Heat and Thermodynamics

$v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$

For same temperature, $v_{\text {rms }} \propto \frac{1}{\sqrt{M}}$

$ \frac{V_{\mathrm{rms}}^{\mathrm{Ar}}}{V_{\mathrm{rms}}^{\mathrm{Cl}}}=\sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}=\sqrt{\frac{70}{40}}=\frac{\sqrt{7}}{2} $

Using $1^{\text {st }}$ law of thermodynamics for electric heater,

$ \begin{aligned} & Q=\Delta U+W \\ & \Rightarrow \quad \frac{d Q}{d t}=\frac{d(\Delta U)}{d t}+\frac{d W}{d t} \\ & \Rightarrow \quad 100 \mathrm{~W}=\frac{d(\Delta U)}{d t}+75 \mathrm{~W} \\ & \therefore \quad \frac{d(\Delta U)}{d t}=25 \mathrm{~W} \end{aligned} $

For the metallic sphere, $\frac{\Delta R}{R}=\alpha \Delta T$

Volume of the sphere, $V=\frac{4}{3} \pi R^3$

$ \therefore \frac{\Delta V}{V}=3 \frac{\Delta R}{R} $

Increase in volume $=\Delta V=3\left(\frac{\Delta R}{R}\right)\left(\frac{4}{3} \pi R^3\right)=4 \pi R^3 \alpha \Delta T$

$ \begin{aligned} &\text { Work done in adiabatic process }\\ &\begin{aligned} & W=\frac{n R \Delta T}{1-\gamma} \\ & \Delta T=-10 K \\ & W=\frac{1 \times 8.3 \times(-10)}{1-\frac{5}{3}} \\ & W=\frac{83 \times 3}{2} \\ & W=124.5 \mathrm{~J} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { The total internal energy for one mole of gas, U }\\ &\begin{aligned} & =\left(\frac{3}{2} k_B T+\frac{3}{2} k_B T+f k_B T\right) N_A \\ & =(3+f) k_B T N_A=(3+f) R T \\ & \therefore \quad C_V=\frac{d U}{d T}=(3+f) R \\ & C_P=C_V+R=(4+f) R \\ & \Rightarrow \quad \frac{C_P}{C_V}=\frac{4+f}{3+f}=\frac{8}{7} \text { (given) } \\ & \Rightarrow \quad f=4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \lambda_A=\frac{1}{2} \lambda_B \\ & \lambda=\frac{1}{\sqrt{2} \pi d^2 n} \\ & d_A=2 d_B \\ & \lambda_A=\frac{1}{\sqrt{2} \pi d_A^2 n_A} \\ & \lambda_B=\frac{1}{\sqrt{2} \pi d_B^2 n_B} \\ & \frac{\lambda_A}{\lambda_B}=\frac{1}{2}=\frac{d_B^2}{d_A^2} \frac{n_B}{n_A} \\ & \frac{n_A}{n_B}=\frac{2 \cdot d_B^2}{d_A^2} \\ & \frac{n_A}{n_B}=\frac{2 \cdot d_B^2}{4 d_B^2}=\frac{1}{2} \end{aligned} $

$ \text { For any cyclic process, change in internal energy } \Delta U=0 $

$ \begin{aligned} &\text { Here, total heat exchange }\\ &\begin{aligned} \Delta Q & =W+\Delta U \\ & =W \\ & =(300-100)(5-2) \\ & =200 \times 3 \\ & =600 \mathrm{~J} \end{aligned} \end{aligned} $

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

Given, Temperature of food material in refrigerator,

$T_2=4^{\circ} \mathrm{C}=273+4=277 \mathrm{~K}$

Temperature of environment, $T_1=15^{\circ} \mathrm{C}$

$=273+15=288 \mathrm{~K}$

$\therefore$ Carnot efficiency, $\eta=1-\frac{T_2}{T_1}=1-\frac{277}{288}=0.038$

For $\mathrm{N}_2$, degree of freedom $f=5$ [For di-atomic gas]

$\therefore \quad C_{V_1}=\frac{f}{2} \cdot R=\frac{5}{2} R$ and $C_{p_1}=C_{V_1}+R=\frac{7}{2} R$

Amount of $\mathrm{N}_2$ gas, $m_1=7 \mathrm{~g}$

$\therefore$ Number of moles of $\mathrm{N}_2$,

$\begin{aligned} & n_1=\frac{m_1}{28}=\frac{7}{28} \\ & n_1=\frac{1}{4} \end{aligned}$

For Ar, degree of freedom, $f=3$

$\begin{aligned} & \therefore \quad C_{V_2}=\frac{3}{2} R \\ & \therefore \quad C_{p_2}=C_{V_2}+R \\ & =\frac{3}{2} R+R=\frac{5}{2} R \\ & \end{aligned}$

Amount of Ar gas, $m_2=20 \mathrm{~g}$

$\therefore$ Number of moles of Ar gas,

$\begin{aligned} n_2 & =\frac{m_2}{40}=\frac{20}{40}=\frac{1}{2} \\ \therefore \quad\left(C_p\right)_{\text {mix }} & =\frac{n_1 C_{p_1}+n_2 C p_2}{n_1+n_2} \\ & =\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{17 R}{6} \\ \left(C_V\right)_{\text {mix }} & =\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \\ & =\frac{\frac{1}{4} \times \frac{5}{2}+\frac{1}{2} \times \frac{3}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{11 R}{6} \\ \therefore \quad\left(\frac{C_p}{C_V}\right)_{\text {mix }} & =\gamma_{\text {mix }}=\frac{\frac{17 R}{6}}{\frac{11R}{6}}=\frac{17}{11} \end{aligned}$

