatomic gas

   A rigid diatomic molecule has $5$ degrees of freedom, so

   $C_V = \frac{5}{2}R = \frac{5}{2}\times 8.31 = 20.775\ \text{J mol}^{-1}\text{K}^{-1}.$ …

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2. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Two vessels are filled with ideal gases A and B and are connected through a pipe of zero volume as shown in figure. The stop cock is opened and the gases are allowed to mix homogeneously and the temperature is kept constant. The partial pressures of A and B respectively (in atm) are (A) 8.0, 5 (B) 9.6, 4 (C) 6.4, 4 (D) 4.8, 2

   ›Reveal solutionSolution

   Each gas expands into the whole combined volume at constant $T$, so its partial pressure is $P_iV_i/(V_A+V_B)$. This gives $4.8$ atm for A and $2.0$ atm for B — option (D).

   Concept

   When the stopcock is opened, gases A and B each spread through the total volume $V_A+V_B$. At constant temperature Boyle's law applies to each gas separately, so its new (partial) pressure is

   $p_i = \frac{P_i\,V_i}{V_A+V_B}.$

   Solution …

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3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Which of the following processes are reversible? I. Vaporization of a liquid at its boiling point. II. Expansion of gas into vacuum. III. Transformation of a solid substance into liquid at its melting point. IV. Neutralization of an acid by a base. (A) I & III (B) II & III (C) II & IV (D) I & IV

   ›Reveal solutionSolution

   Reversibility requires that the process be carried out infinitesimally slowly and that the system and surroundings can be restored exactly to their initial states. Only phase changes at constant temperature (boiling/melting) under equilibrium conditions are reversible; free expansion and neutralization are irreversible.

   The key concept is thermodynamic reversibility. A process is reversible if it can be reversed by an infinitesimal change in a variable (like pressure or temperature) and, when reversed, leaves both the system and the surroundings exactly as they were. This usually means the process occurs through a continuous series of equilibrium states — no friction, no turbulence, no spontaneous mixing, and no finite driving force.

   Let’s examine each process:

   1. I. Vaporization of a liquid at its boiling point.

      At the boiling point, liquid and vapor coexist in equilibrium. If we add heat infinitesimally slowly (e.g., using a large reservoir at a temperature just above the boiling point), the liquid vaporizes while staying at the same temperature and pressure. The process can be reversed by infinitesimally lowering the temperature or removing heat. Both system and surroundings can be restored.

      → Reversible.

   2. II. Expansion of gas into vacuum.

      This is a free expansion (Joule expansion). The gas rushes into an empty space with a finite pressure difference — no equilibrium is maintained. The process is sudden and uncontrolled. To reverse it, you would need to compress the gas back, but that requires work input, and the surroundings would not return to their original state (they would have gained heat from the compression).

      → Irreversible.

   3. III. Transformation of a solid into liquid at its melting point.

      Exactly analogous to vaporization: at the melting point, solid and liquid coexist in equilibrium. Adding heat infinitesimally slowly melts the solid without changing temperature. The process can be reversed by infinitesimally removing heat.

      → Reversible.

   4. IV. Neutralization of an acid by a base. …

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4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Which of the following processes are reversible? I. Vaporization of a liquid at its boiling point. II. Expansion of gas into vacuum. III. Transformation of a solid substance into liquid at its melting point. IV. Neutralization of an acid by a base. (A) I & III (B) I & IV (C) II & III (D) II & IV

   ›Reveal solutionSolution

   A reversible process must be carried out infinitesimally slowly so that the system remains in equilibrium with its surroundings at every stage. Only vaporization at the boiling point and melting at the melting point satisfy this condition; free expansion and neutralization are irreversible.

   The key idea is thermodynamic reversibility: a process is reversible if it can be reversed by an infinitesimal change in a variable (like pressure or temperature) and the system passes through a continuous series of equilibrium states. In practice, this means the driving force must be infinitesimally small, and no dissipative effects (friction, turbulence, spontaneous chemical reaction) occur.

   Let’s examine each process:

   1. I. Vaporization of a liquid at its boiling point.

      At the boiling point, liquid and vapor coexist in equilibrium. If we add heat extremely slowly (through a temperature difference approaching zero), the liquid vaporizes while the temperature and pressure remain constant. The process can be reversed by infinitesimally cooling or compressing. This is a classic example of a reversible phase change.

