Heat and Thermodynamics

Let the temperature $=X$

$ F=K=X $

$F$ is temp in Fahrenheit, $K$ is temperature in Kelvin.

Relation between $F$ and $K$ is

$ \frac{F-32}{9}=\frac{K-273}{5} $

Put $X$ in place for $F, K$

$ \begin{aligned} \frac{X-32}{9} & =\frac{X-273}{5} \\ & =4 X-2457+160=0 \\ X & =574.6 \end{aligned} $

Thus, the temperature is $574.6^{\circ} \mathrm{F}$ or 574.6 K

Given,

Ratio of densities $\left(\frac{\rho_1}{\rho_2}\right)=\frac{5}{6}$

Ratio of specific heat capacities $\left(\frac{C_1}{C_2}\right)=\frac{3}{5}$

Using the formula for heat

$\Delta Q=m C \Delta T[\Delta T \rightarrow$ Change in temperature]

For $1^{\text {st }}$ substance

$ \Delta Q_1=m_1 C_1 \Delta T_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For $2^{\text {nd }}$ substance

$ \begin{aligned} \Delta Q_2 & =m_2 C_2 \Delta T_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \frac{\Delta Q_1}{\Delta Q_2} & =\frac{m_1 C_1}{m_2 C_2} \quad\left[\Delta T_1=\Delta T_2 \text { (given) }\right] \\ \frac{\Delta Q_1}{\Delta Q_2} & =\frac{V_1 \rho_1 C_1}{V_2 \rho_2 C_2} \\ \end{aligned} $

$ \Rightarrow \quad \frac{\frac{\Delta Q_1}{V_1}}{\frac{\Delta Q_2}{V_2}}=\frac{\rho_1 C_1}{\rho_2 C_2}=\frac{5}{6} \times \frac{3}{5} \Rightarrow \frac{\frac{\Delta Q_1}{V_1}}{\frac{\Delta Q_2}{V_2}}=\frac{1}{2} $

Hence, the ratio of heat energy required per unit volume. So that two substance have same temperature is $1 / 2$ or $1: 2$

Given,

Work done by the system = decrease in internal energy

$ d W=-d U $

Using $1{ }^{\text {st }}$ Law of thermodynamics.

$ \begin{aligned} & d Q=d U+d W \\ & d Q=d U-d U \\ & d Q=0 \end{aligned} $

where $Q$ is heat supplied (change in heat).

So, the process in which $d Q$ is zero is called as adiabatic process.

Given,

Mass of molecule $A=m$

number of molecule $A=N$

mean square component of

Velocity of $A$ gas $=v_1^2$

Mass of molecule $B=2 \mathrm{~m}$

number of molecule $B=2 \mathrm{~N}$

Mean square velocity $B=v_2^2$

Mean square velocity is given by $v^2=\frac{3 k T}{m}$

Where $T$ is temperature, $K$ is boltzmann constant.

For gas $v_A^2=v_2^2=\frac{3 k T}{2 m} \quad(m=2 m)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For gas molecule $A$

$ v^2=v_x^2+v_y^2+v_z^2=3 v_x^2 \quad\left(\because v_x^2=v_y^2=v_z^2\right) $

or $v_x^2=\frac{v^1}{3}$

$ \begin{aligned} v_1^2 & =\frac{3 k T}{3 m}=\frac{k T}{m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ \frac{v_1^2}{v_2^2} & =\frac{k T}{m} \times \frac{2 m}{3 k T}=\frac{2}{3} \end{aligned} $

So, the ratio is

$ \frac{v_1}{v_2}=\sqrt{\frac{2}{3}} $

Given, length of rod at $30^{\circ} \mathrm{C}$ is 30 cm

Rise in temperature

$ \begin{aligned} \Delta T & =105^{\circ}-30{ }^{\circ} \mathrm{C} \\ & =75^{\circ} \mathrm{C} \end{aligned} $

Increase in length,

$ \begin{aligned} \Delta l & =0.027 \mathrm{~cm} \\ & =0.027 \times 10^{-2} \mathrm{~m} \end{aligned} $

We know that, $\Delta l=l \alpha \Delta T$

Thus, $0.027 \times 10^{-2}=0.30 \times \alpha \times 75$

$ \begin{aligned} \Rightarrow \quad \alpha & =\frac{0.027 \times 10^{-2}}{0.30 \times 75} \\ & =0.0012 \times 10^{-2} \\ & =12 \times 10^{-6} \rho \mathrm{C} \end{aligned} $

Heat energy required for 10 kg to bring to $0^{\circ} \mathrm{C}$ will be, $Q_1=m s \Delta T$

$ \begin{aligned} & =10 \times 0.5 \times 10 \times 10^3 \\ & =50 \times 10^3 \mathrm{cal} \end{aligned} $

Heat energy required for 10 kg ice at $0^{\circ} \mathrm{C}$ to convert into water at $0^{\circ} \mathrm{C}$ will be.

