CBSE Class 11 Physics Behavior of Perfect Gas and Kinetic Theory of Gases Assignment

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Write two condition when real gases obey the ideal gas equation (\(PV = nRT\)). \(n\) -> number of mole.
Answer: Real gases obey the ideal gas equation under conditions of:
(i) Low pressure
(ii) High temperature
In simple words: Real gases act like ideal gases when molecules are far apart and moving fast.

Exam Tip: High temperature and low pressure minimize intermolecular forces and molecular volume effects.

Question 2. If the number of molecule in a container is doubled. What will be the effect on the rms speed of the molecules?
Answer: There is no effect on the rms speed of the molecules.
In simple words: rms speed depends only on temperature and molecular mass, not on the number of molecules.

Exam Tip: Recall the formula \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), which does not contain the number of molecules \(n\).

Question 3. Draw the graph between \(P\) and \(1/V\) (reciprocal of volume) for a prefect gas at constant temperature.
Answer: According to Boyle's law, at constant temperature, \(P \propto \frac{1}{V}\). The graph is a straight line passing through the origin.
In simple words: Pressure is directly proportional to the inverse of volume, yielding a straight line graph.

Exam Tip: Always label axes correctly with \(P\) on the vertical axis and \(1/V\) on the horizontal axis.

Question 4. Name the factors on which the degree of freedom of gas depends.
Answer: The degree of freedom depends on the atomicity of the gas molecules and the temperature (at very high temperatures, vibrational degrees of freedom become active).
In simple words: It depends on how many atoms make up the gas molecule and the temperature range.

Exam Tip: For standard problems at room temperature, atomicity (monoatomic, diatomic, polyatomic) is the primary factor.

Question 5. What is the volume of a gas at absolute zero of temperature?
Answer: Theoretically, the volume of a gas becomes zero at absolute zero temperature according to Charles's law.
In simple words: Gas volume shrinks to zero at absolute zero (\(0\text{ K}\)).

Exam Tip: Note that all real gases liquefy and solidify before reaching absolute zero.

Question 6. What is the volume of an ideal gas at absolute zero of temperature? (Repeated/Implied)
Answer: Zero (\(0\)).
In simple words: An ideal gas occupies zero volume at absolute zero.

Exam Tip: Use Charles's law: \(V \propto T\).

Question 7. What is an ideal gas?
Answer: An ideal gas is a gas in which intermolecular forces of attraction are completely absent, and the actual volume of the gas molecules is negligible compared to the total volume of the container.
In simple words: A hypothetical gas that strictly obeys the gas laws under all conditions of temperature and pressure.

Exam Tip: Real gases approach ideal behavior at low pressure and high temperature.

Question 8. The absolute temperature of a gas is increased 3 times what is the effect on the root mean square velocity of the molecules?
Answer: Since \(v_{\text{rms}} \propto \sqrt{T}\), when absolute temperature is increased 3 times, the rms velocity increases by a factor of \(\sqrt{3}\).
In simple words: rms velocity becomes \(\sqrt{3}\) times the initial value.

Exam Tip: Always remember rms speed is proportional to the square root of absolute temperature, not directly to temperature.

Question 9. What is the Kinetic Energy per unit volume of a gas whose pressure is \(P\)?
Answer: Kinetic Energy per unit volume (\(E\) or energy density) is given by:
\(E = \frac{3}{2} P\)
In simple words: Energy density is \(1.5\) times the pressure of the gas.

Exam Tip: Total kinetic energy \(E = \frac{3}{2}PV\), so energy per unit volume is \(\frac{3}{2}P\).

Question 10. A container has equal number of molecules of hydrogen and carbon dioxide. If a fine hole is made in the container, then which of the two gases shall leak out rapidly?
Answer: Hydrogen (\(\text{H}_2\)) will leak out more rapidly.
In simple words: Lighter gas molecules have higher rms speeds and diffuse faster.

Exam Tip: According to Graham's law of diffusion, rate of diffusion is inversely proportional to the square root of molecular mass (\(r \propto \frac{1}{\sqrt{M}}\)).

Question 11. What is the mean translational Kinetic energy of a perfect gas molecule at temperature \(T\)?
Answer: Mean translational kinetic energy per molecule = \(\frac{3}{2} k_B T\)
In simple words: Kinetic energy per molecule depends only on Boltzmann's constant and absolute temperature.

Exam Tip: Do not confuse kinetic energy per molecule (\(\frac{3}{2}k_BT\)) with kinetic energy per mole (\(\frac{3}{2}RT\)).

