Thermodynamics And Kinetic Theory – NEET UG PhysicsPractice Questions & PYQs

An adiabatic process is defined by the complete thermal insulation of a system, effectively preventing any thermal energy from crossing its boundaries. Because there is no interaction with the surrounding environment, the net heat transfer remains zero, denoted as $Q = 0$, meaning that any changes in the state of the system are driven entirely by work performed rather than through the gain or loss of heat.

The adiabatic index $\gamma$, defined as the ratio of molar heat capacities $\frac{C_p}{C_v}$, is determined by the degrees of freedom $f$ of the gas molecules according to the relationship $\gamma = 1 + \frac{2}{f}$. Hydrogen acts as a standard rigid diatomic gas with $f = 5$, resulting in $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$. Helium, being a monoatomic gas with $f = 3$, provides a value of $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$. Finally, for the diatomic gas $X$ that includes an additional vibrational mode, the degrees of freedom increase to $f = 7$, yielding a $\gamma$ value of $1 + \frac{2}{7} = \frac{9}{7}$.

The efficiency of an ideal heat engine depends entirely on the absolute temperatures of its heat source and heat sink, a relationship expressed by the formula $\eta = (1 - \frac{T_2}{T_1}) \times 100$. In this context, the heat source is the boiling point of water at 373 K, and the heat sink is the freezing point of water at 273 K. Substituting these values into the efficiency formula provides $\eta = (1 - \frac{273}{373}) \times 100$. Simplifying this expression leads to $\eta = \frac{100}{373} \times 100$, which equals approximately 26.8%.

The First Law of Thermodynamics states that the heat absorbed by a system is utilized for both increasing the internal energy and performing mechanical work against external pressure, which is expressed as the equation $\Delta Q = \Delta U + P\Delta V$. Converting the heat input from calories to joules involves calculating $54 \times 4.18 = 225.72 \text{ J}$.

The mechanical work performed during the expansion is determined by multiplying the constant pressure of $1.013 \times 10^5 \text{ N m}^{-2}$ by the volume change of $167.1 \times 10^{-6} \text{ m}^3$, which results in approximately 16.93 J. By applying the First Law to find the change in internal energy, $\Delta U = \Delta Q - P\Delta V$, the subtraction of $225.72 \text{ J} - 16.93 \text{ J}$ reveals that the internal energy of the sample increases by 208.7 J.

To determine the temperature at which oxygen molecules possess enough energy to break free from Earth's gravity, we must equate their root mean square speed to Earth's escape velocity of 11200 m/s. This relationship is defined by the formula $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$ where $k_B$ represents the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of the molecule. By squaring both sides and rearranging the expression to isolate $T$, we get $T = \frac{m v_{rms}^2}{3 k_B}$ Substituting the given values $m = 2.76 × 10^{-26}$ kg and $k_B = 1.38 × 10^{-23}$ J/K into this equation leads to a temperature of $8.360 × 10^4$ K.

The linear relationship between volume and temperature, passing through the origin, indicates that the gas undergoes an isobaric process, where pressure remains constant. The heat absorbed by a monoatomic gas during such a process is defined by $dQ = n C_p dT$, where the molar heat capacity at constant pressure is $C_p = \frac{5}{2}R$, resulting in $dQ = n \left(\frac{5}{2}R\right) dT$. The work done by the gas is given by $dW = P dV$, and by applying the ideal gas law $PV = nRT$, we substitute $P dV$ with $nR dT$ since the pressure $P$ is constant. The required ratio of work done to heat absorbed is determined by dividing these expressions, $\frac{dW}{dQ} = \frac{nR dT}{n \left(\frac{5}{2}R\right) dT}$, which simplifies to $\frac{2}{5}$.

The total internal energy of a mixture of gases is determined by summing the individual energy contributions from each component, calculated using the relationship $U = \frac{f}{2}nRT$ based on the degrees of freedom $f$ for each gas type. For the diatomic oxygen molecules, excluding vibrational modes leaves 5 degrees of freedom, so 2 moles contribute $\frac{5}{2}(2)RT$, which simplifies to $5RT$. Because argon is a monatomic gas, it possesses only 3 degrees of freedom, meaning its 4 moles contribute $\frac{3}{2}(4)RT$, resulting in $6RT$. Combining these individual contributions yields a total internal energy of $11RT$ for the system.

