Thermodynamics And Kinetic Theory – NEET UG PhysicsPractice Questions & PYQs

Thermodynamic processes are distinguished by which state variable remains invariant, and an isochoric process is defined by the volume being held constant. Keeping pressure constant, however, corresponds to an isobaric process, making the claim in option A a misidentification of the physical condition. Unlike the other statements describing isothermal changes or the adiabatic relation $PV^{\gamma} = \text{constant}$, this specific definition incorrectly equates pressure regulation with isochoric behavior.

Thermodynamics defines internal energy as a state function, meaning its value depends exclusively on the system's current condition rather than the path taken to achieve it. Because a cyclic process requires the system to return to its exact starting point, the final state is equivalent to the initial state. Consequently, there is no net change in the total internal energy, represented by the equation $E = 0$.

The behavior of an ideal gas is governed by the equation $PV = nRT$, where $n$ is the number of moles. By substituting the number of moles with the mass $m$ divided by the molar mass $M$, the equation becomes $PV = \frac{m}{M}RT$. Since density $\rho$ is defined as mass per unit volume $\frac{m}{V}$, rearranging the ideal gas equation allows us to express density as $\rho = \frac{PM}{RT}$. The problem asks us to examine the ratio of density to pressure, which simplifies to $\frac{\rho}{P} = \frac{M}{RT}$. Because the molar mass and the ideal gas constant are fixed, this specific ratio is inversely proportional to the absolute temperature $T$.

This inverse relationship implies that the product of the ratio and the absolute temperature remains constant. At the initial temperature of $10^\circ\text{C}$, which converts to $283\text{ K}$, the ratio is $x$, giving us the constant product $x \cdot 283$. At the second temperature of $110^\circ\text{C}$, or $383\text{ K}$, let the new ratio be $x'$, which results in the product $x' \cdot 383$. Equating these two scenarios gives $x' \cdot 383 = x \cdot 283$, and solving for the new ratio $x'$ yields $x' = \frac{283}{383} x$.

The efficiency of a heat engine is defined by the temperatures of the source and the sink according to the formula $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ and $T_2$ represent the absolute temperatures in Kelvin. Given that the initial efficiency is 1/6, we can determine the relationship between these temperatures as $\frac{T_2}{T_1} = 5/6$.

When the temperature of the sink is reduced by 62 K, the efficiency doubles to 1/3 while the source temperature remains unchanged. Substituting these conditions into the efficiency formula results in $1/3 = 1 - \frac{T_2 - 62}{T_1}$ which simplifies to $\frac{T_2 - 62}{T_1} = 2/3$ By replacing the ratio $\frac{T_2}{T_1}$ with the initial value of 5/6, the equation becomes $\frac{5/6 T_1 - 62}{T_1} = 2/3$ This expression simplifies to $5/6 - 62/T_1 = 4/6$, which further reduces to $62/T_1 = 1/6$. Solving this calculation reveals that the source temperature is 372 K, which corresponds to 99°C on the Celsius scale.

The efficiency of a Carnot engine depends on the absolute temperatures of its hot source and cold sink as defined by the relationship $\eta = 1 - \frac{T_{sink}}{T_{source}}$. Given a sink temperature of $300$ K and an initial efficiency of $40\%$, we can determine the original source temperature by using the equation $\frac{T_{sink}}{T_{source}} = 1 - 0.4$. This simplifies to $\frac{300}{T_{source}} = 0.6$, which leads to an initial source temperature of $T_{source} = 500$ K.

Increasing the efficiency by $50\%$ of its original value means adding $20\%$ to the existing $40\%$, resulting in a new target efficiency of $60\%$. To find the required source temperature for this improved performance, we use the formula again such that $\frac{300}{T'_{source}} = 1 - 0.6$. This calculation gives $\frac{300}{T'_{source}} = 0.4$, indicating that the new source temperature must be $T'_{source} = 750$ K. The required increase is the difference between the new and original temperatures, calculated as $750$ K minus $500$ K, resulting in a temperature rise of $250$ K.

The fundamental relationship connecting the molar specific heats of an ideal gas is expressed by $C_P - C_V = R$. Given that $C_P$ is $\frac{7}{2} R$, we determine $C_V$ by rearranging the formula to $C_V = C_P - R$, which yields $C_V = \frac{7}{2} R - R = \frac{5}{2} R$. Dividing $C_P$ by $C_V$ provides the required ratio, which is calculated as $\frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5}$.

A reversible process is defined by its ability to be undone without leaving any net change in the surroundings, which requires the system to remain in continuous equilibrium by proceeding at an infinitesimally slow rate. While heat transfer through conduction, radiation, and electrical heating involve spontaneous energy dissipation that increases entropy, an isothermal compression can be performed quasistatically. This allows it to function as a reversible step within idealized models like the Carnot cycle, where the energy exchange satisfies the condition $W = Q$ while the temperature remains constant.

The efficiency of an ideal heat engine operating in a Carnot cycle depends exclusively on the absolute temperatures of the hot and cold reservoirs, defined by the relationship $\eta = 1 - \frac{T_2}{T_1}$. Converting the given temperatures from Celsius to Kelvin yields $T_1 = 500$ K and $T_2 = 400$ K, which results in an efficiency of $\eta = 1 - \frac{400}{500} = 0.2$. Since efficiency represents the ratio of work performed to heat absorbed, the amount of heat converted into work is determined by multiplying this fraction by the input heat of $6 \times 10^4$ cal, producing a final result of $1.2 \times 10^4$ cal.