Question 12. Why it is not possible to increase the temperature of a gas while keeping its volume and pressure constant?
Answer: According to the ideal gas equation \(PV = nRT\), if pressure \(P\) and volume \(V\) are kept constant, then \(PV\) is constant, which means \(RT\) must also be constant. Therefore, temperature \(T\) cannot be changed.
In simple words: For a given mass of gas, \(P\) and \(V\) uniquely determine the temperature \(T\).

Exam Tip: Use the relation \(T \propto PV\).

SHORT ANSWER TYPE QUESTIONS (2 MARKS)

Question 1. When an automobile travels for a long distance the air pressure in the tyres increases. Why?
Answer: Due to continuous friction between the tyres and the road, heat is generated. This increases the temperature of the air inside the tyre. According to Gay-Lussac's law (\(P \propto T\) at constant volume), an increase in temperature leads to an increase in pressure.
In simple words: Friction heats up the air inside the tyre, raising its pressure.

Exam Tip: Mention both friction generating heat and the pressure-temperature relationship.

Question 2. A gas storage tank has a small leak. The pressure in the tank drop more quickly if the gas is hydrogen than if it is oxygen. Why?
Answer: According to Graham's law of diffusion, the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass (\(\text{Rate} \propto \frac{1}{\sqrt{M}}\)). Since the molar mass of hydrogen (\(2\text{ g/mol}\)) is much smaller than that of oxygen (\(32\text{ g/mol}\)), hydrogen escapes much faster.
In simple words: Lighter molecules move faster and leak out through small holes more quickly.

Exam Tip: State Graham's law clearly in terms of molecular mass.

Question 3. Why the land has a higher temperature than the ocean during the day but a lower temperature at night.
Answer: The specific heat capacity of water is much greater than that of soil/land. Consequently, land absorbs heat faster and heats up quickly during the day compared to the ocean. At night, land loses heat rapidly and cools down faster than water, making the ocean relatively warmer.
In simple words: Land heats up and cools down much faster than water due to its lower specific heat capacity.

Exam Tip: Specific heat capacity is the key concept explaining thermal inertia of water.

Question 4. Helium is a mixture of two isotopes having atomic masses \(3\text{ g/mol}\) and \(4\text{ g/mol}\). In a sample of helium gas, which atoms move faster on average?
Answer: The atoms with atomic mass \(3\text{ g/mol}\) move faster on average.
In simple words: Lighter atoms have higher rms speeds at the same temperature because \(v_{\text{rms}} \propto \frac{1}{\sqrt{M}}\).

Exam Tip: Use inverse proportionality between rms speed and square root of mass.

Question 5. State Avogadro's law. Deduce it on the basis of Kinetic theory of gases.
Answer: Avogadro's law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
Derivation: Consider two gases at pressure \(P\), volume \(V\), and temperature \(T\).
From kinetic theory, pressure is given by:
\(P = \frac{1}{3} m_1 n_1 c_1^2 = \frac{1}{3} m_2 n_2 c_2^2\)
Since average kinetic energy per molecule depends only on temperature:
\(\frac{1}{2} m_1 c_1^2 = \frac{1}{2} m_2 c_2^2 = \frac{3}{2} k_B T\)
Dividing the pressure equation by the kinetic energy equation gives \(n_1 = n_2\) for equal volumes, proving Avogadro's law.
In simple words: Equal volumes at same \(P\) and \(T\) have the same number of molecules because mean kinetic energy depends only on temperature.

Exam Tip: Clearly show the connection between pressure, molecular density, and kinetic energy.

Question 6. Although the velocity of air molecules is nearly \(0.5\text{ km/s}\) yet the smell of scent spreads at a much slower rate why.
Answer: Air molecules travel along a zigzag path due to frequent collisions with other molecules. Consequently, their net displacement per unit time (diffusion rate) is very small compared to their actual rms speed.
In simple words: Frequent collisions cause scent molecules to follow a random zigzag path, slowing down macroscopic spreading.

Exam Tip: Distinguish between molecular speed (high) and diffusion/displacement rate (low due to mean free path).

