Thermodynamics And Kinetic Theory – NEET UG PhysicsPractice Questions & PYQs

The mean free path of a gas molecule represents the average distance it travels between successive collisions, which is fundamentally limited by the effective collision cross-section of the molecules. This cross-section depends on the square of the molecular diameter, and since the diameter is proportional to the radius, the area scales as $r^2$. Expressing this relationship as $\lambda = \frac{1}{4 \pi r^2 n \sqrt{2}}$, where $n$ is the number density, demonstrates that the mean free path varies inversely with the square of the molecular radius, or $\lambda \propto \frac{1}{r^2}$.

In any thermodynamic cycle plotted on a pressure-volume diagram, the net work done is the algebraic sum of the work performed during each portion of the cycle, where clockwise loops indicate positive work due to expansion and counter-clockwise loops indicate negative work due to compression. For the section represented by triangle AEDA, the system performs positive work calculated as $W_{\text{AEDA}} = +\frac{1}{2} P_{0} V_{0}$. Conversely, the segment represented by triangle BCEB corresponds to a compression process, where the work done is $W_{\text{BCEB}} = -\frac{1}{2} P_{0} V_{0}$. Summing these two contributions yields a net work of $W_{\text{net}} = \frac{1}{2} P_{0} V_{0} - \frac{1}{2} P_{0} V_{0}$, which results in zero.

The behavior of a gas during expansion is dictated by the specific path it follows, with isothermal and adiabatic processes defining the relationship between pressure and volume. During the initial isothermal phase, where volume increases from $V$ to $2V$, the relationship $PV = \text{constant}$ allows us to determine the intermediate pressure using $PV = P_2(2V)$, which simplifies to $P_2 = P/2$.

With the intermediate pressure established, the gas undergoes an adiabatic expansion from $2V$ to $16V$. This process follows the law $PV^\gamma = \text{constant}$, which requires that $P_2V_2^\gamma = P_3V_3^\gamma$ with $\gamma = 5/3$. Substituting the intermediate pressure and volumes into the equation yields $(P/2)(2V)^{5/3} = P_3(16V)^{5/3}$. Isolating $P_3$ produces $P_3 = (P/2) \left(\frac{2V}{16V}\right)^{5/3}$. Simplifying the ratio $2V/16V$ to $1/8$ provides $P_3 = (P/2)(1/8)^{5/3}$. Because $1/8$ is $(1/2)^3$, the exponentiation $(1/2^3)^{5/3}$ resolves to $(1/2)^5$, which equals $1/32$. Multiplying $P/2$ by $1/32$ results in a final pressure of $P/64$.

The fundamental equation governing an adiabatic process is $PV^{\gamma} = \text{constant}$, and we can derive variations relating pressure, volume, and temperature by substituting variables using the ideal gas law, $PV = nRT$. By replacing $P$ with $\frac{nRT}{V}$, we obtain $TV^{\gamma-1} = \text{constant}$, while replacing $V$ with $\frac{nRT}{P}$ yields $P^{1-\gamma} T^{\gamma} = \text{constant}$. Because the term $P^{\gamma} T^{1-\gamma} = \text{constant}$ does not arise from these standard algebraic substitutions, it represents an incorrect relation for an adiabatic process.

Adiabatic processes for ideal gases are governed by specific relationships between pressure and temperature, defined by the adiabatic index $\gamma$, which is the ratio of molar heat capacities $\frac{C_P}{C_V}$. The standard adiabatic law relating these variables is given by the expression $P T^{\frac{\gamma}{1-\gamma}} = \text{constant}$.

Because the problem states that the pressure is proportional to the cube of the temperature, we can express this condition as $P \propto T^3$, or equivalently $P T^{-3} = \text{constant}$. Comparing the exponent of temperature in this specific case to the general adiabatic expression, we set up the equality:

$\frac{\gamma}{1-\gamma} = -3$

Solving for $\gamma$ yields $\gamma = -3 + 3\gamma$, which simplifies to $2\gamma = 3$. This indicates that the ratio $\frac{C_P}{C_V}$ is $\frac{3}{2}$.

The ideal gas equation is given by $PV = nRT$, which can be rearranged to express volume as a function of temperature: $V = (\frac{nR}{P})T$. In a $V-T$ diagram, this indicates that the slope of the line is equal to $\frac{nR}{P}$, demonstrating that the slope is inversely proportional to the pressure. Since the line corresponding to $P_2$ is steeper than the line for $P_1$, the higher slope implies a smaller pressure value, which confirms that $P_2 < P_1$.