Given, for isobaric process,

Work done by a di-atomic gas

By ideal gas equation, at constant pressure,

$p \Delta V=n R \Delta T$ .... (i)

$\therefore$ Work done,

$\begin{aligned} & W=p \Delta V \\ & W=n R \Delta T \quad \text{.... (ii)} \end{aligned}$

At constant pressure, heat given to gas is given by

$Q=n C_p \Delta T$ ..... (iii)

where, $C_p=$ specific heat at constant pressure

From Eqs. (ii) and (iii), we get

$\frac{W}{Q}=\frac{n R \Delta T}{n C_p \Delta T}=\frac{R}{C_p}$

$\begin{aligned} & =\frac{R}{\frac{(f+2) R}{2}} \quad\left[\because C_p=\frac{(F+2) R}{2}\right] \\ \frac{W}{Q} & =\frac{2}{f+2} \quad[\because W=10 \mathrm{~J} \text { (given) }] \\ \frac{10}{Q} & =\frac{2}{5+2} \quad[\text { For di-atomic gas, } f=5] \\ \Rightarrow \quad Q & =35 \mathrm{~J} \end{aligned}$

Temperature of sphere, $T=227^{\circ} \mathrm{C}$

$=273+227=500 \mathrm{~K}$

Radius, $\quad r=2 \mathrm{~m}$

Emissivity, $e=0.8$

$\therefore$ Radiation power of sphere $=$ Radiated energy per second

$\begin{aligned} & =\sigma A e T^4 \\ & =5.67 \times 10^{-8} \times 4 \pi \times 2^2 \times 0.8 \times(500)^4 \\ & =142502.6 \mathrm{~W}=142.5 \mathrm{~kW} \end{aligned}$

Given, initial pressure of ideal gas,

$p_1=1 \text { bar }=1 \mathrm{~N} / \mathrm{m}^2$

Initital volume $V_1=30 \mathrm{~m}^3$

Final volume, $V_2=10 \mathrm{~m}^3$

Initial temperature, $T_1=320 \mathrm{~K}$

Final temperature, $T_2=280 \mathrm{~K}$

Final pressure of gas, $p_2=$ ?

By ideal gas equation,

$\begin{aligned} & \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \Rightarrow \frac{1 \times 30}{320}=\frac{p_2 \times 10}{280} \\ \Rightarrow \quad & p_2=2.625 \mathrm{~bar} \end{aligned}$

There are two vibrational degree of freedom of di-atomic gas molecules at high temperature. Hence, vibrational degree of freedom of a di-atomic gas molecule appears at every high temperature because vibration in gas molecules is directly proportional to the square root of its temperature. Hence, both assertion and reason are true but reason is not the correct explanation of assertion.

A gas can be liquified by applying pressure only when it is cooled below critical temperature. Critical temperature of $\mathrm{NH}_3$ is more than $\mathrm{CO}_2$, i.e., $T_{\mathrm{NH}_3}=405 \mathrm{~K}$ and $T_{\mathrm{CO}_2}=304.1 \mathrm{~K}$.

Therefore, $\mathrm{NH}_3$ is liquified more easily than $\mathrm{CO}_2$. Hence, both assertion and reason are true and Reason is the correct explanation of assertion.

In adiabatic process, heat transfer to the thermodynamic system is zero, i.e., $Q=0$.

Hence, By first law of thermodynamics,

$\begin{array}{ll} & Q=0=\Delta U+W \\ \therefore & W=-\Delta U \end{array}$

So, work done is equal to negative of change in internal energy.

Since, the internal energy is a function of thermodynamic state of the system.

It does not depend on the path, therefore workdone is independent of path in adiabatic process.

Hence, both the assertion and reason are true and reason is the correct explanation of assertion.

Initial and final states are same in all the process.

Hence, $\Delta U=0$ is same for each case.

$\therefore \Delta Q=\Delta W$

Area enclosed by curve with volume.

$\begin{array}{cc} \because & (\text { Area })_1<(\text { Area })_2<(\text { Area })_3 \\\\ \therefore & Q_1< Q_2< Q_3 \end{array}$

For diatomic gas, $\gamma=1.4=\frac{7}{5}$

For adiabatic process, $T V^{\gamma-1}=$ constant

$\begin{aligned} T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ (300) V_1^{\frac{7}{5}-1} & =T_2\left(\frac{V_1}{5}\right)^{\frac{7}{5}-1} \\ \Rightarrow \quad T_2 & =\frac{300 \times V_1^{2 / 5}}{V_1^{\frac{2}{5}} \times\left(\frac{1}{5}\right)^{2 / 5}}=\frac{300}{5^{\frac{-2}{5}}}=300 \times 5^{2 / 5} \\ & =300 \times 1.903=571 \end{aligned}$

Mean kinetic energy of rotating molecules

$=k T=1.38 \times 10^{-23} \times 571=787.98 \times 10^{-23} \mathrm{~J}$

As there is no change in internal energy of the system during as isothermal change. Hence, the energy taken by the gas is utilised by doing work against external pressure.

Hence, whole heat energy gapping to the body at isothermal process converted into work done. Hence, Assertion is wrong.

Internal energy of gas depends only on temperature of gas.