Thermodynamics And Kinetic Theory – NEET UG PhysicsPractice Questions & PYQs

A thermodynamic system's equilibrium state is uniquely defined by its state variables, specifically pressure ($P$), volume ($V$), and temperature ($T$), which depend solely on the system's instantaneous macroscopic condition. In contrast, total heat and work done are classified as path variables, as their values rely on the specific process used to transition between states rather than defining the equilibrium configuration itself.

Based on the ideal gas law equation $PV = nRT$, for a fixed amount of gas at a constant temperature, the volume occupied by the gas is inversely proportional to its pressure. If we observe a vertical line representing a constant temperature cutting through the three curves, the volume $V$ increases as we move from the curve for $P_1$ toward $P_3$. Consequently, because the smallest volume corresponds to the greatest pressure, the pressure values follow the descending order $P_1 > P_2 > P_3$.

A polyatomic gas molecule possesses three translational and three rotational degrees of freedom, alongside $f$ vibrational modes. Summing the energy contributions where each translational and rotational mode holds $\frac{1}{2}RT$ and each vibrational mode accounts for $RT$ due to both kinetic and potential components, the total internal energy is $U = (3+f)RT$. Differentiating this with respect to temperature yields the molar heat capacity at constant volume $C_v = (3+f)R$, and applying Mayer's relation provides the molar heat capacity at constant pressure $C_p = C_v + R = (4+f)R$. Using these expressions to define the ratio $\gamma = \frac{C_p}{C_v} = \frac{4+f}{3+f}$, we can solve for $f$ through algebraic manipulation. Rearranging the equation gives $3\gamma + f\gamma = 4+f$, which simplifies to $f(\gamma - 1) = 4-3\gamma$, resulting in $f = \frac{4-3\gamma}{\gamma-1}$.

Thermal insulation relies on materials that significantly restrict the conduction of heat, a property exemplified by foam due to the presence of trapped air pockets within its structure. By effectively blocking the transfer of thermal energy from the outside, this insulation ensures that the interior of a house remains cooler during the hot summer months. Consequently, the assertion that such a roof causes a room to become hotter is factually incorrect, while the explanation regarding the physical mechanism of the foam insulation is entirely accurate.

Thermodynamic work is determined by the integral $W = \int P \, dV$, which signifies that work depends entirely on changes in volume. For an isochoric process, the volume remains fixed, which mathematically implies $dV = 0$. Since path bc represents an isochoric transition, there is no displacement of volume, and consequently, no work is performed by the gas.

The total pressure of a gaseous mixture is determined by the ideal gas law $PV = nRT$, where $n$ represents the total number of moles of all gas components combined. First, sum the moles of hydrogen and argon to find the total quantity of gas, resulting in $n = 0.2 + 0.3 = 0.5 \text{ mol}$. Before performing the calculation, convert the container volume into SI units so that the pressure is obtained in Pascals; thus, $V = 200 \text{ cm}^3 = 200 \times 10^{-6} \text{ m}^3$.

With all values prepared in standard units, substitute the total moles, the gas constant $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, the temperature $T = 200 \text{ K}$, and the converted volume into the rearranged equation $P = \frac{nRT}{V}$:
$P = \frac{0.5 \times 8.3 \times 200}{200 \times 10^{-6}}$
Simplifying the expression by canceling the $200$ terms in the numerator and denominator leaves $P = 0.5 \times 8.3 \times 10^6 \text{ Pa}$. Executing this final multiplication results in a system pressure of $4.15 \times 10^6 \text{ Pa}$.

The root-mean-square speed of gas molecules is directly proportional to the square root of the absolute temperature, which is expressed by the relationship $v_{\text{rms}} \propto \sqrt{T}$. To begin, we convert the initial temperature of -50 °C into Kelvin by adding 273, giving us 223 K. Since the problem states that the speed increases by 3 times, the new speed becomes 4 times the initial speed, which can be written as $v_2 = 4v_1$.