Identifying thermodynamic processes on a $P-V$ diagram relies on observing the constancy of specific state variables. Process I appears as a vertical line, signifying that volume is fixed, which defines an isochoric process. Process II is the steeper curve characteristic of an adiabatic change, whereas Process III represents an isothermal process where temperature remains constant. A horizontal line in Process IV indicates that pressure does not change, identifying it as an isobaric process.

The root mean square velocity of an ideal gas depends exclusively on its absolute temperature and molar mass, following the relation $V_{\mathrm{RMS}}=\sqrt{\frac{3RT}{M_0}}$. Because this expression does not contain any term related to pressure, variations in pressure are irrelevant to the calculation of the final velocity. Converting the temperatures to Kelvin, we get 300 K and 400 K. Applying the proportional relationship, we obtain the expression $V_1/V_2=\sqrt{T_1/T_2}$, which leads to the equation $200/V_2=\sqrt{300/400}$. Solving this equation provides a final result of $V_2 = 400/\sqrt{3}$ m/s.

The density of an ideal gas, defined as the total mass divided by the volume, is derived from the ideal gas law equation $PV = NkT$, where $N$ is the number of molecules and $k$ is the Boltzmann constant. Dividing both sides by the volume $V$ yields the number density $\frac{N}{V} = \frac{P}{kT}$. Since the total mass of the gas is the product of the number of molecules $N$ and the mass of each molecule $m$, the density $\rho$ is calculated as $\rho = \frac{Nm}{V}$, which simplifies to $\frac{Pm}{kT}$.

Any process following the polytropic form $PV^\alpha = \text{constant}$ has a molar heat capacity defined by the relation $C = C_v + \frac{R}{1 - \alpha}$. For an ideal monatomic gas, the heat capacity at constant volume is $C_v = \frac{3}{2} R$, and the provided equation sets the exponent $\alpha = 3$. Substituting these values into the governing expression gives $C = \frac{3}{2} R + \frac{R}{1 - 3}$, which simplifies to $C = \frac{3}{2} R - \frac{1}{2} R$, resulting in a heat capacity of $C = R$.

A refrigerator functions as a heat pump by extracting heat from its interior and rejecting it into the surrounding room. The ratio of the heat delivered to the room, $Q_H$, relative to the work consumed, $W$, is defined by the coefficient of performance for an ideal heat pump:
$\frac{Q_H}{W} = \frac{T_H}{T_H - T_L}$
Converting the given Celsius temperatures to Kelvin gives the room temperature as $T_H = t_1 + 273$ and the refrigerator internal temperature as $T_L = t_2 + 273$. Substituting these absolute temperatures into the equation results in $\frac{t_1 + 273}{(t_1 + 273) - (t_2 + 273)}$. Simplifying the denominator leads to the final expression of $\frac{t_1 + 273}{t_1 - t_2}$.

The work performed on a gas during compression is geometrically equivalent to the area under the curve on a $p-V$ diagram. Since an adiabatic compression is governed by the relation $PV^\gamma = \text{constant}$ and an isothermal compression follows $PV = \text{constant}$, the adiabatic curve is inherently steeper because the adiabatic exponent $\gamma$ is greater than one. As the volume decreases to half of the initial volume, this steeper slope ensures that the pressure rises more rapidly in the adiabatic case, forcing the adiabatic curve to remain above the isothermal one throughout the process. This difference in position results in a larger area under the adiabatic process curve, signifying that a greater amount of work must be performed on the system to achieve the same final volume compared to the isothermal compression.

Determining the power required for a refrigerator relies on the Coefficient of Performance, which relates the heat extracted to the work input required. This relationship is defined by $\beta = \frac{T_L}{T_H - T_L} = \frac{\text{Heat Extracted}}{\text{Power}}$, where temperatures must be expressed \in Kelvin. Converting the given values, we find the cold reservoir temperature is $277$ K and the hot reservoir is $303$ K, while the heat extraction rate converts to $600 × 4.2 = 2520$ Joules per second. Rearranging the expression to solve for the power input, the calculation becomes $\text{Power} = 2520 \times \frac{303 - 277}{277}$, which equals approximately $236.5$ Watts.

The performance of a refrigerator is fundamentally governed by the ratio of the heat extracted from the cold reservoir to the work required, which for an ideal system is expressed using absolute temperatures. Specifically, the coefficient of performance relates the internal temperature of the freezer to the temperature of the surroundings through the formula $COP = \frac{T_L}{T_H - T_L}$, where temperatures must be measured in Kelvin.