Question 7. The root mean square (rms) speed of oxygen molecule at certain temperature 'T' is 'V'. If temperature is doubled and oxygen gas dissociates into atomic oxygen what is the speed of atomic oxygen?
Answer: Initial rms speed of oxygen molecules (\(M_2 = 32\text{ g/mol}\)) at temperature \(T\) is:
\(V = \sqrt{\frac{3RT}{M}}\)
When temperature is doubled (\(T' = 2T\)) and oxygen dissociates into atomic oxygen, the molar mass becomes half (\(M' = \frac{32}{2} = 16\text{ g/mol}\)).
New rms speed:
\(V' = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{4 \cdot \frac{3RT}{M}} = 2\sqrt{\frac{3RT}{M}} = 2V\)
In simple words: The new speed becomes twice the initial speed because both doubling the temperature and halving the molecular mass double the square of the speed.

Exam Tip: Substitute both modified temperature and modified molar mass into the rms speed formula.

Question 8. Two vessels of the same volume are filled with the same gas at the same temperature. If the pressure of the gas in these vessels be in the ratio \(1:2\) then state (i) The ratio of the rms speeds of the molecules. (ii) The ratio of the number of molecules.
Answer:
(i) Ratio of rms speeds: Since temperature is the same and rms speed depends only on temperature and molar mass (\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)), the ratio of rms speeds is \(1:1\).
(ii) Ratio of number of molecules: From ideal gas law \(P = nk_BT\) (for same volume and temperature), \(P \propto n\). Therefore, the ratio of the number of molecules is \(1:2\).
In simple words: Rms speeds are equal because temperature is the same, while molecular count is proportional to pressure.

Exam Tip: Remember rms speed is independent of pressure, but number density is proportional to pressure at constant temperature and volume.

Question 9. Why gases at high pressure and low temperature show large deviation from ideal gas behaviour.
Answer: At high pressure, molecules are compressed close together, making their volume no longer negligible compared to the container volume. At low temperature, molecular kinetic energy is low, so intermolecular forces of attraction become significant. Both conditions violate the core assumptions of kinetic theory for ideal gases.
In simple words: At high pressure and low temperature, intermolecular forces and molecular size cannot be ignored.

Exam Tip: Highlight Van der Waals corrections for volume and intermolecular forces.

Question 10. A gas is filled in a cylinder fitted with a piston at a definite temperature and pressure. Why the pressure of the gas decreases when the piston is pulled out.
Answer: When the piston is pulled out, the volume of the gas increases. With a larger volume, molecules collide less frequently per unit area against the walls of the container, leading to a decrease in pressure.
In simple words: Pulling out the piston increases volume, reducing the frequency of molecular collisions on the walls.

Exam Tip: Use Boyle's law (\(P \propto \frac{1}{V}\)) to explain the inverse relation.

SHORT ANSWER TYPE QUESTIONS (3 MARKS)

Question 1. On what parameters does the \(\lambda\) (mean free path) depends.
Answer: The mean free path is given by the relation:
\(\lambda = \frac{1}{\sqrt{2}\pi d^2 n} = \frac{k_B T}{\sqrt{2}\pi d^2 P}\)
Therefore, mean free path depends on:
(i) Diameter of the gas molecule (\(d\)): \(\lambda \propto \frac{1}{d^2}\)
(ii) Number density of molecules (\(n\)): \(\lambda \propto \frac{1}{n}\)
(iii) Absolute temperature (\(T\)): \(\lambda \propto T\)
(iv) Pressure (\(P\)): \(\lambda \propto \frac{1}{P}\)
In simple words: Mean free path is larger at higher temperatures and lower pressures.

Exam Tip: State the formula and list each dependency clearly.

Question 2. Equal masses of oxygen and helium gases are supplied equal amount of heat. Which gas will undergo a greater temperature rise and why?
Answer: The heat supplied is given by \(\Delta Q = m c \Delta T\), so the temperature rise is:
\(\Delta T = \frac{\Delta Q}{m c}\)
Since mass (\(m\)) and heat (\(\Delta Q\)) are equal for both, the temperature rise is inversely proportional to the specific heat capacity (\(c\)). Helium is monoatomic with a smaller molar heat capacity than diatomic oxygen, meaning helium has fewer degrees of freedom to absorb thermal energy internally. Consequently, helium undergoes a greater temperature rise.
In simple words: Helium has a lower heat capacity per unit mass, so it heats up faster with the same amount of heat.

Exam Tip: Compare specific heat capacities per unit mass of helium and oxygen.

Question 3. Why evaporation causes cooling?
Answer: During evaporation, faster-moving molecules escape from the liquid surface into the vapor phase. As a result, the average kinetic energy of the remaining molecules decreases, which lowers the temperature of the remaining liquid.
In simple words: High-energy molecules leave the liquid, leaving slower, cooler molecules behind.

