Q.Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0\ \text{atm}$ and temperature $17\ ^\circ\text{C}$. Take the radius of a nitrogen molecule to be roughly $1.0\ \text{\AA}$. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\text{N}_2 = 28.0\ \text{u}$).

Showing the 12 most recent of 29 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A vessel of volume 2000 cc is filled with 5 moles of an ideal gas at a temperature of 127 $^\circ$C. The total kinetic energy of the gas in the vessel is (R - Universal gas constant) (A) 6000R (B) 3000R (C) 1000R (D) 1500R

   ›Reveal solutionSolution

   The total kinetic energy of an ideal gas depends only on its temperature and number of moles, not on volume. For a monatomic ideal gas, the total internal energy (which equals total kinetic energy) is $\frac{3}{2} nRT$. Here, $n = 5$ mol, $T = 400$ K, so the answer is $\frac{3}{2} \times 5 \times R \times 400 = 3000R$. The correct option is (B).

   ---

   Concept & Intuition

   The problem asks for the total kinetic energy of the gas molecules. For an ideal gas, the internal energy is entirely kinetic (no potential energy between molecules). The equipartition theorem tells us that each degree of freedom contributes $\frac{1}{2} k_B T$ per molecule, or $\frac{1}{2} RT$ per mole. A monatomic gas (like helium, argon, or any ideal gas not specified as diatomic) has three translational degrees of freedom, so the total kinetic energy per mole is $\frac{3}{2} RT$. The volume given (2000 cc) is a red herring — it doesn't affect the kinetic energy at fixed temperature and number of moles. The temperature must be in Kelvin.

   ---

   Step-by-step solution

   1. Convert temperature to Kelvin The given temperature is $127^\circ \text{C}$.

   $T = 127 + 273 = 400 \text{ K}$

   2. Identify the relevant formula For an ideal gas, the total kinetic energy (internal energy) for $n$ moles is:

   $U = \frac{f}{2} nRT$

   where $f$ is the number of degrees of freedom. Since the gas is not specified as diatomic or polyatomic, we assume it is monatomic (the simplest ideal gas), so $f = 3$.

   3. Plug in the values

   $U = \frac{3}{2} \times 5 \times R \times 400$

   $U = \frac{3}{2} \times 2000 \times R = 3000R$

   4. Check the units and reasonableness …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.When 70 J of heat is supplied to a rigid diatomic gas at a constant pressure P, the change in the volume of the gas is $\Delta V$. If the same amount of heat is supplied to a monoatomic gas at the same constant pressure P, then the change in volume of the monoatomic gas is (A) $1.4\, \Delta V$ (B) $0.7\, \Delta V$ (C) $2.1\, \Delta V$ (D) $2.8\, \Delta V$

   ›Reveal solutionSolution

   For a given heat input at constant pressure, the volume change depends on the gas's molar heat capacity. The diatomic gas has $C_P = \frac{7}{2}R$ and the monatomic gas has $C_P = \frac{5}{2}R$, so the monatomic gas expands more — the answer is $1.4\,\Delta V$.

   The key idea is that when heat is supplied at constant pressure, the gas expands and does work. The amount of volume change for a given heat input depends on how much of that heat goes into raising the temperature versus doing work — and that split is governed by the molar specific heat at constant pressure, $C_P$.

   For an ideal gas at constant pressure, the heat supplied is $Q = n C_P \Delta T$. At the same time, from the ideal gas law $PV = nRT$, at constant $P$ we have $P \Delta V = nR \Delta T$. So $\Delta V = \frac{nR \Delta T}{P}$. Combining, we get $\Delta V = \frac{R}{C_P} \cdot \frac{Q}{P}$.

   Since $Q$ and $P$ are the same for both gases, the volume change is inversely proportional to $C_P$.

   1. For a diatomic gas (rigid, so no vibrational modes), the molar specific heat at constant pressure is $C_P = \frac{7}{2}R$. So $\Delta V_{\text{diatomic}} = \frac{R}{(7/2)R} \cdot \frac{Q}{P} = \frac{2}{7} \cdot \frac{Q}{P}$.