Applying the ratio $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$, we substitute the known values to obtain the equation $4 = \sqrt{\frac{T_2}{223}}$. Squaring both sides results in $16 = \frac{T_2}{223}$, which leads to a final absolute temperature of 3568 K. Converting this value back into Celsius by subtracting 273 yields a final temperature of 3295 °C.

The efficiency of a Carnot engine depends entirely on the absolute temperatures of the heat source and the heat sink, expressed by the formula $\eta = 1 - \frac{T_2}{T_1}$. It is vital to remember that all temperature inputs for this equation must be in Kelvin, so we first convert the source temperature of $327^{\circ} \text{C}$ to $600 \text{ K}$ by adding 273.

With the efficiency given as $50\%$ or $0.5$, we substitute the known values into the equation $0.5 = 1 - \frac{T_2}{600}$. Solving for the sink temperature gives $\frac{T_2}{600} = 0.5$, which simplifies to $T_2 = 300 \text{ K}$. Converting this absolute temperature back to Celsius by subtracting 273 results in a final sink temperature of $27^{\circ} \text{C}$.

An isobaric process is characterized by a constant pressure, and the work done during such a transformation is determined by the relationship $W = P(V_2 - V_1)$. In this cycle, the process AB takes place at a constant pressure of $P = 3 \times 10^2$ while the volume increases from an initial state of 1 to a final state of 3. Substituting these values into the work formula results in $W = 3 \times 10^2 \times (3 - 1)$, which confirms that the total work done is 600 J.

Oscillating systems are governed by energy principles and the specific forces acting upon them, which dictate the characteristic shapes of their representative graphs. The restoring force of a spring follows Hooke's Law, defined by $F = -kx$, which corresponds to the linear relationship shown in graph (b), while the bob of a pendulum oscillating with air friction exhibits damped motion as seen in graph (a). Conversely, a pendulum swinging without air friction undergoes undamped oscillations represented in graph (c), and the scenario where total mechanical energy remains invariant throughout the motion matches the characteristics of graph (d).

The assumption that intermolecular forces vanish implies that water vapor acts as an ideal gas, making it possible to determine its volume via the ideal gas law $PV = nRT$. To find the number of moles in 4.5 kg of water, divide the total mass of 4500 g by the molar mass of 18 g/mol, resulting in $n = 4500/18 = 250$ mol. Given the standard temperature and pressure conditions where $T = 273$ K and $P = 10^5$ N/m^2, along with the ideal gas constant $R = 8.3$ J/mol K, we rearrange the ideal gas law to solve for volume:

$V = \frac{nRT}{P}$

Substituting these values into the equation gives $V = \frac{250 \times 8.3 \times 273}{10^5}$, which yields a volume of 5.6 m^3.

The behavior of an ideal gas is governed by the relationship $PV = nRT$, which, when substituted into the process equation $PV^2 = C$, reveals that the product $TV$ must remain constant. Since temperature and volume are inversely proportional in this specific process, any expansion where the volume increases necessitates a corresponding drop in temperature to keep their product unchanged. Consequently, whenever the final volume $V_2$ is greater than the initial volume $V_1$, the final temperature $T_2$ will always be lower than the initial temperature $T_1$.

The root mean square speed of gas molecules is defined by the expression $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $R$ represents the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas. Since the temperature is held constant across all vessels, the speed of the molecules depends solely on their molar mass, being inversely proportional to the square root of $M$, and remains entirely independent of the degrees of freedom or the atomicity of the gas. Because Helium possesses the smallest molar mass among the three gases provided, it will necessarily exhibit the highest root mean square speed compared to the heavier Fluorine and Sulfur Hexafluoride molecules.

The steepness of a curve on a pressure-volume diagram reflects how rapidly pressure changes as the volume fluctuates, a property governed by the adiabatic index $\gamma$. By examining the derivatives of these processes, the slope of the adiabatic curve is greater in magnitude than that of the isothermal curve, specifically identified by the relationship $\left| \frac{dP}{dV} \right| = \gamma \frac{P}{V}$ compared to $\left| \frac{dP}{dV} \right| = \frac{P}{V}$. Since the adiabatic index is greater than 1 for an ideal gas, the adiabatic process must result in a steeper descent than the isothermal process, identifying curve 2 as the adiabatic representation.

