Thermodynamics And Kinetic Theory – NEET UG PhysicsPractice Questions & PYQs

For an adiabatic process, the relationship between temperature and volume is governed by the expression $T V^{\gamma-1} = \text{constant}$. The initial temperature must be converted from Celsius to Kelvin, resulting in 300 K. Using the adiabatic relation for two states, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, the final temperature is found by $T_2 = T_1 (\frac{V_1}{V_2})^{\gamma-1}$.

Given $\gamma = \frac{5}{3}$ and the volume ratio $\frac{V_2}{V_1} = \frac{8}{27}$, the ratio $\frac{V_1}{V_2}$ is $\frac{27}{8}$ and the exponent $\gamma - 1$ is $\frac{2}{3}$. Substituting these values into the expression provides the final temperature:
$T_2 = 300 \left( \frac{27}{8} \right)^{2/3} = 300 \left( \left( \frac{3}{2} \right)^3 \right)^{2/3} = 300 \left( \frac{3}{2} \right)^2 = 300 \times \frac{9}{4} = 675 \text{ K}$
The rise in temperature is the difference between the final temperature and the initial temperature, which results in 375 K.

The critical point for a real gas described by the van der Waals equation, $\left(P+\frac{a}{V^{2}}\right)(V-b)=RT$, represents a state where the distinctions between the liquid and gas phases vanish, occurring at an inflection point on the pressure-volume curve. Identifying this point involves expressing pressure as a function of volume, $P=\frac{RT}{V-b}-\frac{a}{V^{2}}$, and finding where both the first derivative, $\frac{\partial P}{\partial V}=-\frac{RT}{(V-b)^{2}}+\frac{2a}{V^{3}}$, and the second derivative, $\frac{\partial^{2} P}{\partial V^{2}}=\frac{2RT}{(V-b)^{3}}-\frac{6a}{V^{4}}$, vanish simultaneously. Solving these simultaneous equations provides the critical volume as $V_{c}=3b$. Substituting this value back into the first derivative condition determines the critical temperature as $T_{c}=\frac{8a}{27Rb}$.

The efficiency of a Carnot engine depends entirely on the absolute temperatures of the hot and cold reservoirs, where temperature must be expressed in Kelvin. Converting the given Celsius temperatures to the Kelvin scale requires adding 273, which sets the hot reservoir temperature at 373 K and the cold reservoir temperature at 250 K. Applying the standard thermodynamic efficiency formula $\eta = \frac{T_{1}-T_{2}}{T_{1}}$ we substitute these Kelvin values to obtain the expression $\frac{373-250}{373}$.

The First Law of Thermodynamics establishes that heat added to a system is consumed by both increasing internal energy and performing mechanical work, as expressed by the equation $\Delta Q = \Delta U + \Delta W$. To determine the external work done, we rearrange this principle to the form $\Delta W = \Delta Q - \Delta U$. With the heat added at 110 J and the internal energy increase at 40 J, the calculation becomes 110 J - 40 J. Consequently, the system performs 70 J of external work.

The internal energy of a system is a state function, meaning the total change in internal energy between two points is independent of the path taken. By applying the First Law of Thermodynamics, defined as $\Delta U = Q - W$, we can determine the change for the process from A to C by calculating the total heat added and the total work done. The process from A to B is isochoric, so the volume remains constant and the work done is zero. For the isobaric process from B to C, the work done is calculated using $W = P_B(V_C - V_A)$, which results in 240 J. The system absorbs 600 J of heat in the first process and 200 J in the second, making the total heat added 800 J. Subtracting the total work done from the total heat added gives a change in internal energy of 560 J.

The first law of thermodynamics establishes the principle of conservation of energy by stating that the heat added to a system equals the sum of the change in its internal energy and the work done by the system, expressed as $\Delta Q = \Delta U + W$. In an adiabatic process, the system is perfectly insulated so that no heat exchange occurs with the surroundings, meaning $\Delta Q = 0$. By substituting this zero value into the first law equation, the relationship simplifies to $0 = \Delta U + W$, which mathematically rearranges to $\Delta U = -W$.

Since the gas undergoes a complete cyclic process, the final state is identical to the initial state, which means the change in internal energy is $\Delta U = 0$. According to the first law of thermodynamics, this condition implies that the total heat exchanged is equal to the net work performed by the system, which corresponds to the area enclosed by the rectangle on a pressure-volume graph.