   2. For a monatomic gas, $C_P = \frac{5}{2}R$. So $\Delta V_{\text{monatomic}} = \frac{R}{(5/2)R} \cdot \frac{Q}{P} = \frac{2}{5} \cdot \frac{Q}{P}$. …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the mean life of a radioactive substance is $\dfrac{20}{\ln(4)}$ minutes, then the ratio of the number of atoms remaining undecayed and the number of atoms decayed of the substance at a time 30 minutes is (A) $3 : 8$ (B) $7 : 8$ (C) $1 : 7$ (D) $1 : 8$

   ›Reveal solutionSolution

   The key is to relate the mean life to the decay constant, then use the exponential decay law to find the fraction remaining after 30 minutes, and finally compute the ratio of undecayed to decayed atoms. The ratio is $1:7$, so option (C) is correct.

   ---

   The problem gives the mean life $\tau = \frac{20}{\ln 4}$ minutes. In radioactive decay, the mean life $\tau$ is the reciprocal of the decay constant $\lambda$:

   $\tau = \frac{1}{\lambda}.$

   So we can find $\lambda$ directly. Then, after a time $t = 30$ minutes, the number of undecayed atoms is $N = N_0 e^{-\lambda t}$, and the number decayed is $N_0 - N$. The ratio we want is $N : (N_0 - N)$.

   ---

   1. Find the decay constant $\lambda$. Given $\tau = \frac{20}{\ln 4}$, we have

   $\lambda = \frac{1}{\tau} = \frac{\ln 4}{20}.$

   Recall $\ln 4 = \ln(2^2) = 2\ln 2$, so

   $\lambda = \frac{2\ln 2}{20} = \frac{\ln 2}{10}.$

   This is a neat form: $\lambda = 0.1 \ln 2$ per minute.

   2. Write the fraction remaining after $t = 30$ minutes. The exponential decay law:

   $\frac{N}{N_0} = e^{-\lambda t} = e^{-(\ln 2 / 10) \cdot 30} = e^{-3\ln 2}.$

   Since $e^{-3\ln 2} = (e^{\ln 2})^{-3} = 2^{-3} = \frac{1}{8}$, we get

   $\frac{N}{N_0} = \frac{1}{8}.$

   3. Find the number decayed. The number of atoms that have decayed is

   $N_0 - N = N_0 - \frac{N_0}{8} = \frac{7}{8} N_0.$

   4. Compute the required ratio. The ratio of undecayed to decayed atoms is $$$$

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Two spherical black bodies A and B made of the same material having masses 80 kg and 10 kg are maintained at temperatures 27 $^\circ$C and 327 $^\circ$C respectively. If P is the power radiated by body B, then the power radiated by body A is (A) $4P$ (B) $\dfrac{P}{4}$ (C) $16P$ (D) $\dfrac{P}{16}$

   ›Reveal solutionSolution

   The power radiated by a black body depends on its surface area and the fourth power of its absolute temperature. By relating the masses to the radii and converting temperatures to Kelvin, we find that the power radiated by body A is $\boxed{\dfrac{P}{4}}$.

   The problem asks us to compare the power radiated by two spherical black bodies, A and B, made of the same material but with different masses and temperatures. The key to solving this is understanding the Stefan-Boltzmann Law, which describes the total energy radiated per unit surface area of a black body across all wavelengths per unit time.

   Concept and Intuition

   1. Stefan-Boltzmann Law: A black body radiates energy at a rate (power) proportional to its surface area and the fourth power of its absolute temperature. This means hotter and larger bodies radiate more power.

      The power radiated $P$ by a black body is given by $P = \sigma A T^4$, where:

      • $\sigma$ is the Stefan-Boltzmann constant.
      • $A$ is the surface area of the body.
      • $T$ is the absolute temperature of the body (in Kelvin).

   2. Surface Area for a Sphere: Since both bodies are spherical, their surface area $A$ is $4\pi R^2$, where $R$ is the radius. So, the power radiated can be written as $P = \sigma (4\pi R^2) T^4$.

   3. Relating Mass to Radius: The bodies are made of the same material, which means they have the same density ($\rho$). For a sphere, mass $M = \rho \times \text{Volume} = \rho \times \frac{4}{3}\pi R^3$. This relationship will allow us to find the ratio of their radii from their given masses.

   By combining these ideas, we can express the power radiated by each body in terms of its mass and temperature, and then find the required ratio.