$ \begin{aligned} Q_2 & =m L \\ & =10 \times 10^3 \times 80 \\ & =8 \times 10^5 \mathrm{cal} \end{aligned} $

Total heat $Q=Q_1+Q_2$

$ \begin{aligned} & =50 \times 10^3+800 \times 10^3 \\ & =850 \times 10^3 \mathrm{cal} \\ & =357 \times 10^4 \mathrm{~J} \end{aligned} $

We know that, ${ }^{\circ} \mathrm{C}=\left({ }^{\circ} \mathrm{F}-30\right) \times \frac{5}{9}$

Given ${ }^{\circ} \mathrm{F}=2 \times\left({ }^{\circ} \mathrm{C}\right)$

Thus, ${ }^{\circ} \mathrm{C}=\left(2^{\circ} \mathrm{C}-32\right) \times \frac{5}{9}$

$ \begin{aligned} & { }^{\circ} \mathrm{C} \times \frac{9}{5}=2{ }^{\circ} \mathrm{C}-32 \\ & 2^{\circ} \mathrm{C}-{ }^{\circ} \mathrm{C} \times \frac{9}{5}=32 \\ \Rightarrow & \frac{{ }^{\circ} \mathrm{C}}{5}=32 \Rightarrow{ }^{\circ} \mathrm{C}=160^{\circ} \end{aligned} $

Hence, Fahrenheit $\left({ }^{\circ} \mathrm{F}\right)=2^{\circ} \mathrm{C}$

$ =2 \times 160=320^{\circ} \mathrm{F} $

The heat $Q$ is converted into the internal energy and work. According to the first low of thermodynamics,

$ \begin{aligned} & \Delta Q=\Delta U+\Delta W \\ & n c_p d T=n c_V d T+W \end{aligned} $

The fraction of heat converted into

$ \begin{aligned} & \text { internal energy } \frac{\Delta U}{\Delta Q}=\frac{\Delta Q-\Delta W}{\Delta Q} \\ & =\frac{n c_p d T-\left(n c_p d T-n c_V d T\right)}{n c_p d T} \\ & =\frac{n c_V d T}{n c_p d T}=\frac{c_V}{c_p}=\frac{3}{5} \quad\left(\because r=\frac{5}{3}\right) \end{aligned} $

Therefore percentage of heat energy used

$ =\frac{3}{5} \times 100 \%=60 \% $

$ \begin{aligned} &\text { Average energy of an oscillator is }\\ &\begin{aligned} \langle E\rangle & =k_B T \\ & =1.38 \times 10^{-23} \times 300 \\ & =4.14 \times 10^{-21} \mathrm{~J} \end{aligned} \end{aligned} $

We know that, ideal gas equation is given gy

$ \begin{gathered} p V=n R T \\or\,\,\,\frac{1}{T}=\frac{n R}{p V} \end{gathered} $

Since, $p$ is constant

$ p d V=n R d T \Rightarrow \frac{d V}{d T}=\frac{n R}{p} $

coefficient of volume expansion will be

$ \begin{aligned} \alpha_V & =\frac{d V}{V d T}=\frac{n R}{p V}=\frac{1}{T} \\ \Rightarrow \quad \alpha_V & =\frac{1}{300 \mathrm{~K}} \quad\left(\because 23^{\circ} \mathrm{C}=300 \mathrm{~K}\right) \\ & =33 \times 10^{-4} \mathrm{~K}^{-1} \end{aligned} $

Given, temperature,

$ T_1=60^{\circ} \mathrm{C}=273+60=333 \mathrm{~K} $

Let initial mass be $M_1=m$ and air escaped be $M_2=M-\frac{M}{4}=\frac{3 M}{4}$

If pressure and volume of gas remain constant.

Then, $M T=$ Constant

$ \begin{aligned} \frac{T_1}{T_2} & =\frac{M_1}{M_2}=\left(\frac{M}{\frac{3 M}{4}}\right)=\frac{4}{3} \\ \Rightarrow \quad T_2 & =\frac{4}{3} T=\frac{4}{3} \times 333 \\ & =444 \mathrm{~K}=171^{\circ} \mathrm{C} \end{aligned} $

Given mass of water, $m_w=100 \mathrm{~g}$

Change in temperature, $(\Delta T)=90^{\circ} \mathrm{C}- 24^{\circ} \mathrm{C}=66^{\circ} \mathrm{C}$

From the principle of calorimetry

Heat gain by water $=$ Heat lost by steam

$ \begin{aligned} & m_w S \Delta T \\ = & m_{\text {steam }}(540)+\left(m_{\text {steam }}\right)+\left(m_{\text {steam }}(100-90)\right. \\ \Rightarrow & 100 \times 1(66)=m(540)+m(10) \\ \Rightarrow & 550 m=100 \times 66 \\ \Rightarrow \quad & m=12 \mathrm{~g} \end{aligned} $

For an isothermal process, the temperature, will constant Volume is increasing therefore, pressure will decrease i.e., $p \propto \frac{1}{V}$

In chamber $A, \Delta p_A=\left(p_A\right)_i-\left(p_A\right)_f$

$ \begin{aligned} & =\frac{n_A R T}{V}-\frac{n_A R T}{2 V} \\ & =\frac{n_A R T}{2 V} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