      → Reversible.

   2. II. Expansion of gas into vacuum.

      Here, a gas expands against zero external pressure (free expansion). There is no opposing force, so the gas rushes in uncontrollably. The intermediate states are not equilibrium states (pressure and temperature are not uniform). Moreover, you cannot reverse it by an infinitesimal change — you would need to compress the gas back, which requires work. This is a prototypical irreversible process.

      → Irreversible.

   3. III. Transformation of a solid into liquid at its melting point. …

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5. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The efficiency of a Carnot's heat engine is 25%. If the absolute temperature of the sink is increased by 10%, then the efficiency of the engine is (A) 27.5% (B) 37.5% (C) 17.5% (D) 21.5%

   ›Reveal solutionSolution

   The efficiency of a Carnot engine depends on the absolute temperatures of its hot source and cold sink. When the sink temperature increases, the temperature difference between the source and sink decreases relative to the source temperature, leading to a reduction in efficiency. The new efficiency is $\boxed{17.5\%}$.

   A Carnot engine is an idealized heat engine that operates on the reversible Carnot cycle. It represents the most efficient possible heat engine operating between two given temperature reservoirs. Its efficiency is determined solely by the absolute temperatures of the hot reservoir (source) and the cold reservoir (sink).

   The fundamental concept is that a heat engine converts heat energy into mechanical work. The maximum possible efficiency for this conversion is achieved when the process is reversible, as in a Carnot engine. The efficiency is limited by the second law of thermodynamics, which states that it's impossible to convert all heat into work; some heat must always be rejected to a colder reservoir.

   The efficiency $\eta$ of a Carnot engine is given by:

   $\eta = 1 - \frac{T_2}{T_1}$

   where $T_1$ is the absolute temperature of the hot source and $T_2$ is the absolute temperature of the cold sink. Both temperatures must be expressed in Kelvin.

   This formula shows that efficiency increases as the ratio $T_2/T_1$ decreases. This means a larger temperature difference between the source and sink, or a lower sink temperature, leads to higher efficiency. Conversely, if the sink temperature increases, the ratio $T_2/T_1$ increases, and the efficiency decreases.

   Let's apply this to the problem:

   1. Determine the initial ratio of sink to source temperatures.

      We are given that the initial efficiency of the Carnot engine is 25%. Let $T_1$ be the absolute temperature of the source and $T_2$ be the absolute temperature of the sink.

      Using the efficiency formula:

      $\eta = 1 - \frac{T_2}{T_1}$

      $0.25 = 1 - \frac{T_2}{T_1}$

      Rearranging the equation to find the ratio $\frac{T_2}{T_1}$:

      $\frac{T_2}{T_1} = 1 - 0.25$

      $\frac{T_2}{T_1} = 0.75$

   2. Calculate the new sink temperature.

      The absolute temperature of the sink is increased by 10%. Let the new sink temperature be $T_2'$.

      $T_2' = T_2 + 10\% \text{ of } T_2$

      $T_2' = T_2 + 0.10 T_2$

      $T_2' = 1.1 T_2$

   3. Calculate the new efficiency. …

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6. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Nitrogen and oxygen of volumes one litre each at atmospheric pressure are mixed in a closed container of volume three litres. If the process is isothermal, the ratio of final pressure of the mixture and atmospheric pressure is (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{2}$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   Each gas expands from $1\ \text{L}$ to the full $3\ \text{L}$, so its partial pressure falls to $\tfrac{1}{3}$ atm; the mixture pressure is $\tfrac{1}{3}+\tfrac{1}{3} = \tfrac{2}{3}$ atm, giving a ratio of $\tfrac{2}{3}$.

   Given: $V_{N_2} = 1\ \text{L}$ and $V_{O_2} = 1\ \text{L}$, each initially at atmospheric pressure $P_0$, mixed into a closed container of $V = 3\ \text{L}$ at constant temperature (isothermal).