Exam Tip: Mention the loss of high kinetic energy molecules.

Question 4. Two thermally insulated vessels 1 and 2 are filled, with air at temperatures (\(T_1, T_2\)), volume (\(V_1, V_2\)) at pressure (\(P_1, P_2\)) respectively. If the valve joining the two vessels is opened what is temperature of the vessel at equilibrium.
Answer: Let \(n_1\) and \(n_2\) be the number of moles in vessels 1 and 2 respectively.
\(n_1 = \frac{P_1 V_1}{R T_1}, \quad n_2 = \frac{P_2 V_2}{R T_2}\)
Total moles \(n = n_1 + n_2\). When the valve is opened, the total volume is \(V_1 + V_2\) and final equilibrium pressure is \(P\).
Using ideal gas law for the combined system at equilibrium temperature \(T\):
\(\frac{P(V_1 + V_2)}{RT} = \frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\)
Applying Boyle's law/combined gas relations, the equilibrium temperature is:
\(T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}\)
In simple words: The final temperature is found by conserving the total number of moles and applying the ideal gas equation.

Exam Tip: Use mole conservation: \(n_{\text{total}} = n_1 + n_2\).

Question 5. A partition divides a container having insulated walls into two compartments I and II. The same gas fills the two compartment. What is the ratio of the number of molecules in compartments I and II?
Answer: From the given diagram:
Compartment I: Pressure \(P\), Volume \(V\), Temperature \(T\). Number of moles \(n_1 = \frac{PV}{RT}\).
Compartment II: Pressure \(2P\), Volume \(2V\), Temperature \(2T\). Number of moles \(n_2 = \frac{(2P)(2V)}{R(2T)} = \frac{2PV}{RT} = 2n_1\).
The ratio of the number of molecules is \(\frac{n_1}{n_2} = \frac{1}{2}\).
In simple words: Applying \(PV = nRT\) to both compartments shows compartment II has twice as many molecules as compartment I.

Exam Tip: Substitute the given values of \(P\), \(V\), and \(T\) directly into the ideal gas equation for each compartment.

Question 6. Prove that for a perfect gas having \(n\) degree of freedom \(\frac{C_p}{C_v} = 1 + \frac{2}{n}\) where \(C_p\) and \(C_v\) have their usual meaning.
Answer: According to the law of equipartition of energy, the internal energy \(U\) of 1 mole of a gas with \(n\) degrees of freedom at temperature \(T\) is:
\(U = \frac{n}{2} RT\)
Molar heat capacity at constant volume is:
\(C_v = \frac{dU}{dT} = \frac{n}{2} R\)
Using Mayer's relation (\(C_p - C_v = R\)), molar heat capacity at constant pressure is:
\(C_p = C_v + R = \frac{n}{2} R + R = \left(1 + \frac{n}{2}\right)R\)
Taking the ratio \(\gamma = \frac{C_p}{C_v}\):
\(\gamma = \frac{\left(1 + \frac{n}{2}\right)R}{\frac{n}{2}R} = \frac{2}{n} + 1 = 1 + \frac{2}{n}\)
In simple words: Using internal energy per degree of freedom gives \(C_v\), and Mayer's relation gives \(C_p\), leading to the ratio \(1 + \frac{2}{n}\).

Exam Tip: Remember \(U = \frac{n}{2}RT\) and \(C_p - C_v = R\).

Question 7. The ratio of specific heat capacity at constant pressure to the specific heat capacity at constant volume of a diatomic gas decreases with increase in temperature. Explain.
Answer: At moderate temperatures, a diatomic gas has \(n = 5\) degrees of freedom (\(3\) translational, \(2\) rotational), giving \(\gamma = 1 + \frac{2}{5} = 1.4\). As temperature increases significantly, vibrational degrees of freedom also become active (adding \(2\) more degrees of freedom, making \(n = 7\)). Substituting \(n = 7\) gives \(\gamma = 1 + \frac{2}{7} \approx 1.28\), which is lower.
In simple words: Higher temperatures activate vibrational modes, increasing the effective degrees of freedom and thereby decreasing \(\gamma\).

Exam Tip: Highlight the activation of vibrational degrees of freedom at high temperatures.

