Heat and Thermodynamics

First, consider the equivalent thermal resistance $R_{\text{eq}}$ of the series arrangement:

$ R_{\text{eq}} = R_1 + R_2 + R_3 $

Calculating each resistance:

$ R_1 = \frac{1}{2KA}, \quad R_2 = \frac{1}{KA}, \quad R_3 = \frac{1}{2KA} $

Summing these gives:

$ R_{\text{eq}} = \frac{1}{2KA} + \frac{1}{KA} + \frac{1}{2KA} = \frac{2}{KA} $

In a series arrangement, the rate of heat flow is constant. Thus:

$ \frac{3T - T_1}{R_1} = \frac{3T - T}{R_{\text{eq}}} $

Substituting the resistances:

$ \frac{(3T - T_1) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $

Solving the equation yields:

$ 6T - 2T_1 = T \implies 2T_1 = 5T \implies T_1 = \frac{5T}{2} \quad \ldots (1) $

Next, examine the heat flow rate in the third section:

$ \frac{T_2 - T}{R_3} = \frac{3T - T}{R_{\text{eq}}} $

Substitute:

$ \frac{(T_2 - T) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $

Solving gives:

$ 2T_2 - 2T = T \implies 2T_2 = 2T + T \implies T_2 = \frac{3T}{2} \quad \ldots (2) $

By substituting equations (1) and (2) into the ratio:

$ \frac{T_1}{T_2} = \frac{5T/2}{3T/2} = \frac{5}{3} $

Thus, the ratio $\frac{T_1}{T_2}$ is $\frac{5}{3}$.

To find the equilibrium pressure of the system after the partition is removed, we can apply the principle of conservation of moles (or the ideal gas law in combined volumes).

Initially, we have:

Chamber 1: $ P_1 = 1 \, \text{atm}, \, V_1 = 2 \, \text{L} $

Chamber 2: $ P_2 = 2 \, \text{atm}, \, V_2 = 3 \, \text{L} $

The total pressure after the partition is removed and the gases mix can be calculated using:

$ P_{\text{final}} = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2} $

Substituting the given values:

$ P_{\text{final}} = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{2 + 6}{5} = \frac{8}{5} $

This gives us:

$ P_{\text{final}} = 1.6 \, \text{atm} $

Therefore, the equilibrium pressure when the gases mix is $ 1.6 \, \text{atm} $.

To find the mass of oxygen withdrawn from the cylinder, we start by calculating the number of moles left in the cylinder after some oxygen is withdrawn. We use the ideal gas law in the form:

$ n = \frac{PV}{RT} $

Substituting the given values:

$ P = 11 $ atm converts to $ 11 \times 1.01 \times 10^5 \, \text{N/m}^2 $

$ V = 30 \, \text{liters} = 30 \times 10^{-3} \, \text{m}^3 $

$ R = \frac{100}{12} \, \text{J/mol K} $

$ T = 27^\circ \text{C} = 300 \, \text{K} $

We calculate the moles after oxygen has been withdrawn:

$ n = \frac{12 \times 1.01 \times 10^5 \, \text{N/m}^2 \times 30 \times 10^{-3} \, \text{m}^3}{\left(\frac{100}{12}\right) \times 300} $

Simplifying the expression:

$ n = \frac{12 \times 1.01 \times 12}{10} = 14.54 \, \text{moles} $

Next, determine the moles of oxygen removed:

$ \text{Moles removed} = 18.20 - 14.54 = 3.656 \, \text{moles} $

Finally, convert the moles removed into mass:

$ \text{Mass removed} = 3.656 \times 32 = 116.99 \, \text{g} = 0.116 \, \text{kg} $

Thus, the mass of the oxygen withdrawn from the cylinder is approximately 0.116 kg.

The problem involves two gases, $A$ and $B$, at the same pressure in separate cylinders with movable pistons of radii $r_A$ and $r_B$. Equal amounts of heat are supplied to both cylinders reversibly under constant pressure, displacing the pistons by 16 cm for gas $A$ and 9 cm for gas $B$. Given that the change in their internal energy is the same, we need to find the ratio $\frac{r_A}{r_B}$.