   Partial pressure of each gas (Boyle's law): each gas now occupies the whole $3\ \text{L}$:

   $P_{N_2} = \frac{P_0 V_{N_2}}{V} = \frac{P_0 \times 1}{3} = \frac{P_0}{3}, \qquad P_{O_2} = \frac{P_0 \times 1}{3} = \frac{P_0}{3}$ …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.An open vessel containing air was heated from $27\,^\circ\mathrm{C}$ to $727\,^\circ\mathrm{C}$. Some air was expelled. What is the fraction of air remaining in the vessel? (Assume air as an ideal gas) (A) $\dfrac{1}{10}$ (B) $\dfrac{7}{10}$ (C) $\dfrac{3}{10}$ (D) $\dfrac{9}{10}$

   ›Reveal solutionSolution

   When an open vessel containing an ideal gas is heated, its pressure and volume remain constant, causing the number of moles of gas to decrease proportionally to the inverse of the absolute temperature. The fraction of air remaining is $\boxed{\dfrac{3}{10}}$.

   The problem describes an open vessel containing air, which is then heated. We need to determine the fraction of air that remains in the vessel after heating. This scenario is a classic application of the ideal gas law, where certain parameters are held constant while others change.

   Concept and Intuition

   The behavior of an ideal gas is described by the ideal gas law:

   $PV = nRT$

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

   Let's break down what happens in this specific situation:

   • Open Vessel: An "open vessel" means it is exposed to the atmosphere. Therefore, the pressure inside the vessel remains constant and equal to the external atmospheric pressure throughout the heating process.
   • Vessel: The vessel itself has a fixed capacity, so its volume ($V$) remains constant.
   • Heating: The temperature ($T$) of the air inside the vessel increases.
   • Air Expelled: Since the pressure and volume are constant, and the temperature increases, the ideal gas law ($PV = nRT$) tells us that the number of moles ($n$) must decrease. This decrease in moles corresponds to air being expelled from the vessel as it heats up and expands.

   Our goal is to find the fraction of air remaining, which is the ratio of the final number of moles ($n_2$) to the initial number of moles ($n_1$), i.e., $\frac{n_2}{n_1}$.

   Step-by-step Solution

   1. Convert temperatures to absolute scale (Kelvin).

      The ideal gas law requires temperature to be in Kelvin.

      Initial temperature, $T_1 = 27\,^\circ\mathrm{C} = 27 + 273 = 300\,\mathrm{K}$.

      Final temperature, $T_2 = 727\,^\circ\mathrm{C} = 727 + 273 = 1000\,\mathrm{K}$.

      Watch out

      Always convert Celsius temperatures to Kelvin when using the ideal gas law or any gas law involving temperature ratios. Failing to do so is a common mistake.

   2. Apply the ideal gas law to the initial state.

      Let $n_1$ be the initial number of moles of air in the vessel.

      $P V = n_1 R T_1$ …

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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.When $Q_1$ amount of heat is supplied to a monatomic gas, the work done by the gas is $W$. When $Q_2$ amount of heat is supplied to a diatomic gas, the work done by the gas is $2W$. Then $Q_1 : Q_2 =$ (A) $2:3$ (B) $3:5$ (C) $5:7$ (D) $5:14$

   ›Reveal solutionSolution

   For an ideal gas undergoing a process, the work done relates to heat supplied through the degrees of freedom. Using $W = Q - \Delta U$ and the first law with specific heat ratios, we find $Q_1 : Q_2 = 5:14$.

   Concept and Intuition

   When heat is supplied to an ideal gas, some energy goes into increasing the internal energy (temperature) and some goes into doing work (expansion). The split between these depends on the degrees of freedom of the gas molecules, which differs between monatomic and diatomic gases.

   For any process, the first law of thermodynamics tells us:

   $Q = \Delta U + W$

   The key insight is that for a given amount of work $W$, different gases require different amounts of heat $Q$ because they have different heat capacities.