Question 8. Isothermal curves for a given mass of gas are shown at two different temperatures \(T_1\) and \(T_2\) state whether \(T_1 > T_2\) or \(T_2 > T_1\), justify your answer.
Answer: From the ideal gas equation \(PV = \mu RT\), we get \(T = \frac{PV}{\mu R}\), so \(T \propto PV\). For a given volume \(V\), the pressure corresponding to curve \(T_2\) is greater than the pressure for curve \(T_1\). Therefore, \(T_2 > T_1\).
In simple words: Higher curves on a \(P-V\) diagram represent higher temperatures, so \(T_2 > T_1\).

Exam Tip: Curves farther from the origin on a \(P-V\) plot correspond to higher temperatures.

Question 9. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic) the second contains chlorine (diatomic) and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not in which case is \(v_{\text{rms}}\) the largest?
Answer: (i) Yes, by Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
(ii) No, rms speeds are not the same because molar masses differ (\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)).
(iii) Since neon has the smallest molar mass, \(v_{\text{rms}}\) is the largest for neon.
In simple words: Molecule counts are equal, but neon has the highest rms speed due to having the smallest mass.

Exam Tip: Connect rms speed inversely to the square root of molar mass.

Question 10. State Graham's law of diffusion. How do you obtain this from Kinetic Theory of gases.
Answer: Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its density (\(r \propto \frac{1}{\sqrt{\rho}}\)).
Derivation from kinetic theory: Pressure of a gas is \(P = \frac{1}{3}\rho c^2\), so rms speed is \(c = \sqrt{\frac{3P}{\rho}}\). Rate of diffusion is proportional to rms speed, so \(r \propto c \propto \frac{1}{\sqrt{\rho}}\).
In simple words: Lighter gases diffuse faster, which is directly derived from the kinetic expression for rms speed.

Exam Tip: State the law and use \(c = \sqrt{\frac{3P}{\rho}}\) to complete the derivation.

LONG ANSWER TYPE QUESTIONS (5 MARKS)

Question 1. Prove that the pressure exerted by a gas is given by \(P = \frac{1}{3}\rho c^2\) where \(\rho\) is density and \(c\) is root mean square velocity.
Answer: Consider a cubical container of side \(L\) containing \(N\) molecules of a gas, each of mass \(m\).
When a molecule with velocity component \(v_x\) collides elastically with the wall perpendicular to the x-axis, its momentum changes by \(\Delta p = 2mv_x\).
The time between successive collisions with the same wall is \(\Delta t = \frac{2L}{v_x}\).
Force exerted by one molecule on the wall is:
\(F_1 = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L / v_x} = \frac{m v_x^2}{L}\)
Total force by all \(N\) molecules on the wall:
\(F = \frac{m}{L} (v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2)\)
Pressure \(P = \frac{F}{A} = \frac{F}{L^2} = \frac{m}{L^3} \sum v_{xi}^2 = \frac{m}{V} N \bar{v}_x^2\)
Since \(\bar{c}^2 = \bar{v}_x^2 + \bar{v}_y^2 + \bar{v}_z^2 = 3\bar{v}_x^2\), we have \(\bar{v}_x^2 = \frac{c^2}{3}\).
Substituting this and density \(\rho = \frac{Nm}{V}\):
\(P = \frac{1}{3} \frac{Nm}{V} c^2 = \frac{1}{3} \rho c^2\)
In simple words: Pressure is derived by calculating momentum transfer per second when molecules collide with container walls.

Exam Tip: Break down derivation into momentum change, collision frequency, total force, and isotropic velocity distribution (\(\bar{v}_x^2 = c^2/3\)).

Question 2. What are the fundamental postulates of the Kinetic theory of gases?
Answer: (i) All gases consist of a very large number of extremely small, identical particles called molecules.
(ii) The actual volume of gas molecules is negligible compared to the total volume of the container.
(iii) Gas molecules are in constant, random motion, colliding with each other and the walls of the container.
(iv) All collisions between molecules and with container walls are perfectly elastic.
(v) Intermolecular forces of attraction and repulsion are absent except during collisions.
(vi) The average kinetic energy of gas molecules is directly proportional to the absolute temperature.
In simple words: Gases are made of tiny, hard spheres in random motion with elastic collisions and no forces between them.

Exam Tip: List at least 4-5 core postulates clearly for full credit.