First, we apply the first law of thermodynamics:

$ \Delta Q = \Delta U + P \Delta V $

Since $\Delta Q$ and $\Delta U$ are the same for both gases, their work done $W_A$ and $W_B$ is also equal. Therefore:

$ (P \Delta V)_A = (P \Delta V)_B $

With constant pressure $P$, the relationship becomes:

$ A_A d_A = A_B d_B $

where $A$ is the area of the piston. We know:

$ \pi r_A^2 d_A = \pi r_B^2 d_B $

Simplifying, the ratio of the radii is:

$ \frac{r_A}{r_B} = \left(\frac{d_B}{d_A}\right)^{\frac{1}{2}} = \left(\frac{9}{16}\right)^{\frac{1}{2}} $

Calculating the square root gives:

$ \frac{r_A}{r_B} = \frac{3}{4} $

Let's analyze the given statements to determine the correct answer.

Assertion $\mathbf{A}$: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.

Reason $\mathbf{R}$: The layer of foam insulation prohibits heat transfer, as it contains air pockets.

Explanation:

Foam insulation generally acts as an effective thermal barrier because it contains numerous air pockets that impede heat transfer. The primary function of foam insulation is to reduce the amount of heat that penetrates into or escapes from a building, thereby maintaining a more constant internal temperature. If a house has a concrete roof overlaid with foam, the foam will work to keep the interior cooler in summer by prohibiting the transfer of heat from the outside into the room. Hence, the foam insulation should actually help in keeping rooms cooler during summer, not hotter.

Therefore, the Assertion $\mathbf{A}$ is false because foam insulation typically helps in keeping rooms cooler during summer. Meanwhile, the Reason $\mathbf{R}$ is true because air pockets within the foam insulation impede heat transfer.

Based on this analysis, the correct option is:

Option B: A is false but R is true.

The correct answer is Option D: A, C and D only. Here's why:

The equilibrium state of a thermodynamic system is defined by a set of conditions where the system is in a stable, unchanging state. These conditions include:

• Pressure (A): A measure of the force exerted by the system per unit area. At equilibrium, the pressure within the system is constant and uniform.
• Temperature (C): A measure of the average kinetic energy of the particles within the system. At equilibrium, the temperature throughout the system is constant and uniform.
• Volume (D): The amount of space occupied by the system. At equilibrium, the volume remains constant unless external forces act upon the system.

Let's examine why the other options are incorrect:

• Total heat (B): While heat transfer can occur during a process, the total heat content of a system is not a defining characteristic of its equilibrium state. Equilibrium is about the absence of net change, not the total amount of energy present.
• Work done (E): Work is a process, not a state variable. It describes the energy transfer during a change in the system, not its equilibrium condition.

In summary, pressure, temperature, and volume are the key state variables that define the equilibrium state of a thermodynamic system because they are constant and uniform under these conditions.

A polygamic gas has 3 translational, 3 rotational and $f$ vibration modes

$\begin{aligned} & U=\frac{3}{2} k_B T+\frac{3}{2} k_B T+f k_B T \\ & U=(3+f) k_B T \\ & C_V=(3+f) R \\ & C_p=(4+f) R \\ & \frac{C_p}{C_v}=\frac{4+f}{3+f}=\gamma \\ & 4+f=3 \gamma+f \gamma \\ & 4-3 \gamma=f(\gamma-1) \Rightarrow f=\frac{4-3 \gamma}{\gamma-1} \end{aligned}$

Path $b c$ is an isochoric process.

$\therefore \quad$ Work done by gas along path $b c$ is zero.

At same temperature, curve with higher volume corresponds to lower pressure.

$\begin{aligned} & V_3>V_2>V_1 \\ & \Rightarrow P_1>P_2>P_3 \end{aligned}$

(We draw a straight line parallel to volume axis to get this)

$\mathrm{AB}$ is isobaric process

$\begin{aligned} & \mathrm{W}_{\mathrm{AB}}=\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \mathrm{W}_{\mathrm{AB}}=3 \times 10^2(3-1) \\\\ & \mathrm{W}_{\mathrm{AB}}=3 \times 100 \times 2 \\\\ & \mathrm{~W}_{\mathrm{AB}}=600 \mathrm{~J} \end{aligned}$