   Step-by-Step Solution

   1. Establish the relationship for a constant pressure process

   While the problem doesn't explicitly state the process type, we'll assume an isobaric (constant pressure) process, which is the most common scenario where both heat and work are significant. For such a process:

   • Work done: $W = P\Delta V = nR\Delta T$
   • Heat supplied: $Q = nC_p\Delta T$

   2. Find the ratio of heat to work for each gas

   From the equations above:

   $\frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R}$

   Using the relation $C_p = C_V + R$ and $C_V = \frac{f}{2}R$ where $f$ is degrees of freedom:

   $C_p = \frac{f}{2}R + R = \frac{f+2}{2}R$

   Therefore:

   $\frac{Q}{W} = \frac{f+2}{2}$

   3. Apply to the monatomic gas

   For monatomic gas: $f = 3$ …

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9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.At 400 K, an ideal gas is enclosed in a $0.5\,\mathrm{m}^3$ vessel at a pressure of $203\,\mathrm{kPa}$. What is the change in temperature required (in K), if it occupies a volume of $0.2\,\mathrm{m}^3$ under a pressure of $304\,\mathrm{kPa}$? (Nearest integer) (A) 240 (B) 160 (C) 120 (D) 80

   ›Reveal solutionSolution

   Applying $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ gives $T_2\approx239.6\ \text{K}$, so the temperature must change by $|239.6-400|\approx160\ \text{K}$ - option (B).

   For a fixed amount of ideal gas, $\dfrac{PV}{T}=nR$ is constant, giving the combined gas law:

   $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

   Given: $P_1=203\ \text{kPa},\ V_1=0.5\ \text{m}^3,\ T_1=400\ \text{K}$; $P_2=304\ \text{kPa},\ V_2=0.2\ \text{m}^3$.

   Step 1 - Solve for $T_2$.

   $T_2=T_1\cdot\frac{P_2V_2}{P_1V_1}=400\times\frac{304\times0.2}{203\times0.5}=400\times\frac{60.8}{101.5}$

   $T_2=400\times0.599=239.6\ \text{K}$ …

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10. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.At 1 bar pressure and 373K, the enthalpy change for the vapourisation of 1 mol of water is $41~\text{kJ mol}^{-1}$. The change in internal energy for the same change under the same conditions (in kJ mol$^{-1}$) is (R = $8.3~\text{JK}^{-1}\text{mol}^{-1}$, Assume water vapour as an ideal gas) (A) $-37.9$ (B) $+3.1$ (C) $+37.9$ (D) $+379$

    ›Reveal solutionSolution

    The key is to relate enthalpy change ($\Delta H$) to internal energy change ($\Delta U$) using $\Delta H = \Delta U + \Delta n_g RT$. For vaporisation, $\Delta n_g = +1$, so $\Delta U = \Delta H - RT$. Substituting values gives $\Delta U \approx +37.9~\text{kJ mol}^{-1}$, so the correct option is (C).

    The problem asks for the change in internal energy ($\Delta U$) when 1 mole of water vaporises at 1 bar and 373 K, given the enthalpy change ($\Delta H = 41~\text{kJ mol}^{-1}$). The relationship between $\Delta H$ and $\Delta U$ for a process involving gases is:

    $\Delta H = \Delta U + \Delta (PV)$

    For an ideal gas, $PV = nRT$, so at constant temperature and pressure, $\Delta (PV) = (\Delta n_g) RT$, where $\Delta n_g$ is the change in the number of moles of gas. This is because the volume change of liquids/solids is negligible compared to gases.

    Intuition: When water vaporises, it expands greatly against the atmosphere, doing work. Enthalpy includes this $PV$ work energy, while internal energy does not. So $\Delta U$ will be smaller than $\Delta H$ by the amount of work done.

    ---

    1. Determine $\Delta n_g$

       The reaction is: $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$

       • Moles of gas before: 0
       • Moles of gas after: 1 So $\Delta n_g = 1 - 0 = +1$.

    2. Apply the formula

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

    Rearranging:

    $\Delta U = \Delta H - \Delta n_g RT$

    3. Plug in the values
       • $\Delta H = 41~\text{kJ mol}^{-1} = 41000~\text{J mol}^{-1}$
       • $R = 8.3~\text{J K}^{-1} \text{mol}^{-1}$
       • $T = 373~\text{K}$
       • $\Delta n_g = 1$

    $\Delta U = 41000 - (1)(8.3)(373)$

    Calculate $8.3 \times 373$:

    $8.3 \times 373 = 8.3 \times (370 + 3) = 3071 + 24.9 = 3095.9 \approx 3096~\text{J mol}^{-1}$