In chamber $B, \Delta p_B=\left(p_B\right)_i-\left(p_B\right)_t$

$ =\frac{n_B R T}{V}-\frac{n_B R T}{2 V}=\frac{n_B R T}{2 V} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

According to question, $\frac{\Delta p_A}{\Delta p_B}=\frac{2}{3}$

$ \Rightarrow \quad \frac{1}{1.5}=\frac{n_A}{n_B}=\frac{2}{3} \frac{m_A / M}{m_B / M}=\frac{2}{3} $

or   $ 3 m_A=2 m_B $

Initial Setup :

First rod: length $L$, coefficient of expansion $2\alpha$

Second rod: length $2L$, coefficient of expansion $\alpha$

Composite Rod:

Total initial length = $L + 2L = 3L$

Expansion Calculation:

Expansion of the First Rod:

$ \Delta L_1 = L(2\alpha)\Delta T $

Expansion of the Second Rod:

$ \Delta L_2 = 2L(\alpha)\Delta T $

Total Expansion:

Combine the expansions:

$ \Delta L_{\text{total}} = 2L\alpha\Delta T + 2L\alpha\Delta T = 4L\alpha\Delta T $

Final Step:

Given that the total increase in length of the composite rod is the sum of the individual expansions, equate the change to find the overall coefficient of linear expansion, $\alpha_{\text{total}}$:

$ 4L\alpha\Delta T = 3L\alpha_{\text{total}}\Delta T $

Solving for $\alpha_{\text{total}}$:

$ \alpha_{\text{total}} = \frac{4}{3}\alpha $

Thus, the coefficient of linear expansion of the composite rod is $\frac{4}{3}\alpha$.

For a gas with a constant temperature, we examine the relationship between the volume ($V$) and pressure ($p$). Specifically, we are interested in the slope of a graph where $\log_e V$ is plotted on the X-axis, and $\log_e p$ is plotted on the Y-axis.

Given the ideal gas equation:

$ pV = nRT $

Taking the natural logarithm of both sides, we have:

$ \log_e(pV) = \log_e(nRT) $

This expands to:

$ \log_e p + \log_e V = \log_e(nRT) $

Since $nRT$ is a constant (as both the amount of gas and the temperature are constant), we can rearrange the equation as follows:

$ \log_e p = -\log_e V + \log_e(nRT) $

This equation is in the form of a straight line, where:

$ Y = mx + c $

By comparing:

$ \log_e p = (-1)\log_e V + \log_e(nRT) $

We identify that the slope ($m$) is $-1$.

Therefore, the slope of the graph where $\log_e V$ is on the X-axis and $\log_e p$ is on the Y-axis is $-1$.

Given:

Initial temperature of the ideal gas, $ T_0 = 127^{\circ} \mathrm{C} = 400 \, \mathrm{K} $

Initial volume, $ V_0 $

Final volume, $ V_1 = \frac{8}{27} V_0 $

Heat capacity ratio, $ \gamma = \frac{5}{3} $

The process described is adiabatic, meaning no heat is exchanged with the surroundings. For an adiabatic process involving an ideal gas, the relation $ T V^{\gamma-1} = \text{constant} $ holds.

Initial State:

$ T_0 V_0^{\gamma-1} = T_0 V_0^{5/3-1} \quad \text{(equation i)} $

Final State:

$ T_1 V_1^{\gamma-1} = T_1 \left(\frac{8}{27} V_0\right)^{5/3-1} \quad \text{(equation ii)} $

By dividing equation (i) by equation (ii), we have:

$ \frac{T_0}{T_1} = \frac{V_1^{\gamma-1}}{V_0^{\gamma-1}} = \left(\frac{8}{27}\right)^{5/3-1} = \left(\frac{8}{27}\right)^{2/3} $

This simplifies to:

$ \frac{T_0}{T_1} = \frac{4}{9} $

Solving for $ T_1 $:

$ T_1 = \frac{400 \times 9}{4} = 900 \, \mathrm{K} $

The rise in temperature is then calculated as:

$ T_1 - T_0 = 900 \, \mathrm{K} - 400 \, \mathrm{K} = 500 \, \mathrm{K} $

Given an insulating cylinder containing 4 moles of an ideal diatomic gas, when heat $ Q $ is supplied, 2 moles of the gas molecules dissociate. The temperature of the gas remains constant. We need to find the value of $ Q $.

Initially, the gas has 4 moles of an ideal diatomic gas. When supplied with heat, 2 moles dissociate into monoatomic gas, effectively converting 2 moles of diatomic gas into 4 moles of monoatomic gas.

Degrees of Freedom

Diatomic gas: $ f = 5 $

Monoatomic gas: $ f = 3 $

Internal Energy Calculation

The internal energy $ E $ for gas is given by:

$ E = n \times \frac{f}{2} \times RT $

Where $ n $ is the number of moles, $ f $ is the degree of freedom, $ R $ is the universal gas constant, and $ T $ is the temperature.