The pressure exerted by an ideal gas in a container is determined by the kinetic theory relationship
$P = \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2$
where $P$ is directly proportional to both the mass of the molecules $m$ and the square of their speed $v_{\text{rms}}^2$. When the mass of each molecule is halved to $\frac{m}{2}$ and the speed is doubled to $2 v_{\text{rms}}$, the new pressure $P'$ depends on the modified values as follows:
$P' = \frac{1}{3} \frac{N}{V} \left( \frac{m}{2} \right) (2 v_{\text{rms}})^2 = \frac{1}{3} \frac{N}{V} \left( \frac{m}{2} \right) (4 v_{\text{rms}}^2) = 2 \left( \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2 \right)$
By substituting the original pressure expression into this result, it becomes clear that the resultant pressure is $2 P$.

The First Law of Thermodynamics states that the heat absorbed by a system is equal to the sum of the change in internal energy and the work done, represented by the equation $Q = \Delta U + W$. A fundamental property of internal energy is that it is a state function, meaning the change in internal energy between two specific states depends only on those endpoints, regardless of the path taken. Therefore, the change in internal energy for the path $adc$ is identical to the change in internal energy for the path $abc$.

For the first path, the change in internal energy is determined by subtracting the work of 20 J from the heat of 50 J:
$\Delta U = 50 - 20 = 30$
This value of 30 J represents the change in internal energy for the transition between states $a$ and $c$. Because this change is constant, the First Law applied to the path $abc$, where 36 J of heat is absorbed, allows us to calculate the unknown work:
$36 = 30 + W$
$W = 6$
The work obtained along the path $abc$ is 6 J.

Heating a monatomic gas such as Helium at a constant volume involves energy transfer dependent on the number of moles and the molar heat capacity at constant volume. Helium has a molar heat capacity of $C_V = \frac{3}{2} R$ due to its monatomic nature. With a molar mass of 4 g/mol, a 1 g sample of Helium consists of $n = \frac{1}{4}$ moles. Applying the heat energy formula $\Delta Q = n C_V \Delta T$, we substitute the known values to obtain $\Delta Q = (\frac{1}{4}) (\frac{3}{2} R) (T_2 - T_1)$, which simplifies to $\frac{3}{8} R (T_2 - T_1)$. Using the relation $R = N_a k_B$, the energy required to raise the temperature is expressed as $\frac{3}{8} N_a k_B (T_2 - T_1)$.

The efficiency of a Carnot engine depends exclusively on the absolute temperatures of the heat source and the heat sink, defined by the formula $\eta = 1 - \frac{T_{sink}}{T_{source}}$. For the first engine in the series, the efficiency is given by $\eta_A = 1 - \frac{T}{T_1}$, while the second engine operates between $T$ and $T_2$, resulting in an efficiency of $\eta_B = 1 - \frac{T_2}{T}$.

To ensure both engines have equal efficiency, we set these two expressions equal to each other, which gives $1 - \frac{T}{T_1} = 1 - \frac{T_2}{T}$. Subtracting 1 from both sides and simplifying the signs leads to the relation $\frac{T}{T_1} = \frac{T_2}{T}$. Cross-multiplying the terms yields $T^2 = T_1 T_2$, confirming that the intermediate temperature must be $\sqrt{T_1 T_2}$.

In any cyclic thermodynamic process, the system returns to its initial state, meaning the total change in internal energy is zero. According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$, the net heat exchange must equal the total work done during the cycle. On a pressure-volume diagram, the magnitude of this work corresponds to the area enclosed by the cycle. With dimensions of $P$ and $2V$, the area enclosed is $2PV$. Consequently, the heat rejected by the gas is $2PV$.

Internal energy is a state function, meaning it depends exclusively on the beginning and end points of a process rather than the path taken; since all three processes start at point A and end at point B, the change in internal energy is identical for every path, giving $\Delta U_1 = \Delta U_2 = \Delta U_3$. According to the first law of thermodynamics, heat absorbed is defined as $Q = \Delta U + W$, where the work done $W$ is equivalent to the area under the P-V curve. Observing that the area under path 1 is greater than path 2, which is greater than path 3, we establish that $W_1 > W_2 > W_3$. Consequently, since the internal energy change is constant while the work done is greatest for the first process and least for the third, the heat absorbed follows the relationship $Q_1 > Q_2 > Q_3$.

For an ideal gas undergoing an isothermal process, the relationship $PV = \text{constant}$ dictates a rectangular hyperbola on the P-V diagram where pressure decreases as volume increases from $V$ to $3V$. The subsequent isobaric process, occurring at the constant final pressure reached during the expansion, is represented by a horizontal line segment returning from volume $3V$ to $V$ at that reduced pressure level. This specific sequence of a curved downward path followed by a horizontal return path characterizes the process depicted in option D.