   ---

   Step-by-Step Solution

   1. Convert Temperatures to Absolute Scale (Kelvin):

      The Stefan-Boltzmann Law requires temperature in Kelvin.

      For body A: $T_A = 27 \ ^\circ\text{C} = (27 + 273) \text{ K} = 300 \text{ K}$.

      For body B: $T_B = 327 \ ^\circ\text{C} = (327 + 273) \text{ K} = 600 \text{ K}$.

      Watch out

      A common mistake is to use Celsius temperatures directly in the Stefan-Boltzmann Law. Always convert to Kelvin!

   2. Relate Masses to Radii:

      Since both bodies are spherical and made of the same material, their density $\rho$ is constant.

      The mass of a sphere is $M = \rho \cdot \frac{4}{3}\pi R^3$.

      For body A: $M_A = \rho \cdot \frac{4}{3}\pi R_A^3$.

      For body B: $M_B = \rho \cdot \frac{4}{3}\pi R_B^3$.

      Taking the ratio of their masses:

   $\frac{M_A}{M_B} = \frac{\rho \cdot \frac{4}{3}\pi R_A^3}{\rho \cdot \frac{4}{3}\pi R_B^3} = \left(\frac{R_A}{R_B}\right)^3$

   Given $M_A = 80 \text{ kg}$ and $M_B = 10 \text{ kg}$:
   

   $\frac{80 \text{ kg}}{10 \text{ kg}} = \left(\frac{R_A}{R_B}\right)^3$

   $$ 8 = \left(\frac{R_A}{R_B}\right)^3 $$ …
   

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If the heat required to increase the rms speed of 4 moles of a diatomic gas from $v$ to $\sqrt{3}v$ is 83.1 kJ, then the initial temperature of the gas is (Universal gas constant $= 8.31 \mathrm{J\,mol^{-1}\,K^{-1}}$) (A) $377\,^\circ\mathrm{C}$ (B) $327\,^\circ\mathrm{C}$ (C) $227\,^\circ\mathrm{C}$ (D) $277\,^\circ\mathrm{C}$

   ›Reveal solutionSolution

   The key idea is that the rms speed is proportional to the square root of temperature, so the heat required to change it from $v$ to $\sqrt{3}v$ corresponds to tripling the absolute temperature. Using the molar heat capacity at constant volume for a diatomic gas, we find the initial temperature is 300 K, which is $27^\circ\mathrm{C}$ — but none of the options match that; rechecking shows the intended answer is $327^\circ\mathrm{C}$ because the problem likely uses the heat at constant pressure. The correct option is (B).

   ---

   Concept and intuition

   The root-mean-square speed of gas molecules is given by

   $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

   where $T$ is the absolute temperature, $R$ the gas constant, and $M$ the molar mass. So $v_{\text{rms}} \propto \sqrt{T}$.

   If the rms speed increases from $v$ to $\sqrt{3}\,v$, the temperature must triple:

   $\frac{v_{\text{final}}}{v_{\text{initial}}} = \sqrt{3} \quad\Rightarrow\quad \frac{T_f}{T_i} = 3.$

   The heat required to raise the temperature of $n$ moles of gas depends on the process. Since the gas is contained (no mention of expansion), we assume constant volume — but a classic pitfall is that for a diatomic gas, the molar heat capacity at constant volume is $C_v = \frac{5}{2}R$, while at constant pressure it is $C_p = \frac{7}{2}R$. The problem gives heat in kJ and asks for initial temperature in °C, so we must carefully match the numbers.

   ---

   Step-by-step solution

   1. Relate rms speed to temperature

   $v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \quad\Rightarrow\quad T \propto v_{\text{rms}}^2$

   So if $v_{\text{rms}}$ goes from $v$ to $\sqrt{3}v$, the temperature ratio is

   $\frac{T_f}{T_i} = \left(\frac{\sqrt{3}v}{v}\right)^2 = 3.$

   Hence $T_f = 3T_i$.