Initial Internal Energy $ E_i $

For 4 moles of diatomic gas:

$ E_i = 4 \times \frac{5}{2} \times RT = 10RT $

Final Internal Energy $ E_f $

After dissociation, we have:

2 moles of diatomic gas:

$ 2 \times \frac{5}{2} \times RT = 5RT $

4 moles of monoatomic gas:

$ 4 \times \frac{3}{2} \times RT = 6RT $

Total final internal energy:

$ E_f = 5RT + 6RT = 11RT $

Change in Internal Energy

The change in internal energy is:

$ \Delta E = E_f - E_i = 11RT - 10RT = RT $

Thus, the value of $ Q $ required to maintain constant temperature during the dissociation process is $ RT $.

Let's analyze the problem step by step.

For the object with mass $2m$:

Initial temperature: $T$

Final temperature: $2T$

Temperature change: $\Delta T = 2T - T = T$

The amount of heat required is given by:

$Q = \text{mass} \times c \times \Delta T = 2m \cdot c \cdot T$

So, we have:

$Q = 2m c T$

For the object with mass $m$:

Initial temperature: $2T$

Let the final temperature be $T_f$

Temperature change: $\Delta T = T_f - 2T$

The same heat $Q$ is supplied, so:

$Q = m \cdot c \cdot (T_f - 2T)$

Since both expressions for $Q$ are equal, set them equal to each other:

$2m c T = m c (T_f - 2T)$

Cancel the common factors $m$ and $c$ (assuming they are non-zero):

$2T = T_f - 2T$

Now, solve for $T_f$:

$ \begin{aligned} T_f &= 2T + 2T \\ &= 4T \end{aligned} $

Thus, when the same heat is supplied to the object of mass $m$, its temperature raises to $4T$.

The correct answer is Option C: $4T$.

To convert temperatures between the Celsius scale and the new X scale, we can assume a linear relationship. Here's how to determine the conversion:

Recognize the given points:

Melting point of ice: $0^\circ \text{C} = 20^\circ \text{X}$

Boiling point of water: $100^\circ \text{C} = 110^\circ \text{X}$

Express the relationship in the form:

$X = a + b \cdot C,$

where $a$ is the offset and $b$ is the scale factor.

Substitute the known temperatures:

For $0^\circ \text{C}:$

$20 = a + b \cdot 0 \quad \Rightarrow \quad a = 20.$

For $100^\circ \text{C}:$

$110 = 20 + b \cdot 100.$

Solve for $b$:

$110 - 20 = 100b \quad \Rightarrow \quad 90 = 100b \quad \Rightarrow \quad b = \frac{90}{100} = 0.9.$

Write the complete conversion formula:

$X = 20 + 0.9 \cdot C.$

Convert $40^\circ \text{C}$ to the new X scale:

$X = 20 + 0.9 \times 40 = 20 + 36 = 56.$

Therefore, on the new scale, $40^\circ \text{C}$ is indicated as $56^\circ \text{X}.$ This corresponds to Option B.

In an isobaric process, the pressure of the system remains constant. For such a process, the heat supplied to a diatomic ideal gas can be expressed as:

$ \Delta Q = n C_p \Delta T $

Where:

$ n $ represents the number of moles of the ideal gas.

$ C_p $ is the specific heat at constant pressure.

$ \Delta T $ is the change in temperature.

The work done during an isobaric process is calculated using:

$ \Delta W = p \Delta V $

Substituting the ideal gas law $ pV = nRT $ into the work done expression, we get:

$ \Delta W = p(V_2 - V_1) = nR(T_2 - T_1) = nR \Delta T $

The ratio of the work done $\Delta W$ to the heat supplied $\Delta Q$ is:

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

For a diatomic gas, the specific heat at constant pressure is:

$ C_p = \frac{7}{2} R $

Substituting this into the ratio, we find:

$ \frac{W}{Q} = \frac{R}{\frac{7}{2} R} = \frac{2}{7} $

Therefore, the percentage of the heat supplied that is converted into work is:

$ = \frac{2}{7} \times 100 \approx 28.6\% $

For any gas molecule, the translational degrees of freedom are determined by its ability to move in three-dimensional space. This means:

Translational degrees of freedom = 3 (movement along the x, y, and z axes).

For rotational degrees of freedom:

In a linear molecule (like a polyatomic linear gas molecule), rotation about the axis of the molecule doesn't contribute significantly because the moment of inertia about that axis is negligible.

Thus, only the rotations about the two axes perpendicular to the molecular axis matter.

So, rotational degrees of freedom = 2.

Now, the ratio of translational to rotational degrees of freedom is:

$\frac{3 \text{ (translational)}}{2 \text{ (rotational)}} = 3:2.$

Hence, the correct answer is:

Option D: $3:2$

Ring just slip over rod when area of hole is equals to area of cross-section of rod.