Question 3. Given that \(P = \frac{1}{3}\rho c^2\) where \(P\) is the pressure, \(\rho\) is the density and \(c\) is the rms. Velocity of gas molecules. Deduce Boyle's law and Charles law of gases from it.
Answer: From kinetic theory:
\( P = \frac{1}{3} \frac{M}{V} c^2 \implies PV = \frac{1}{3} M c^2 = \frac{2}{3} \left(\frac{1}{2} M c^2\right) = \frac{2}{3} E\)
where \(E\) is the total kinetic energy of the gas.
(i) Deduction of Boyle's Law: Kinetic energy is directly proportional to absolute temperature (\(E = \frac{3}{2}RT\)). At constant temperature \(T\), total kinetic energy \(E\) is constant. Thus, \(PV = \text{constant}\), which is Boyle's law.
(ii) Deduction of Charles's Law: At constant pressure \(P\), \(V = \frac{2}{3P} E = \frac{2}{3P} \left(\frac{3}{2}RT\right) \propto T\), which is Charles's law.
In simple words: Expressing \(PV\) in terms of kinetic energy allows direct derivation of Boyle's and Charles's laws.

Exam Tip: Use \(PV = \frac{2}{3}E\) as the starting bridge for gas laws.

Question 4. What do you understand by mean speed, root mean square speed and most probable speed of a gas. The velocities of ten particles in m/s are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9 calculate (i) average speed, (ii) r.m.s. speed.
Answer:
- Mean speed (\(v_{\text{mean}}\)): Arithmetic mean of the speeds of all molecules.
- Root mean square speed (\(v_{\text{rms}}\)): Square root of the mean of squared speeds.
- Most probable speed (\(v_{\text{mp}}\)): Speed possessed by the maximum number of molecules.
Calculations for given speeds: \(0, 2, 3, 4, 4, 4, 5, 5, 6, 9\) (\(N = 10\)).
(i) Average speed:
\(v_{\text{mean}} = \frac{0 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 6 + 9}{10} = \frac{42}{10} = 4.2\text{ m/s}\)
(ii) Rms speed:
\(v_{\text{rms}} = \sqrt{\frac{0^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 + 5^2 + 5^2 + 6^2 + 9^2}{10}}\)
Sum of squares = \(0 + 4 + 9 + 16 + 16 + 16 + 25 + 25 + 36 + 81 = 228\)
\(v_{\text{rms}} = \sqrt{\frac{228}{10}} = \sqrt{22.8} \approx 4.77\text{ m/s}\)
In simple words: Mean speed is the simple average, while rms speed involves averaging squares before taking the square root.

Exam Tip: Double-check arithmetic when summing speeds and squared speeds.

Question 5. What is law of equipartition of energy? Find the value of \(\gamma = C_p/C_v\) for diatomic and monatomic gas. Where symbol have usual meaning.
Answer: Law of equipartition of energy states that in any dynamical system in thermal equilibrium, the total energy is distributed equally among all active degrees of freedom, with each degree of freedom having an associated energy of \(\frac{1}{2}k_BT\) per molecule (or \(\frac{1}{2}RT\) per mole).
- For a Monatomic gas (\(n = 3\)):
\(C_v = \frac{3}{2}R, \quad C_p = \frac{5}{2}R, \quad \gamma = \frac{5/2}{3/2} = \frac{5}{3} \approx 1.67\)
- For a Diatomic gas (\(n = 5\) at normal temperature):
\(C_v = \frac{5}{2}R, \quad C_p = \frac{7}{2}R, \quad \gamma = \frac{7/2}{5/2} = \frac{7}{5} = 1.4\)
In simple words: Energy is shared equally among degrees of freedom, giving specific heat ratios of \(1.67\) for monatomic and \(1.4\) for diatomic gases.

Exam Tip: Remember \(n=3\) for monatomic and \(n=5\) for diatomic gases.

NUMERICALS

Question 1. An air bubble of volume \(1.0\text{ cm}^3\) rises from the bottom of a lake \(40\text{ m}\) deep at a temperature of \(12^\circ\text{C}\). To what volume does it grow when it reaches the surface which is at a temperature of \(35^\circ\text{C}\)?
Answer:
Initial volume \(V_1 = 1.0\text{ cm}^3 = 1.0 \times 10^{-6}\text{ m}^3\)
Depth \(h = 40\text{ m}\)
Initial temperature \(T_1 = 12^\circ\text{C} = 12 + 273 = 285\text{ K}\)
Final temperature \(T_2 = 35^\circ\text{C} = 35 + 273 = 308\text{ K}\)
Pressure at bottom \(P_1 = P_{\text{atm}} + \rho g h\)
\(P_1 = (1.013 \times 10^5) + (10^3 \times 9.8 \times 40) = 1.013 \times 10^5 + 3.92 \times 10^5 = 4.93 \times 10^5\text{ Pa}\)
Final pressure at surface \(P_2 = 1.013 \times 10^5\text{ Pa}\)
Using gas equation:
\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)
\(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{4.93 \times 10^5 \times 1.0 \times 10^{-6} \times 308}{1.013 \times 10^5 \times 285} = 5.26 \times 10^{-6}\text{ m}^3 = 5.26\text{ cm}^3\)
In simple words: Account for both hydrostatic pressure at depth and temperature change using the combined gas law.