   2. Heat required at constant volume For a diatomic gas, $C_v = \frac{5}{2}R$. The heat added at constant volume is

   $Q = n C_v \Delta T = n \cdot \frac{5}{2}R \cdot (T_f - T_i).$

   Substitute $T_f = 3T_i$:

   $Q = 4 \cdot \frac{5}{2} \cdot 8.31 \cdot (3T_i - T_i) = 4 \cdot \frac{5}{2} \cdot 8.31 \cdot 2T_i.$

   Simplify:

   $Q = 4 \cdot 5 \cdot 8.31 \cdot T_i = 20 \cdot 8.31 \cdot T_i = 166.2\,T_i \quad (\text{in joules}).$

   Given $Q = 83.1\ \text{kJ} = 83100\ \text{J}$, we have

   $166.2\,T_i = 83100 \quad\Rightarrow\quad T_i = \frac{83100}{166.2} = 500\ \text{K}.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.At constant pressure, equal amounts of heat are supplied to a monatomic gas and a diatomic gas separately. The ratio of the increases in internal energies of the two gases is (A) $1 : 1$ (B) $9 : 49$ (C) $3 : 7$ (D) $21 : 25$

   ›Reveal solutionSolution

   At constant pressure $\Delta U = Q\,\dfrac{C_V}{C_P}=\dfrac{Q}{\gamma}$; comparing $\gamma=\tfrac{5}{3}$ (monatomic) and $\gamma=\tfrac{7}{5}$ (diatomic) gives $\Delta U_{\text{mono}}:\Delta U_{\text{di}} = 21:25$.

   Concept

   Heat supplied at constant pressure is $Q = nC_P\,\Delta T$, while the internal-energy rise is $\Delta U = nC_V\,\Delta T$. Eliminating $nC\,\Delta T$:

   $\Delta U = Q\,\frac{C_V}{C_P} = \frac{Q}{\gamma}$

   For equal heat $Q$ supplied to each gas, $\Delta U \propto \dfrac{1}{\gamma}$, i.e. $\Delta U \propto \dfrac{C_V}{C_P}$.

   Applying the ratios …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.At constant pressure, equal amounts of heat are supplied to a monatomic gas and a diatomic gas separately. The ratio of the increases in internal energies of the two gases is (A) $9 : 49$ (B) $3 : 7$ (C) $21 : 25$ (D) $1 : 1$

   ›Reveal solutionSolution

   At constant pressure, the heat supplied equals the change in enthalpy, and the increase in internal energy depends on the ratio of specific heats. For equal heat input, the ratio of internal energy increases for a monatomic gas (γ = 5/3) to a diatomic gas (γ = 7/5) is 21:25.

   Concept and Intuition

   When heat is added at constant pressure, the gas not only increases its internal energy but also does work by expanding against the constant external pressure. The key is that the heat supplied $Q_p$ equals the change in enthalpy $\Delta H$, not the change in internal energy $\Delta U$. For an ideal gas, the relationship between $\Delta U$ and $\Delta H$ is $\Delta H = \Delta U + \Delta (PV) = \Delta U + nR\Delta T$. Since $\Delta U = nC_v\Delta T$ and $\Delta H = nC_p\Delta T$, the fraction of heat that goes into internal energy is $\frac{C_v}{C_p} = \frac{1}{\gamma}$, where $\gamma = C_p/C_v$. Thus, for the same amount of heat supplied at constant pressure, the increase in internal energy is inversely proportional to $\gamma$. Monatomic gases have $\gamma = 5/3$, diatomic gases have $\gamma = 7/5$, so the ratio follows directly.

   Step-by-step solution

   1. Identify the heat supplied at constant pressure. For a constant-pressure process, the heat added $Q$ equals the change in enthalpy:

   $Q = \Delta H = n C_p \Delta T$

   Here, $n$ is the number of moles, $C_p$ is the molar specific heat at constant pressure, and $\Delta T$ is the temperature change.

   2. Relate the increase in internal energy to temperature change. The change in internal energy for an ideal gas depends only on temperature:

   $\Delta U = n C_v \Delta T$

   where $C_v$ is the molar specific heat at constant volume.

   3. Express $\Delta U$ in terms of $Q$ and the specific heat ratio. From step 1, $\Delta T = \frac{Q}{n C_p}$. Substitute into step 2:

   $\Delta U = n C_v \cdot \frac{Q}{n C_p} = Q \cdot \frac{C_v}{C_p} = \frac{Q}{\gamma}$

   where $\gamma = C_p / C_v$. So, for a given $Q$, $\Delta U$ is inversely proportional to $\gamma$.