$\Rightarrow A_{\text {ring }}$ at $T_2$ temperature $=A_{\text {rod }}$ at $T_2$ temperature

$ \Rightarrow\left[A_0(1+\beta \Delta T)\right]_{\text {ring }}=\left[A_0(1+\beta \Delta T)\right]_{\text {rod }} $

Here,

$ \begin{aligned} & A_0(\text { ring })=9.98 \mathrm{~cm}^2 \\ & A_0(\text { rod })=10 \mathrm{~cm}^2 \\ & \beta_{\text {ring }}=2 \times 17 \times 10^{-6} \mathrm{C}^{-1} \quad(\because \beta=2 \alpha) \\ & \beta_{\text {rod }}=2 \times 11 \times 10^{-6} \mathrm{C}^{-1} \end{aligned} $

So, we have from Eq. (i),

$ \begin{aligned} & 9.98\left(1+2 \times 17 \times 10^{-6} \cdot \Delta T\right) \\ & \quad=10\left(1+2 \times 11 \times 10^{-6} \cdot \Delta T\right) \\ & \text { or } \quad 998+2 \times 998 \times 17 \times 10^{-6} \Delta T \\ & \quad=1000+2 \times 1000 \times 11 \times 10^{-6} \Delta T \\ & \Rightarrow \quad 2 \times 10^{-6} \Delta T(17 \times 998-1000 \times 11) \\ & \quad=1000-998 \\ & \Rightarrow \quad 2 \times 10^{-6} \Delta T(5966)=2 \\ & \Rightarrow \quad \Delta T=\frac{10^6}{5966}=167.61^{\circ} \mathrm{C} \end{aligned} $

Consider statement II, water is formed due to lowering of melting point by increase of pressure (not temperature), when skater skates on ice, increased pressure causes ice to melt and allows skater to move smoothly.

When pressure is released water formed freezes again.

So, statement II is incorrect but statements I and III are correct.

Specific heat of liquids is inversely proportional to the density.

Density of water is maximum at $4^{\circ} \mathrm{C}$.

So, its specific heat is minimum at $4{ }^{\circ} \mathrm{C}$.

Hence, option (c) is incorrect.

In cyclic process, $\Delta U=0$

So, by first law,

$ \begin{aligned} \Delta Q & =\Delta U+\Delta W \\or\,\,\, \Delta Q & =\Delta W \end{aligned} $

As cycle given is clockwise so, work done is positive. Hence, $\Delta Q$ is positive that means heat is absorbed.

Now, $\Delta Q=\Delta W=$ Area of cycle

$ \begin{aligned} & =\frac{1}{2} \times(2 V-V) \times(2 p-p) \\ & =\frac{1}{2} p V \end{aligned} $

Hence, heat released

$ \begin{aligned} & =- \text { Heat absorbed } \\ & =-\frac{1}{2} p V \end{aligned} $

Process is given isothermal, so process equation is $p_1 V_1=p_2 V_2$

$ \begin{array}{lc} \text { Given, } V_2=2 V_1 \\ \Rightarrow p_1 \times V_1=p_2 \times 2 V_1 \\ \Rightarrow p_2=p_1 / 2 \\ \text { but } p_2=p^{\prime} \text { and } p_1=p \\ \therefore p^{\prime}=p / 2 \end{array} $

So, new pressure is $1 / 2$ of initial pressure.

Change in length of combined rod $=$ Change in length of $\operatorname{rod} A+$ Change in length of rod B

$ \begin{aligned}So,\,\, \Delta l & =\Delta l_A+\Delta l_B \\ & =l_A \alpha_A \Delta T+l_B \alpha_B \Delta T \end{aligned} $

$ \begin{aligned}Here,\,\, \Delta T & =T_f-T_i=230-30=200^{\circ} \mathrm{C} \\ l_A & =l_B=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m} \\ \alpha_A & =2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} \text { and } \\ \alpha_B & =1 \times 10^{-50}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

$ \begin{aligned}So,\,\, \Delta l= & 50 \times 10^{-2} \times 2 \times 10^{-5} \times 200 \\ & +50 \times 10^{-2} \times 1 \times 10^{-5} \times 200 \\ = & 3 \times 10^{-3} \mathrm{~m}=3 \mathrm{~mm} \end{aligned} $

Given, $10 \%$ of total PE of water is used up in heating

$ \begin{array}{cc} \Rightarrow & \text { Heat absorbed }=10 \% \text { of PE } \\ \Rightarrow & m \times s \times \Delta T=\frac{10}{100} \times m g h \\ \Rightarrow & \Delta T=\frac{g \times h}{10 \times s} \\ \Rightarrow & \Delta T=\frac{10 \times 20}{10 \times 4000}=0.5 \times 10^{-2} \\ \Rightarrow & \Delta T=0.005^{\circ} \mathrm{C} \end{array} $

Hence, difference in temperature of water at top and bottom $=0.005^{\circ} \mathrm{C}$

Zeroth law of thermodynamics states about thermal equilibrium. i.e. when two systems are in thermal equilibrium with a third system, then both systems are in equilibrium also with each other.