The relationship between pressure, volume, and temperature for an ideal gas is governed by the ideal gas law, expressed as $PV = nRT$. Here, $n$ represents the number of moles of the gas, calculated by dividing the total mass of the substance by its molar mass. Oxygen gas has a molar mass of 32 g/mol, so with a given mass of 5 g, the number of moles becomes $n = \frac{5}{32}$. Substituting this value into the ideal gas equation yields the result $PV = \left(\frac{5}{32}\right)RT$.

For an ideal gas undergoing an adiabatic process, the first law of thermodynamics dictates that any work performed by the system causes an equivalent decrease in its internal energy because there is no heat exchange with the surroundings. This energy balance is defined by $\Delta U = -W$, where the change in internal energy relates to the temperature difference through the expression $\Delta U = n C_v \Delta T$, and the molar heat capacity at constant volume is given by $C_v = \frac{R}{\gamma - 1}$.

With the provided values of $n=1$, $W = 6R$, and $\gamma = 5/3$, the equation for the process becomes $1 \cdot \frac{R}{(5/3) - 1} (T_f - T) = -6R$. Simplifying the adiabatic term leads to $1.5R (T_f - T) = -6R$. Dividing both sides of the equation by $1.5R$ leaves $T_f - T = -4$, which confirms that the final temperature of the gas is $T - 4$ kelvin.

The performance of a Carnot heat engine is defined by the temperatures of its hot and cold reservoirs, which must be expressed in Kelvin to calculate efficiency. Converting the given temperatures of $227^{\circ}\text{C}$ and $127^{\circ}\text{C}$ results in an absolute temperature of $500\text{ K}$ for the source and $400\text{ K}$ for the sink. The efficiency of the cycle is determined by the formula $\eta = 1 - \frac{T_2}{T_1}$, which equates to $1 - \frac{400}{500}$, or $0.2$. Since efficiency is also the ratio of work output to heat input, we set $\frac{W}{Q_1} = 0.2$. With $Q_1$ given as $6\text{ kcal}$, multiplying this by the efficiency yields $W = 0.2 \times 6$, confirming the heat converted into work is $1.2\text{ kcal}$.

The performance of a Carnot engine is fundamentally governed by the absolute temperatures of the heat source and sink, expressed by the equation $\eta = 1 - \frac{T_{sink}}{T_{source}}$. Given that the engine starts with an efficiency of 50% and a sink temperature of 500 K, the constant source temperature is found by solving the following expression:
$0.5 = 1 - \frac{500}{T_{source}}$
This simplifies to $0.5 = \frac{500}{T_{source}}$, confirming that the source temperature is 1000 K.

To find the new sink temperature required to reach an efficiency of 60% while keeping the source temperature fixed at 1000 K, we set up the equation again:
$0.6 = 1 - \frac{T'_{sink}}{1000}$
Rearranging the terms results in $\frac{T'_{sink}}{1000} = 1 - 0.6$, which simplifies to 0.4. Multiplying this by 1000 gives the new sink temperature as 400 K.

The second law of thermodynamics establishes that no heat engine can operate with an efficiency greater than an ideal Carnot engine working between the same source and sink temperatures, defined by the formula $\eta = 1 - \frac{T_2}{T_1}$. Converting the source and sink temperatures to 400 K and 300 K respectively, the maximum possible efficiency is 25%. Because the scientist claims an efficiency of 26%, which surpasses this fundamental thermodynamic ceiling, the device described is impossible.

The adiabatic index, denoted by $\gamma$, is related to the degrees of freedom of a gas molecule, denoted by $f$, through the standard thermodynamic expression $\gamma = 1 + \frac{2}{f}$. By subtracting 1 from both sides of this equation, we obtain $\gamma - 1 = \frac{2}{f}$, and inverting this relationship allows us to solve for the degrees of freedom as
$f = \frac{2}{\gamma - 1}$

The efficiency of a Carnot engine is defined by $\frac{W}{Q} = 1 - \frac{T_2}{T_1}$, where $T_1$ and $T_2$ are the source and sink temperatures in Kelvin. Initially, $\frac{W}{Q} = \frac{1}{6}$, which implies that the ratio $\frac{T_2}{T_1} = \frac{5}{6}$. Reducing the sink temperature by 62 leads to a doubled efficiency of $\frac{1}{3}$, satisfying the equation $1 - \frac{T_2 - 62}{T_1} = \frac{1}{3}$. Substituting $\frac{T_2}{T_1} = \frac{5}{6}$ into this expression gives $1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{3}$, which simplifies to $\frac{62}{T_1} = \frac{1}{6}$. Consequently, $T_1$ equals 372 Kelvin, which converts to $99^{\circ}\text{C}$, and $T_2$ equals 310 Kelvin, which converts to $37^{\circ}\text{C}$.

In a pressure-volume diagram, the net work performed during a thermodynamic cycle corresponds to the area enclosed by the closed loop, where a clockwise direction indicates that the gas does positive work. We determine the magnitude of this work by calculating the area of the triangular region formed by points A, B, and C. The base of the triangle spans the change in volume, calculated as $(7 - 2) \times 10^{-3}$ m³, and the height represents the change in pressure, which is $(6 - 2) \times 10^5$ Pa.

Applying the geometric area formula for a triangle, $W = \frac{1}{2} \times \text{base} \times \text{height}$, the calculation becomes $W = \frac{1}{2} \times (5 \times 10^{-3}) \times (4 \times 10^5)$. Multiplying these terms results in a final value of 1000 J.