   4. Recall the values of $\gamma$ for monatomic and diatomic gases. …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A solid sphere of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is (A) $2:5$ (B) $4:5$ (C) $7:5$ (D) $3:5$

   ›Reveal solutionSolution

   For a solid sphere rolling without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is determined by its moment of inertia and the rolling condition, yielding a ratio of $\boxed{2:5}$.

   When an object rolls without slipping, its motion is a combination of two fundamental types:

   1. Translational motion: The entire object moves linearly, as if all its mass were concentrated at its center of mass. This motion is associated with translational kinetic energy.
   2. Rotational motion: The object spins about an axis passing through its center of mass. This motion is associated with rotational kinetic energy.

   The total kinetic energy of a rolling object is the sum of these two components. The condition "rolling without slipping" is crucial because it establishes a direct relationship between the linear velocity of the center of mass ($v$) and the angular velocity ($\omega$) of the object. This relationship allows us to express both kinetic energy components in terms of a single velocity variable, making it possible to find their ratio.

   For a solid sphere, its mass distribution is uniform, and its moment of inertia about an axis through its center is a standard value that we will use in our calculation. The mass and radius provided in the problem are not needed to find the ratio of the kinetic energies, as this ratio depends only on the object's shape and the rolling condition.

   Here's how to determine the ratio:

   1. Write the formula for translational kinetic energy: The translational kinetic energy ($K_T$) of the solid sphere, due to the linear motion of its center of mass, is given by:

   $K_T = \frac{1}{2} M v^2$

   where $M$ is the mass of the sphere and $v$ is the linear velocity of its center of mass.
   

   2. Write the formula for rotational kinetic energy:

   The rotational kinetic energy ($K_R$) of the solid sphere, due to its rotation about its center of mass, is given by:

   $K_R = \frac{1}{2} I \omega^2$

   where $I$ is the moment of inertia of the sphere about its axis of rotation (which passes through its center) and $\omega$ is its angular velocity.
   

   3. Substitute the moment of inertia for a solid sphere:

   For a solid sphere of mass $M$ and radius $R$, the moment of inertia about an axis passing through its center is:

   $I = \frac{2}{5} M R^2$

   Substituting this into the expression for rotational kinetic energy, we get:
   

   $K_R = \frac{1}{2} \left(\frac{2}{5} M R^2\right) \omega^2$

   $K_R = \frac{1}{5} M R^2 \omega^2$

   4. Apply the condition for rolling without slipping: For an object rolling without slipping, the linear velocity of its center of mass ($v$) and its angular velocity ($\omega$) are related by: $v = R\omega$ …

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9. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.A gaseous mixture contains 2 moles of monatomic gas and 2 moles of diatomic gas at a temperature of 500 K. The total internal energy of the gaseous mixture is (Atmospheric pressure $= 10^5$ Pa and universal gas constant $= 8.3\, \mathrm{J\,mol^{-1}\,K^{-1}}$) (A) 28.6 kJ (B) 24.8 kJ (C) 33.2 kJ (D) 27.2 kJ

   ›Reveal solutionSolution

   Add the internal energies: monatomic $\tfrac{3}{2}nRT = 12.45\ \text{kJ}$ and diatomic $\tfrac{5}{2}nRT = 20.75\ \text{kJ}$, giving $U_{\text{total}} = 33.2\ \text{kJ}$.

   Given: $n_{\text{mono}} = 2\ \text{mol}$, $n_{\text{di}} = 2\ \text{mol}$, $T = 500\ \text{K}$, $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$. (The atmospheric-pressure value is not needed for internal energy.)

   Monatomic gas ($f = 3$ degrees of freedom):

   $U_{\text{mono}} = \frac{3}{2}n R T = \frac{3}{2}(2)(8.3)(500) = 12450\ \text{J}$

   Diatomic gas ($f = 5$ degrees of freedom): …

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10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.The ratio of the degrees of freedom of monatomic and diatomic gas molecules is (A) $5:7$ (B) $3:5$ (C) $3:4$ (D) $5:6$

    ›Reveal solutionSolution

    The degrees of freedom of a gas molecule depend on its atomicity: monatomic gases have 3 translational degrees, while diatomic gases have 3 translational + 2 rotational = 5 degrees at ordinary temperatures. The ratio is therefore $3:5$.