Relation between volume and temperature in adiabatic expansion is

$ T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} $

Here, $T_2=\frac{T_1}{2}$ and $V_2=2 V_1$

$ \begin{aligned} \text { So, } & & T_1 V_1^{\gamma-1} & =\frac{T_1}{2}\left(2 V_1\right)^{\gamma-1} \\ \Rightarrow & & 1 & =\frac{1}{2} \cdot 2^{\gamma-1} \\ \Rightarrow & & 2 & =2^{\gamma-1} \\ \Rightarrow & & \gamma-1 & =1 \\ \text { or } & & \gamma & =2 \end{aligned} $

In an isobaric process,

$ \Delta Q=\Delta U+\Delta W \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Here, $\Delta Q=700 \mathrm{~J}$

Gas is given diatomic so, $C_p=\frac{7}{2} R$ and

$ C_V=\frac{5}{2} R $

Heat absorbed by gas is

$ \Delta Q=n C_p \Delta T $

$ \begin{aligned} & 700=n\left(\frac{7}{2} R\right) \cdot \Delta T \\ \Rightarrow & n \Delta T=\frac{700 \times 2}{7 R}=\frac{200}{R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

So, change in internal energy of gas,

$ \begin{aligned} \Delta U & =n C_V \Delta T=C_V n \Delta T \\ & =\frac{5}{2} R \times \frac{200}{R}=500 \mathrm{~J} \end{aligned} $

[from Eq. (ii)]

So, from Eq. (i), work done is

$ \begin{aligned} \Delta W & =\Delta Q-\Delta U \\ & =700-500 \\ & =200 \mathrm{~J} \end{aligned} $

Given, mass of $\mathrm{CO}_2, m=176 \mathrm{~g}$

Atomic mass of $\mathrm{CO}_2$,

$ M=12+16 \times 2=44 $

∴ Number of mole of $\mathrm{CO}_2$,

$ n=\frac{m}{M}=\frac{176}{44}=4 $

Change in temperature,

$ \Delta T=30-0=30^{\circ} \mathrm{C}=30 \mathrm{~K} $

Heat absorbed, $Q=3600 \mathrm{~J}$

We know that,

$ \begin{aligned} Q & =n s \Delta T \\ \Rightarrow \quad 3600 & =4 \times s \times 30 \\ \Rightarrow \quad s & =\frac{3600}{30 \times 4} \\ & =30 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned} $

Latent heat of water, $L_w=80 \mathrm{cal} / \mathrm{g}$ Latent heat of ether, $L_e=90 \mathrm{cal} / \mathrm{g}$ Let $m$ be the mass of ether required, ∴ Heat loss by water $=$ Heat gain by ether

$ \begin{aligned} \Rightarrow & & m_w L_w & =m . L_e \\ \Rightarrow & & 5 \times 80 & =m \times 90 \\ \Rightarrow & & m & =\frac{400}{90}=4.4 \mathrm{~g} \end{aligned} $

which is closest to option (b).

In thermodynamic system, heat and work are not state variables because they depend on the path of connecting states, thus, heat and work are treated as a path functions in thermodynamics. According to first law of thermodynamics,

$ \begin{aligned} & & Q & =\Delta U+W \\ \Rightarrow & & Q-W & =\Delta U \end{aligned} $

Thus, heat $Q$ and work $W$ are modes of energy transfer to a system resulting in change in its internal energy.

In isothermal expansion,

$ \begin{aligned} & p_1 V_1=p_2 V_2 \\ \Rightarrow & p_0 V_1=p_2 .8 V_1 \\ \Rightarrow & p_2=p_0 / 8 \end{aligned} \quad\left[\because V_2=8 V_1\right] $

Now, gas is slowly and adiabatically compressed back to its initial volume. Hence, for this adiabatic compression,

$ \begin{aligned} & p_1=p_0 / 8, V_1=8 \mathrm{~V} \\ & p_2=?, V_2=V \end{aligned} $

∴ Using adiabatic relation,

$ \begin{aligned} & & p_1 V_1^\gamma & =p_2 V_2^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot(8 V)^\gamma & =p_2(V)^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma \cdot V^\gamma & =p_2 V^{\prime} \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma & =p_2 \\ \Rightarrow & & p_2 & =\frac{p_0}{8} \cdot 8^\gamma \\ & & & =\frac{p_0}{8} \cdot(8)^{\frac{4}{3}} \left[\because \gamma=\frac{4}{3}\right]\\ & & & =\frac{p_0}{8} \cdot 16 \\ \Rightarrow & & p_2 & =2 p_0 \end{aligned} $

$ \begin{aligned} &\text { ∴ Average kinetic energy per molecule, }\\ &\begin{array}{ll} & K=\frac{3}{2} k T=\frac{3}{2} \frac{R}{N} T=\frac{3}{2} \frac{p V}{N} \,\,\,\,\,\,\,[\because p V=R T]\\ \Rightarrow & \quad K \propto p \\ \because & \frac{K_2}{K_1}=\frac{p_2}{p_1} \\ \Rightarrow & \frac{K_2}{K_1}=\frac{2 p_0}{p_0} \\ \Rightarrow & \frac{K_2}{K_1}=2 \end{array} \end{aligned} $