Root mean square speed of a gas molecule is defined by the expression $\sqrt{\frac{3RT}{M}}$, while the pressure exerted by an ideal gas is established from kinetic theory as $\frac{1}{3}nm\bar{v}^2$. The average kinetic energy of a single molecule corresponds to $\frac{3}{2}k_BT$, and for one mole of a diatomic gas possessing five degrees of freedom, the total internal energy calculates to $\frac{5}{2}RT$. These fundamental physical relationships identify that (A) corresponds to (Q), (B) corresponds to (P), (C) corresponds to (S), and (D) corresponds to (R).

Newton's Law of Cooling establishes that the cooling rate is proportional to the difference between the average temperature of an object during an interval and the surrounding temperature. This principle is expressed as $\frac{T_{\text{initial}} - T_{\text{final}}}{\Delta t} = K \left( \frac{T_{\text{initial}} + T_{\text{final}}}{2} - T_{\text{surroundings}} \right)$, where $K$ is the proportionality constant. During the first cooling interval, the coffee cools from 90°C to 80°C in $t$ minutes with a room temperature of 20°C, which leads to the relation $\frac{90 - 80}{t} = K \left( \frac{90 + 80}{2} - 20 \right)$. This simplifies to $\frac{10}{t} = 65K$, identifying the constant as $K = \frac{2}{13t}$. For the second cooling phase where the coffee cools from 80°C to 60°C, the average temperature is 70°C, leading to the equation $\frac{80 - 60}{t_1} = K \left( \frac{80 + 60}{2} - 20 \right)$, which simplifies to $\frac{20}{t_1} = 50K$. Substituting the value of $K$ found from the first case results in $\frac{20}{t_1} = 50 \left( \frac{2}{13t} \right)$, which further reduces to $\frac{20}{t_1} = \frac{100}{13t}$. Solving for the new time duration $t_1$ gives the final result of $t_1 = \frac{13}{5} t$.

To determine the density of a gas, we rely on the ideal gas equation $PV = nRT$. Since the number of moles is mass divided by molar mass and density is defined as mass per unit volume, we can rearrange the expression to find density using $\rho = \frac{PM}{RT}$.

Before substituting, convert all values to SI units to ensure consistency. The pressure is 249 kPa, which is $249 \times 10^3$ Pa. The molar mass of hydrogen gas is $2 \times 10^{-3}$ kg/mol, and the temperature of 27°C must be converted to 300 K.

Substituting these values into the formula gives $\rho = \frac{(249 \times 10^3)(2 \times 10^{-3})}{8.3 \times 300}$. The numerator simplifies to 498 and the denominator to 2490, which results in a density of 0.2 kg/m^3.

The mean free path represents the average distance a gas molecule travels between successive collisions with other particles, serving as a key indicator of molecular interaction frequency. This distance is inversely proportional to both the number density of the gas and the effective collision cross-section, which is directly related to the square of the molecular diameter $d$. By accounting for the relative velocities of moving gas molecules, the standard derivation introduces a factor of $\sqrt{2}$ in the denominator, leading to the expression $\frac{1}{\sqrt{2} n \pi d^2}$.

An adiabatic process is defined by the complete absence of heat transfer between a thermodynamic system and its external environment, which is symbolically represented as $Q = 0$. In this scenario, the entire apparatus is thermally insulated, which prevents any thermal energy from entering or leaving the system as the gas expands into the evacuated cylinder. This absolute lack of heat exchange during the sudden opening of the stop cock identifies the process as strictly adiabatic.

The behavior of an ideal gas is governed by the direct relationship between thermal energy and the motion of its constituent particles. Because the internal energy of the system is tied to temperature, it is expressed by the following equation:

$U = \frac{F}{2} n R T$

Since temperature is fundamentally a measure of the average kinetic energy of these particles, increasing it forces the molecules to move with greater velocity, which directly causes an increase in the kinetic energy of the gas within the container.