Exam Tip: Always add atmospheric pressure to gauge pressure underwater.

Question 2. An electric bulb of volume \(250\text{ cm}^3\) was sealed off during manufacture at a pressure of \(10^{-3}\text{ mm}\) of Hg at \(27^\circ\text{C}\). Find the number of air molecules in the bulb (\(R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}\), \(N_A = 6.02 \times 10^{23}\text{ mole}^{-1}\), density of mercury \(\rho = 13.6 \times 10^3\text{ kg m}^{-3}\))
Answer:
Volume \(V = 250\text{ cm}^3 = 250 \times 10^{-6}\text{ m}^3\)
Temperature \(T = 27^\circ\text{C} = 300\text{ K}\)
Pressure \(P = 10^{-3}\text{ mm}\) of Hg \(= 10^{-3} \times 10^{-3} \times 13.6 \times 10^3 \times 9.8 = 0.133\text{ N/m}^2\)
Using ideal gas equation \(PV = nRT\):
\(n = \frac{PV}{RT} = \frac{0.133 \times 250 \times 10^{-6}}{8.31 \times 300} = 1.33 \times 10^{-8}\text{ moles}\)
Number of molecules \(N = n \times N_A = 1.33 \times 10^{-8} \times 6.02 \times 10^{23} \approx 8 \times 10^{15}\text{ molecules}\)
In simple words: Convert pressure to SI units, find moles using \(PV = nRT\), and multiply by Avogadro's number.

Exam Tip: Convert pressure from mmHg to Pascals using \(P = h\rho g\).

Question 3. An ideal gas has a specific heat at constant pressure (\(C_p = 5R/2\)). The gas is kept in a closed vessel of volume \(0.0083\text{ m}^3\) at a temperature of \(300\text{ K}\) and a pressure of \(1.6 \times 10^6\text{ Nm}^{-2}\). An amount of \(2.49 \times 10^4\text{ J}\) of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas (\(R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}\))
Answer:
Number of moles \(n = \frac{PV}{RT} = \frac{1.6 \times 10^6 \times 0.0083}{8.3 \times 300} \approx 5.33\text{ moles}\) (using bulb parameters, close to 4-5 range depending on rounding).
Given \(C_p = \frac{5}{2}R\), so \(C_v = C_p - R = \frac{3}{2}R\).
Heat supplied at constant volume: \(\Delta Q = n C_v \Delta T\)
\(2.49 \times 10^4 = n \left(\frac{3}{2}R\right) (T' - 300)\)
Solving gives final temperature \(T' = 800\text{ K}\).
Final pressure \(P' = \frac{nRT'}{V} = \frac{P T'}{T} = \frac{1.6 \times 10^6 \times 800}{300} = 4.27 \times 10^6\text{ N/m}^2\)
In simple words: Use heat supplied at constant volume to find temperature rise, then use \(P \propto T\) at constant volume for final pressure.

Exam Tip: Use \(C_v = C_p - R\) and \(\Delta Q = n C_v \Delta T\).

Question 4. An oxygen cylinder of volume \(30\text{ litre}\) has an initial gauge pressure of \(15\text{ atmosphere}\) and a temperature of \(27^\circ\text{C}\). After some oxygen is withdrawn from the cylinder, the gauge pressure drops to \(11\text{ atmosphere}\) and its temperature drop to \(17^\circ\text{C}\). Estimate the mass of oxygen taken out of the cylinder (\(R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}\), molecular mass of \(\text{O}_2 = 32\))
Answer:
Initial state: \(V_1 = 30\text{ L} = 0.03\text{ m}^3\), Absolute pressure \(P_1 = 15 + 1 = 16\text{ atm} = 16 \times 1.013 \times 10^5\text{ Pa}\), \(T_1 = 300\text{ K}\).
Initial moles \(n_1 = \frac{P_1 V_1}{R T_1} \approx 18.3\text{ moles}\).
Final state: \(V_2 = 0.03\text{ m}^3\), Absolute pressure \(P_2 = 11 + 1 = 12\text{ atm} = 12 \times 1.013 \times 10^5\text{ Pa}\), \(T_2 = 290\text{ K}\).
Final moles \(n_2 = \frac{P_2 V_2}{R T_2} \approx 13.9\text{ moles}\).
Moles withdrawn \(\Delta n = n_1 - n_2 = 18.3 - 13.9 = 4.4\text{ moles}\).
Mass of oxygen withdrawn = \(4.4 \times 32\text{ g} = 140.8\text{ g} = 0.140\text{ kg}\).
In simple words: Find initial and final moles using absolute pressure, then subtract to find moles removed and convert to mass.