Given, atomic mass of Ne,

$ M_1=20.2 \mathrm{u} $

Atomic mass of He ,

$ \begin{aligned} M_2 & =4.0 \mathrm{u} \\ T_2 & =-33^{\circ} \mathrm{C} \\ & =(-33+273) \mathrm{K} \\ & =240 \mathrm{~K} \end{aligned} $

We know, $v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}$

$ \begin{array}{ll} \text { Since, }\left(V_{\text {rms }}\right) \mathrm{He} \\ \Rightarrow \sqrt{\frac{3 R T_2}{M_2}}=\sqrt{\frac{3 R T_1}{M_1}} \\ \Rightarrow \frac{T_2}{M_2}=\frac{T_1}{M_1} \\ \Rightarrow \frac{240}{4}=\frac{T_1}{20.2} \\ \Rightarrow T_1=\frac{20.2 \times 240}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, =1212 \mathrm{~K} \end{array} $

The volume of metal at $32^{\circ} \mathrm{C}$.

$ \begin{aligned} V_{32 \mathrm{C}} & =\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\ & =\frac{(49-39) g}{1.2 \times g}=8.33 \mathrm{~cm}^3 \end{aligned} $

Similarly, volume of metal at $42^{\circ} \mathrm{C}$.

$ V_{42 \mathrm{C}}=\frac{(49-40) \mathrm{g}}{1 \mathrm{~g}}=9 \mathrm{~cm}^3 $

Now, $V_{42 \mathrm{C}}=V_{32 \mathrm{C}}\left[1+\gamma\left(t_2-t_1\right)\right]$

$ \begin{aligned} \Rightarrow \quad \gamma & =\frac{V_{42{ }^{\circ} \mathrm{C}}-V_{322^{\circ} \mathrm{C}}}{V_{322^{\circ} \mathrm{C}}\left(t_2-t_1\right)}=\frac{9-8.33}{8.33(42-32)} \\ & =\frac{0.66}{83.3}=0.0079 /{ }^{\circ} \mathrm{C} \\ \alpha & =\frac{\gamma}{3}=\frac{0.0079}{3}=\frac{7.9}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \\ & \simeq \frac{8}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \end{aligned} $

Given, base area of cooking pot,

$ A=0.2 \mathrm{~m}^2 $

Thickness, $l=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$

Boiling rate, $R=\frac{m}{t}=3 \mathrm{~kg} / \mathrm{min}=\frac{1}{20} \mathrm{~kg} / \mathrm{s}$ Thermal conductivity,

$ K=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1} $

Heat of evaporation, $L=2 \times 10^6 \mathrm{~J} / \mathrm{kg}$

The amount of heat flowing into water through the base of cooking pot,

$ Q=\frac{K A\left(T_1-T_2\right) t}{l} $

where, $T_1=$ Temperature of the cooking pot and

$ T_2=\text { Boiling point of water } . $

Heat required to boil water, $Q^{\prime}=m L$

$ \begin{aligned} & \frac{K A\left(T_1-T_2\right)}{l} t=m L \\ & \begin{aligned} \Rightarrow T_1-T_2 & =\frac{m L l}{K A t}=\left(\frac{m}{t}\right) \frac{L l}{K A} \\ & =\frac{1}{20} \times \frac{2 \times 10^6 \times 2 \times 10^{-2}}{120 \times 0.2}=83^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

In given process

Pressure $\propto$ Volume $\Rightarrow p=k V$ work done in process,

$ \begin{aligned} \Delta W & =\int_{V_0}^{3 V_0} \rho d V=\int_{V_0}^{3 V_0} k V \cdot d V \\ & =k\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0} \\ & =\frac{k}{2}\left\{9 V_0^2-V_0^2\right\} \\ & =k \cdot 4 V_0^2=\frac{p_0}{V_0} \cdot 4 V_0^2 \end{aligned} $

$or\,\, \Delta W=4 p_0 V_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)$

As gas is monoatomic, $C_V=\frac{3}{2} R$

Change in internal energy of gas,

$ \Delta U=n C_v\left(T_f-T_i\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(ii)$

Now, from $p V=n R T$;

$ T_i=\frac{p_0 V_0}{n R} $

and $\quad T_f=\frac{9 p_0 V_0}{n R}$

So, from Eq. (ii)

$ \Delta U=n \cdot \frac{3}{2} R \cdot \frac{8 p_0 V_0}{n R}=12 p_0 V_0 $

By $\Delta Q=\Delta U+\Delta W$, we have Heat absorbed

$ \Delta Q=12 p_0 V_0+4 p_0 V_0=16 p_0 V_0 $

Given, $p_1=p, V_1=V, T_1=T$

$ p_2=p, V_2=4 V, T_2=\frac{T}{2} $

For the adiabatic relation of perfect gas,

$ \begin{aligned} & & T V^{\gamma-1} & =\text { constant } \\ & \Rightarrow & T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ & \Rightarrow & T V^{\gamma-1} & =\frac{T}{2}(4 V)^{\gamma-1} \end{aligned} $