The area is determined by multiplying the pressure difference of 200 Pa by the volume difference of 3 m³. Using the formula $W = \Delta P \times \Delta V$, the total work done evaluates to 600 J. Because the net heat supplied is equal to the net work done in any cyclic process, the total heat supplied to the gas is 600 J.

Kinetic theory defines the pressure of an ideal gas as $P = \frac{1}{3} \rho v_{rms}^2$, where $\rho$ represents the gas density and $v_{rms}$ represents the root-mean-square speed of the molecules. The kinetic energy per unit volume, denoted by $E$, is defined by the expression $E = \frac{1}{2} \rho v_{rms}^2$. By rearranging the pressure equation to incorporate this energy term, we can express the pressure as
$P = \frac{2}{3} E$
This confirms that the pressure exerted by an ideal gas is equal to two-thirds of its kinetic energy per unit volume.

A thermodynamic state function depends solely on the current state of a system-characterized by properties like internal energy $U$, enthalpy $H$, and Gibbs energy $G$-rather than the history of how the system reached that point. Because these properties are path-independent, their values are defined entirely by the state of the system. In contrast, work done $w$ is a path function, meaning the amount of energy transferred depends explicitly on the specific process or route taken during the transformation.

When a metallic sphere undergoes a temperature increase $\Delta T$, its dimensions expand linearly according to the coefficient $\alpha$, leading to a change in radius given by $\Delta R = R \alpha \Delta T$. The volume of a sphere is defined as $V = \frac{4}{3} \pi R^3$, and the small change in volume $\Delta V$ relates to the change in radius through the derivative $\frac{dV}{dR} = 4 \pi R^2$. Multiplying the surface area by the radial expansion provides the total increase in volume:

$\Delta V = 4 \pi R^2 (R \alpha \Delta T) = 4 \pi R^3 \alpha \Delta T$

The kinetic theory of gases dictates that the average kinetic energy of gas molecules is determined by the equation $KE_{avg} = \frac{3}{2}k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature. At a temperature of 0 K, this relationship requires the average kinetic energy to be exactly zero because the substitution of $T=0$ into the expression results in zero. In contrast, vibrational energy persists even at this limit due to quantum zero-point effects, density remains defined by the ratio of mass to volume, and potential energy is determined solely by intermolecular separation, none of which vanish at absolute zero.

The degrees of freedom quantify the independent modes through which a gas molecule can store thermal energy, encompassing translational and rotational motion. Every gas molecule inherently possesses 3 translational degrees of freedom representing movement along the three spatial axes. A triatomic molecule, specifically a non-linear one such as water vapor, exhibits an additional 3 degrees of rotational freedom because it is capable of rotating about all three spatial axes. Since vibrational modes are typically inactive at room temperature, the total degrees of freedom for such a gas is calculated as the sum of its translational and rotational components, which is $3 + 3 = 6$.

The pressure exerted by water vapour in a closed container at equilibrium is determined strictly by the temperature of the system. This saturation vapour pressure is an intensive property and does not vary based on the volume of the container or the specific amount of liquid available, provided that some liquid remains in contact with the vapour. Because both containers share the same temperature, the system requires that $P_A = P_B$, leading to a ratio of 1:1.

At its heart, the First Law of Thermodynamics is simply the universal principle of the conservation of energy applied specifically to thermodynamic systems. It dictates that the change in the internal energy of a system, denoted as $\Delta U$, is defined by the heat supplied to the system, $Q$, minus the work done by the system, $W$, expressed by the equation $\Delta U = Q - W$. Because this relationship confirms that energy is neither created nor destroyed but merely transformed between heat, work, and internal energy, it serves as a direct result of the principle of conservation of energy.

The First Law of Thermodynamics acts as an accounting principle for energy, asserting that energy cannot be created or destroyed, only transferred. When heat is added to a system, that energy must either increase the system's internal energy or be consumed as work performed on its surroundings. Mathematically, this relationship for energy transfer rates is expressed as:

$\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$

In this scenario, the system receives energy at a rate of 100 W, while simultaneously spending energy to perform work at a rate of 75 J/s. Since 1 J/s is equivalent to 1 W, the system uses 75 W of its input to do work, leaving the remaining 25 W to be stored as an increase in the internal energy.