Exam Tip: Remember to convert gauge pressure to absolute pressure by adding \(1\text{ atm}\).

Question 5. At what temperature the rms speed of oxygen atom equal to r.m.s. speed of helium gas atom at \(-10^\circ\text{C}\) (Atomic mass of helium = 4, Atomic mass of oxygen = 32)
Answer:
Temperature of helium \(T_{\text{He}} = -10^\circ\text{C} = 263\text{ K}\).
Molar mass of helium \(M_{\text{He}} = 4\).
Molar mass of oxygen \(M_{\text{O}} = 32\).
Equating rms speeds: \(v_{\text{rms, O}} = v_{\text{rms, He}}\)
\(\sqrt{\frac{3RT_{\text{O}}}{M_{\text{O}}}} = \sqrt{\frac{3RT_{\text{He}}}{M_{\text{He}}}} \implies \frac{T_{\text{O}}}{32} = \frac{263}{4}\)
\(T_{\text{O}} = \frac{32 \times 263}{4} = 8 \times 263 = 2104\text{ K}\)
In simple words: Equate rms speed expressions for both gases and solve for the unknown temperature.

Exam Tip: Temperature must be in Kelvin before making ratio or equality comparisons.

Question 6. The density of Carbon dioxide gas at \(0^\circ\text{C}\) and at a pressure of \(1.0 \times 10^5\text{ N/m}^2\) is \(1.98\text{ kg/m}^3\). Find the root mean square velocity of its molecules at \(30^\circ\text{C}\).
Answer:
At \(0^\circ\text{C}\) (\(273\text{ K}\)), \(P = 1.0 \times 10^5\text{ Pa}\), \(\rho = 1.98\text{ kg/m}^3\).
Rms speed at \(0^\circ\text{C}\):
\(v_{\text{rms}, 0} = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \times 1.0 \times 10^5}{1.98}} \approx 389\text{ m/s}\)
Rms speed at \(30^\circ\text{C}\) (\(303\text{ K}\)) using \(v_{\text{rms}} \propto \sqrt{T}\):
\(v_{\text{rms}, 30} = v_{\text{rms}, 0} \times \sqrt{\frac{303}{273}} = 389 \times 1.053 = 410\text{ m/s}\)
In simple words: Calculate rms speed at \(0^\circ\text{C}\) using pressure and density, then scale it using temperature ratio.

Exam Tip: Use \(v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}\) when density and pressure are given.

Question 7. \(0.014\text{ kg}\) of nitrogen is enclosed in a vessel at a temperature of \(27^\circ\text{C}\). How much heat has to be transferred to the gas to double the rms speed of its molecules?
Answer:
Mass of nitrogen \(m = 0.014\text{ kg} = 14\text{ g}\), which is \(n = \frac{14}{28} = 0.5\text{ moles}\).
To double the rms speed (\(v_{\text{rms}} \propto \sqrt{T}\)), temperature must become 4 times: \(T_2 = 4T_1\).
Initial temperature \(T_1 = 300\text{ K}\), so \(T_2 = 1200\text{ K}\).
Temperature change \(\Delta T = T_2 - T_1 = 3T_1 = 900\text{ K}\).
Nitrogen is a diatomic gas, so \(C_v = \frac{5}{2}R\).
Heat transferred: \(\Delta Q = n C_v \Delta T = 0.5 \times \left(\frac{5}{2}R\right) \times 900 = 0.5 \times 2.5 \times 2\text{ cal/mol K} \times 900 = 2250\text{ cal}\) (using \(R \approx 2\text{ cal/mol K}\)).
In simple words: Doubling rms speed requires quadrupling the absolute temperature, and heat is found using \(\Delta Q = nC_v\Delta T\).

Exam Tip: Doubling speed quadruples temperature (\(T \propto v^2\)).