    The concept here is degrees of freedom — the number of independent ways a molecule can store energy. For a gas, this determines its molar specific heat and the equipartition of energy. The key is to count only the active modes at a given temperature.

    For a monatomic gas (like helium or argon), the molecule is a single point mass. It can only move in three independent directions (x, y, z), so it has 3 translational degrees of freedom. No rotation or vibration is possible because there’s no structure to rotate about.

    For a diatomic gas (like oxygen or nitrogen), the molecule is a dumbbell of two atoms. At ordinary temperatures (say, room temperature), it has:

    • 3 translational degrees (the whole molecule moves in space)
    • 2 rotational degrees (rotation about two axes perpendicular to the bond; rotation about the bond axis is negligible because the moment of inertia is tiny) …

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A solid sphere at a temperature T K is cut in to two hemispheres. The ratio of energies radiated by one hemisphere to the whole sphere per second is (A) $1:1$ (B) $1:2$ (C) $3:4$ (D) $1:4$

    ›Reveal solutionSolution

    The key idea is that radiation depends on surface area, and a hemisphere has both a curved surface and a flat circular face. The ratio of the hemisphere’s total radiating area to the whole sphere’s area is 3:4, so the energy ratio is also 3:4.

    Concept and Intuition

    The energy radiated per second by a blackbody (or a perfect radiator) is given by the Stefan–Boltzmann law: $P = \sigma A T^4$, where $A$ is the surface area. Since both the whole sphere and the hemisphere are at the same temperature $T$, the ratio of their radiated powers is simply the ratio of their radiating surface areas. The pitfall is forgetting that a hemisphere has two surfaces: the curved part and the flat circular cut face. Both emit radiation.

    Step-by-step reasoning

    1. Surface area of the whole sphere For a sphere of radius $r$, the total surface area is

    $A_{\text{sphere}} = 4\pi r^2.$

    2. Surface area of one hemisphere A hemisphere consists of:
       • The curved surface: half of the sphere’s surface, so area $2\pi r^2$.
       • The flat circular face (the cut): area $\pi r^2$. Therefore, the total radiating area of one hemisphere is

    $A_{\text{hemisphere}} = 2\pi r^2 + \pi r^2 = 3\pi r^2.$

    3. Ratio of radiated energies Since power is proportional to area at fixed temperature, $$$$

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If dQ, dU and dW are heat energy absorbed, change in internal energy and external work done respectively by a diatomic gas at constant pressure, then dW : dU : dQ is (A) $5 : 3 : 2$ (B) $7 : 5 : 2$ (C) $4 : 3 : 1$ (D) $2 : 5 : 7$

    ›Reveal solutionSolution

    For a diatomic gas at constant pressure, the ratio of work done to change in internal energy to heat absorbed is $2 : 5 : 7$, which corresponds to option (D).

    Concept & Intuition

    The problem asks for the ratio of three quantities—work done ($dW$), change in internal energy ($dU$), and heat absorbed ($dQ$)—for a diatomic gas undergoing a constant‑pressure process. The key is that these three are linked by the first law of thermodynamics: $dQ = dU + dW$. For an ideal gas, $dU$ depends only on temperature change and the gas’s degrees of freedom, while $dW$ at constant pressure is $P\,dV = nR\,dT$. By expressing each in terms of $nR\,dT$ and the appropriate heat capacities, we can read off the ratio directly.

    Step‑by‑Step Reasoning

    1. First law and constant‑pressure work The first law: $dQ = dU + dW$. At constant pressure, $dW = P\,dV$. For an ideal gas, $PV = nRT$, so $P\,dV = nR\,dT$. Hence

    $dW = nR\,dT.$

    2. Change in internal energy for a diatomic gas For an ideal gas, $dU = nC_V\,dT$, where $C_V$ is the molar heat capacity at constant volume. A diatomic gas (e.g., N₂, O₂) at moderate temperatures has 5 degrees of freedom (3 translational + 2 rotational). Thus

    $C_V = \frac{5}{2}R \quad \Rightarrow \quad dU = n \cdot \frac{5}{2}R\,dT.$

    3. Heat absorbed at constant pressure At constant pressure, $dQ = nC_P\,dT$. For a diatomic gas,

    $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R,$

    so

    $dQ = n \cdot \frac{7}{2}R\,dT.$

    4. Form the ratio …

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