Since the coefficient of performance is 5 and the freezer temperature is $-20^\circ C$, we first express the cold temperature in Kelvin as $T_L = -20 + 273 = 253 K$. Inserting this into the equation $5 = \frac{253}{T_H - 253}$ allows us to isolate the unknown surroundings temperature by multiplying the denominator across to get $5T_H - 1265 = 253$. Adding 1265 to 253 yields $1518$, and dividing by 5 results \in $T_H = 303.6 K$. Converting back to the Celsius scale by subtracting 273 leads to $30.6^\circ C$, which is approximately $31^\circ C$.

A Carnot engine's efficiency $\eta$ is directly linked to the coefficient of performance $\beta$ of the same system when operating as a refrigerator through the relation $\beta = \frac{1 - \eta}{\eta}$. Substituting the efficiency of 0.1 into this expression gives $\beta = \frac{1 - 0.1}{0.1} = 9$. Since the coefficient of performance represents the ratio of heat extracted from the cold reservoir $Q_2$ to the work input $W$, defined by the formula $\beta = \frac{Q_2}{W}$, we find the energy absorbed by rearranging to $Q_2 = \beta × W$. Substituting the given work value of 10 J, we calculate the energy as $Q_2 = 9 × 10 = 90$ J.

The First Law of Thermodynamics for a complete cyclic process implies that the net heat absorbed by the system must equal the net work done, as the total change in internal energy over a closed loop is zero. On a pressure-volume diagram, the net work performed is equal to the area enclosed by the cycle, which is calculated as:
$W = \frac{1}{2} \times (2 \times 10^{-3}) \times (4 \times 10^4) = 40$
This results in 40 J of work. Since the sum of heat exchanges around the cycle follows the equation $Q_{AB} + Q_{BC} + Q_{CA} = W$, we can substitute the known values where 400 J is added in process AB and 100 J is added in process BC to get $400 + 100 + Q_{CA} = 40$. Simplifying this gives $Q_{CA} = -460$ J, representing the heat released in the return path from C to A. Because the heat absorbed along path AC is the negative of the heat in path CA, the system absorbs 460 J of heat in process AC.

The ideal gas law can be rearranged to express pressure in terms of density and molecular weight as $P = \frac{\rho RT}{M}$, where $\rho$ is the density, $R$ is the gas constant, $T$ is the temperature, and $M$ is the molecular weight. Since the temperature is constant for both vessels, the relationship between these variables allows us to determine the ratio of molecular weights as $\frac{M_A}{M_B} = \left(\frac{\rho_A}{\rho_B}\right) \left(\frac{P_B}{P_A}\right)$. Substituting the given values, where the density ratio $\frac{\rho_A}{\rho_B}$ is $1.5$ or $\frac{3}{2}$ and the pressure ratio $\frac{P_B}{P_A}$ is $\frac{1}{2}$, we find that $\frac{M_A}{M_B} = \left(\frac{3}{2}\right) \left(\frac{1}{2}\right) = \frac{3}{4}$.

The work performed on an ideal gas during compression is equivalent to the total area enclosed under the curve on a $P-V$ diagram. When comparing the compression from an initial volume $V_0$ to $\frac{V_0}{2}$, the adiabatic process results in a steeper pressure increase than isothermal or isobaric processes, while an isochoric process involves no volume change and therefore no work. This sharper rise in pressure ensures that the adiabatic path covers the largest area on the graph, indicating that this specific process requires the maximum work input to reduce the gas to the final volume.

Internal energy for an ideal gas depends solely on temperature change, which is defined by the formula $\Delta U = n C_V \Delta T$. By applying the ideal gas law $PV = nRT$, the change in internal energy can be expressed directly through pressure and volume coordinates as:
$\Delta U = \frac{P_2V_2 - P_1V_1}{\gamma - 1}$
For a diatomic gas, the adiabatic index $\gamma$ is $1.4$, meaning the denominator $\gamma - 1$ simplifies to $2/5$. From the provided transition path, the net change in the product of pressure and volume is $-8 \times 10^3$. Inserting these values into the formula, $\Delta U = \frac{-8 \times 10^3}{2/5}$, results in a total change of $-20 \text{ kJ}$.