The behavior of an ideal gas during an adiabatic process is fundamentally described by the relationship $PV^{\gamma} = \text{constant}$. Because the ideal gas law dictates that volume is directly proportional to temperature and inversely proportional to pressure, expressed as $V \propto \frac{T}{P}$, we can replace volume in the adiabatic equation to evaluate the pressure-temperature dependence. Substituting this proportionality leads to $P \left(\frac{T}{P}\right)^{\gamma} = \text{constant}$, which simplifies to the following relationship:
$P^{1-\gamma} T^{\gamma} = \text{constant}$

The behavior of the gas upon removing the partition is determined by the principle of conservation of the product of pressure and volume, which relies on the total amount of gas remaining constant. The sum of the individual pressure-volume products before mixing must equal the product of the final equilibrium pressure and the total combined volume of the two chambers.

Applying the relationship $P_1 V_1 + P_2 V_2 = P(V_1 + V_2)$, we substitute the given parameters where the first chamber contributes 1 atm multiplied by 2 litres, and the second chamber contributes 2 atm multiplied by 3 litres. This leads to the equation:
$1(2) + 2(3) = P(2 + 3)$
Simplifying this calculation results in 8 on the left side and $5P$ on the right side. Dividing 8 by 5 yields a final equilibrium pressure of 1.6 atm.

The ideal gas law, represented as $PV = nRT$, describes the relationship between the pressure, volume, temperature, and quantity of gas in a closed system. Because gauge pressure measures the internal pressure relative to the environment, the absolute pressure is the sum of the gauge pressure and the atmospheric pressure. Given an initial gauge pressure of 11 atm and an atmospheric pressure of 1 atm, the absolute pressure of the oxygen is 12 atm.

To determine the amount of gas remaining, we rearrange the ideal gas equation as $n = \frac{PV}{RT}$. Using an absolute pressure $P$ of $12 \times 1.01 \times 10^5$ Pa, a volume $V$ of $30 \times 10^{-3}$ m$^3$, a temperature $T$ of 300 K, and the gas constant $R = \frac{100}{12}$ J mol$^{-1}$ K$^{-1}$, the final amount of oxygen is calculated as $n = 14.54$ moles. The number of moles withdrawn is the difference between the initial 18.20 moles and the final 14.54 moles, which is 3.656 moles. Multiplying this amount by the molar mass of 32 g/mol, the total mass removed is $3.656 \times 32 = 116.99$ g, which is approximately 0.116 kg.

According to the first law of thermodynamics, the heat supplied to a system equals the sum of the change in internal energy and the work done, given by the expression:
$\Delta Q = \Delta U + P \Delta V$
Since the heat supplied and the change in internal energy are the same for both gases, the work done $P \Delta V$ must be equal for both systems. Because the pressure is constant and identical in both cases, the volume change must be equal, leading to the relation $A_A d_A = A_B d_B$. By substituting the area of the circular pistons as $\pi r^2$, the equation becomes:
$\pi r_A^2 d_A = \pi r_B^2 d_B$
Solving for the ratio of the radii by substituting the given displacements of 16 cm and 9 cm results in:
$\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

In a series arrangement of thermal conductors at steady state, the rate of heat flow must be uniform across every rod because the energy passing through one section cannot accumulate at the junctions. The thermal resistance of a rod is defined as $R = \frac{L}{kA}$, where $L$ is the length, $k$ is the thermal conductivity, and $A$ is the cross-sectional area. Since all rods are identical in dimensions, we can define a standard resistance $R_0 = \frac{L}{KA}$ for the middle rod, making the resistance of the outer rods $\frac{R_0}{2}$ because their thermal conductivity is $2K$. Consequently, the total resistance of the series combination is $R_{eq} = \frac{R_0}{2} + R_0 + \frac{R_0}{2} = 2R_0$. With a total temperature difference of $3T - T = 2T$ across the system, the constant heat flow rate is $\frac{2T}{2R_0} = \frac{T}{R_0}$.

To find the temperature at the left junction $T_1$, we equate the heat flow through the first rod to this total rate, which gives $\frac{3T - T_1}{R_0 / 2} = \frac{T}{R_0}$. Simplifying this equation yields $3T - T_1 = \frac{T}{2}$, resulting in $T_1 = \frac{5}{2}T$. Similarly, determining the temperature at the right junction $T_2$ requires equating the heat flow through the third rod to the total rate, expressed as $\frac{T_2 - T}{R_0 / 2} = \frac{T}{R_0}$. This simplifies to $T_2 - T = \frac{T}{2}$, leading to $T_2 = \frac{3}{2}T$. The ratio of the two junction temperatures is therefore $\frac{T_1}{T_2} = \frac{\frac{5}{2}T}{\frac{3}{2}T} = \frac{5}{3}$.