$ \begin{aligned} & P_{\text {mix }}=\frac{\left(\mu_1+\mu_2\right) \mathrm{RT}_{\text {mix }}}{\mathrm{V}_{\text {mix }}} \\ & =\frac{(0.2+0.3) \times 8.3 \times 200}{200 \times 10^{-6}} \\ & =\frac{0.5 \times 8.3 \times 200}{200 \times 10^{-6}} \\ & \mathrm{P}_{\text {mix }}=4.15 \times 10^6 \mathrm{P}_{\mathrm{a}} \end{aligned} $

$v_{\mathrm{rms}} \propto \sqrt{T}$

$\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$

$=$ let initial speed is $\mathrm{v}$

As speed is increased by $3$ times so final speed become $4 v$

$\Rightarrow \frac{\mathrm{v}}{4 \mathrm{v}}=\sqrt{\frac{223}{\mathrm{~T}}}$

$\mathrm{T}=3568 \mathrm{~K}$

So temp. in ${ }^{\circ} \mathrm{C}=3568-273=3295^{\circ} \mathrm{C}$

Efficiency of carnot engine

$ \begin{aligned} & \% \eta=\left(1-\frac{T_{\text {sink }}}{\mathrm{T}_{\text {source }}}\right) \times 100 \\ & T_{\text {source }}=327^{\circ} \mathrm{C}=600 \mathrm{~K} \\ & 50=\left(1-\frac{\mathrm{T}_{\text {sink }}}{600}\right) \times 100 \\ & \frac{1}{2}=1-\frac{\mathrm{T}_{\text {sink }}}{600} \\ & \mathrm{~T}_{\text {Sink }}=300 \mathrm{~K} \end{aligned} $

So temp. of sink is ${ }^{\circ} \mathrm{C}=300-2763=27^{\circ} \mathrm{C}$

We know,

$PV = nRT$

Given, $P{V^2} = $ constant

$\Rightarrow$ If ${P_1} > {P_2} \Rightarrow {V_1} < {V_2}$

$P{\left( {{{nRT} \over P}} \right)^2} = $ constant

${P^{ - 1}}{T^2} = $ constant

$ \Rightarrow P \propto {T^2}$

i.e., if ${P_1} > {P_2} \Rightarrow {T_1} > {T_2}$

Also, $\left( {{{nRT} \over V}} \right){V^2} = $ constant

$TV = $ constant

$\Rightarrow$ If ${V_2} > {V_1}$ then ${T_2} < {T_1}$

We know in conduction, rate of flow of heat $q = {{ - K.A.\Delta T} \over {\Delta x}}$

As it is a case of steady state heat transfer

$ \Rightarrow {{ - {K_{cu}} \times A \times [100 - {T_j}]} \over l} = {{ - {K_{steel}} \times A \times [{T_j} - 0]} \over l}$

Where T_j is temperature of the junction

$ \Rightarrow 385(100 - {T_j}) = 50({T_j})$

$ \Rightarrow {{100} \over {{T_j}}} - 1 = {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1 + {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1.1298$

$ \Rightarrow {T_j} = 88.5^\circ C$

All three vessels have equal volume and same temperature and pressure.

From ideal gas equation

$PV = nRT$

$nR = {{PV} \over T}$

$n = {{PV} \over {RT}} = $ constant

So, here all three vessels contains equal number of moles and number of gas molecules.

Now, ${v_{rms}} = \sqrt {{{3RT} \over M}} \Rightarrow {v_{rms}} \propto {1 \over {\sqrt M }}$

Here, rms speed of helium is the largest.

${\left( {{{dP} \over {dV}}} \right)_{adiabatic}} = - \gamma P$

${\left( {{{dP} \over {dV}}} \right)_{isothermal}} = -P$

${\left( {{{dP} \over {dV}}} \right)_{adiabatic}} > {\left( {{{dP} \over {dV}}} \right)_{isothermal}}$

From ideal gas equation

$PV = nRT$

$\left[ {n = {{mass\,of\,water} \over {mol.\,wt.}} = {{4.5 \times {{10}^3}} \over {18}}} \right]$

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

At $STP \Rightarrow T = 273\,K$

$P = {10^5}$ N/m²

$V = {{4.5 \times {{10}^3}} \over {18}} \times {{8.3 \times 273} \over {{{10}^5}}} = 5.66$ m³