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to it minus the work done by the system, which is expressed as $\Delta U = Q - W$. Because the process is isothermal for an ideal gas, the temperature remains constant, meaning the internal energy does not change, or $\Delta U = 0$, which simplifies the relationship to show that the heat added equals the work done, or $Q = W$. Given the specific condition that the gas performs -150 J of work against its surroundings, the work done by the system is effectively +150 J. Since the heat supplied is equivalent to the work done, 150 J of heat must have been added to the gas to facilitate this expansion.

Entropy change during a phase transition at a fixed temperature is determined by the ratio of the absorbed heat to the absolute temperature in Kelvin. First, determine the total heat energy by multiplying the mass of the ice by the latent heat of fusion, converting 1 kg to 1000 g to match the latent heat of 80 cal/g, which gives a total of $Q = 1000 \text{ g} \times 80 \text{ cal/g} = 8 \times 10^4 \text{ cal}$. Next, convert the melting point of ice from 0°C to the absolute temperature of 273 K. The final change in entropy is found by dividing the total heat by this temperature:

$\Delta S = \frac{8 \times 10^4 \text{ cal}}{273 \text{ K}} \approx 293 \text{ cal/K}$

In an adiabatic process, the relationship between pressure $P$ and temperature $T$ is governed by the equation $\frac{T^{\gamma}}{P^{\gamma-1}} = \text{constant}$. This expression allows us to determine the final state of the gas by comparing its initial and final conditions, provided we convert the temperatures into absolute Kelvin values. Specifically, the initial temperature is 300 K and the final temperature is 1200 K.

Using the adiabatic relation to solve for the final pressure $P_f$ leads to the equation $P_f = P_i \left( \frac{T_f}{T_i} \right)^{\frac{\gamma}{\gamma-1}}$. Substituting the initial pressure of 2 atm, the given temperatures, and $\gamma = 1.4$ into this relationship, we obtain $P_f = 2 \times \left( \frac{1200}{300} \right)^{\frac{1.4}{1.4-1}}$. Simplifying the ratio inside the parenthesis yields $P_f = 2 \times (4)^{3.5}$. This can be rewritten as $P_f = 2 \times (2^2)^{3.5} = 2 \times 2^7 = 2^8$, resulting in a final pressure of 256 atm.

The first law of thermodynamics, which states that the heat supplied to a system is equal to the sum of the change in internal energy and the work done by the system, is expressed as $\Delta Q = \Delta U + \Delta W$. In an adiabatic process, the system is thermally isolated from its surroundings, meaning there is no heat exchange, so $\Delta Q = 0$. Substituting this into the first law equation results in $0 = \Delta U + \Delta W$, which rearranges to the form $\Delta U = -\Delta W$. This demonstrates that in an adiabatic process, the internal energy change must be exactly opposite to the work done by the system, whereas in an isothermal process, the internal energy remains constant, implying $\Delta Q = \Delta W$.

Adiabatic processes follow the relationship $PV^{\gamma} = \text{constant}$, implying that the condition $P_1V_1^{\gamma} = P_2V_2^{\gamma}$ must hold for any ideal gas. For a monatomic gas, the adiabatic index is defined as $\gamma = \frac{5}{3}$. Substituting the given volume reduction of $V_2 = \frac{V_1}{8}$ into the relation yields $P_2 = P_1 \left( \frac{V_1}{V_1/8} \right)^{\frac{5}{3}}$, which simplifies to $P_2 = P_1 \times 8^{\frac{5}{3}}$. Evaluating this expression leads to the final pressure of $P_2 = 32P_1$.

The fundamental relationship governing an ideal gas dictates that the difference between the molar heat capacities at constant pressure and constant volume is equivalent to the universal gas constant, represented by $C_{P,m} - C_{V,m} = R$. Since specific heat is defined as the heat capacity per unit mass rather than per mole, one must normalize the molar values using the molecular weight $M$. By dividing these molar heat capacities by the molecular weight, we derive the specific heats, yielding $C_P = \frac{C_{P,m}}{M}$ and $C_V = \frac{C_{V,m}}{M}$. Consequently, subtracting these two specific heat values results in $C_P - C_V = \frac{C_{P,m} - C_{V,m}}{M}$, which simplifies to the final expression $C_P - C_V = R/M$.

The First Law of Thermodynamics dictates that energy absorbed as heat is utilized to either increase internal energy or perform external work, expressed by the equation:

$dQ = dU + dW$

Converting 2 kcal of heat into joules provides $2000 \times 4.2 = 8400$ J of energy.

Substituting the heat and work values into the thermodynamic equation allows for the determination of the internal energy change. Because the system performs 500 J of work, the relationship is $8400 = dU + 500$, and solving $dU = 8400 - 500$ reveals an internal energy change of 7900 J.