    So:

    $$$$

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11. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If 2.5 moles of an ideal gas at a certain temperature are allowed to expand isothermally and reversibly from an initial volume of $2\,\mathrm{dm}^3$ to $20\,\mathrm{dm}^3$, the work done by the gas is $-16.5\,\mathrm{kJ}$. The temperature (in K) of the gas is (Round off to the nearest value) $(R = 8.314\,\mathrm{J\,K}^{-1}\mathrm{mol}^{-1})$ (A) 445 (B) 245 (C) 345 (D) 745

    ›Reveal solutionSolution

    For an isothermal reversible expansion of an ideal gas, the work done is $W = -nRT \ln(V_f/V_i)$. Given $W = -16.5\,\mathrm{kJ}$, $n = 2.5$, $V_i = 2\,\mathrm{dm}^3$, $V_f = 20\,\mathrm{dm}^3$, and $R = 8.314\,\mathrm{J\,K^{-1}mol^{-1}}$, solving gives $T \approx 345\,\mathrm{K}$, so the correct option is (C).

    Concept & Intuition

    When an ideal gas expands isothermally and reversibly, the temperature stays constant, so the internal energy doesn’t change ($\Delta U = 0$). By the first law, $\Delta U = Q + W$, so $Q = -W$. The work done by the gas is the negative of the work done on the gas. For a reversible process, the external pressure equals the gas pressure at every step, so we integrate $P\,dV$ using the ideal gas law $P = nRT/V$. The result is a clean logarithmic expression. The negative sign in the given work ($-16.5\,\mathrm{kJ}$) tells us the gas does positive work on the surroundings (expanding), so the system loses energy.

    Step-by-step solution

    1. Write the formula for reversible isothermal work For an ideal gas expanding reversibly and isothermally:

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

    Here $W$ is the work done by the gas (negative if the gas does work on surroundings, as in expansion).

    2. Plug in the known values

       • $n = 2.5$ mol
       • $R = 8.314\ \mathrm{J\,K^{-1}mol^{-1}}$
       • $V_i = 2\ \mathrm{dm}^3$
       • $V_f = 20\ \mathrm{dm}^3$
       • $W = -16.5\ \mathrm{kJ} = -16500\ \mathrm{J}$

       So:

    $-16500 = - (2.5)(8.314)\, T \ln\left(\frac{20}{2}\right)$

    3. Simplify the logarithm

    $\frac{20}{2} = 10 \quad \Rightarrow \quad \ln(10) \approx 2.302585$

    4. Cancel the negative signs Both sides are negative, so:

    $16500 = (2.5)(8.314)(2.302585)\, T$

    5. Compute the coefficient First, $2.5 \times 8.314 = 20.785$ …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.A certain mass of a gas was brought from state A to B by following three different paths, namely 1, 2 and 3, respectively. Which of the following relations is correct for the work done? (A) $W_1 = W_2 = W_3$ (B) $W_1 < W_2 < W_3$ (C) $W_1 > W_2 > W_3$ (D) $W_1 = W_3 < W_2$

    ›Reveal solutionSolution

    Work in a thermodynamic process is the area under the $P$-$V$ curve. For the same initial and final states, the path with the largest area under it does the most work. Here, path 2 encloses the largest area, path 1 the smallest, so $W_1 < W_2 < W_3$.

    The key idea is that work done by a gas during a process is not a state function — it depends on the path taken between the same two states. In a $P$-$V$ diagram, the work done by the gas equals the area under the curve (the integral $\int P \, dV$). So comparing work done along different paths is simply a matter of comparing the areas under those curves.

    Let’s examine the three paths from state A to state B.

    1. Path 1 is a straight line from A to B. The area under it is the area of a trapezoid (or a triangle plus rectangle, depending on the shape). It is the smallest area among the three.

    2. Path 2 goes first at constant pressure (isobaric expansion) from A to some intermediate point, then at constant volume (isochoric) to B. The area under this path is a rectangle — the pressure during expansion is higher than the average pressure along path 1, so the area is larger.

    3. Path 3 goes first at constant volume from A to some intermediate point, then at constant pressure to B. The area under this path is also a rectangle, but the constant pressure during expansion is the final pressure at B, which is lower than the pressure in path 2’s expansion. So the area is between that of path 1 and path 2. …

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