$ \begin{aligned} \Rightarrow & & V^{\gamma-1} & =\frac{1}{2} \cdot(4)^{\gamma-1} \cdot V^{\gamma-1} \\ \Rightarrow & & 1 & =\frac{1}{2}(4)^{\gamma-1} \Rightarrow 2=(4)^{\gamma-1} \\ \Rightarrow & & 2 & =2^{2(\gamma-1)} \Rightarrow 2(\gamma-1)=1 \\ \Rightarrow & & \gamma-1 & =1 / 2 \\ \Rightarrow & & \gamma & =3 / 2 \end{aligned} $

$ \begin{aligned} &\text { Work done in adiabatic expansion, }\\ &\begin{aligned} W & =\frac{n R}{\gamma-1}\left[T_1-T_2\right] \\ & =\frac{n R}{\frac{3}{2}-1}[T-T / 2] \\ & =2 n R \times \frac{T}{2}=n R T \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad W=p V=\alpha p V \quad & \text { (given) } \\ & {[\because p V=n R T] } \end{aligned} $

$ \therefore \quad \alpha=1 $

$ \text { } \begin{aligned} \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\sqrt{\frac{3 R T}{M_{\mathrm{H}_2}}}=\sqrt{\frac{3 R \times 20}{2}} \\ & =\sqrt{30 R} \\ \left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} & =\sqrt{\frac{3 R T^{\prime}}{M_{\mathrm{O}_2}}}=\sqrt{\frac{3 R T^{\prime}}{32}} \end{aligned} $

According to question,

$ \begin{aligned} & & \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} \\ \Rightarrow & & \sqrt{30 R} & =\sqrt{\frac{3 R T^{\prime}}{32}} \Rightarrow 30 R=\frac{3 R T^{\prime}}{32} \\ \Rightarrow & & 30 \times 32 & =3 T^{\prime} \Rightarrow T^{\prime}=320 \mathrm{~K} \end{aligned} $

$ \begin{aligned} &\alpha=2 \times 10^{-5} \mathrm{~K}^{-1} \Rightarrow T=230^{\circ} \mathrm{C}\\ &\text { We know, } L=L_0[1+\alpha \Delta T]\\ &\begin{aligned} & =5\left[1+2 \times 10^{-5} \times(230-30)\right] \\ L & =5\left[1+\left(2 \times 10^{-5} \times 200\right)\right]=5.02 \mathrm{~cm} \end{aligned} \end{aligned} $

We know, $Q=m c \Delta T$

$ \begin{aligned} 2100 & =m \times 900 \times 2 \\ \Rightarrow \quad m & =\frac{2100}{1800}=1167 \mathrm{~kg} \end{aligned} $

which is nearest to option (a).

The given situation is shown below.

Work done in a cyclic process $=$ Area inside the cycle $A B C$

∴ Work done $=$ Area of $\triangle A B C$

$ \begin{aligned} & =\frac{1}{2} \times\left(2 V_1-V_1\right)\left(3 p_1-p_1\right) \\ & =\frac{1}{2} \times V_1 \times 2 p_1=p_1 V_1 \end{aligned} $

From 1st law of thermodynamics,

$ \begin{aligned} & \Delta Q=p \Delta V+n C_V \Delta T \\ = & p \Delta V+n \frac{R}{\gamma-1} \Delta T=p \Delta V\left[1+\frac{1}{\gamma-1}\right] \\ \Rightarrow & \Delta Q=200\left[1+\frac{1}{\gamma-1}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

We know, $\frac{C_p}{C_V}=\gamma$

$ \begin{aligned} & \Rightarrow \quad \frac{C_p}{C_p-R}=\gamma \quad\left(\because C_p-C_V=R\right) \\ & \Rightarrow \quad \gamma=\frac{\frac{7}{2} R}{\frac{7}{2} R-R}=\frac{\frac{7}{5}}{\frac{5}{2}}=\frac{7}{5} \end{aligned} $

Put $\gamma=\frac{7}{5}$ is Eq. (i)

$ \begin{array}{ll} \Delta Q & =200\left[1+\frac{1}{\left(\frac{7}{5}-1\right)}\right] \\ \Rightarrow & \Delta Q=200\left[1+\frac{5}{2}\right] \\ \Rightarrow & \Delta Q=200 \times \frac{7}{2}=700 \mathrm{~J} \end{array} $

Statement I is false because gas thermometers are more sensitive than liquid thermometers. This is because gas expands more than a liquid.

Statement II is true because $R / N_A=k_B$ where, $k_B=$ Boltzmann Constant

$R=$ Universal gas constant

$N_A=$ Avogadro number

Statement III is true because

$ \begin{array}{r} \text { Density }=\frac{\text { mass }}{\text { volume }} \\   d=m / V ...(i)\end{array} $

Now, by ideal gas equation

$ p V=n R T $

$p V=\frac{m}{M} R T$, where $M$ is the molar mass of gas.

$ \begin{aligned} & \therefore \quad m=\frac{p V M}{R T} \\ & \text { Put } m=\frac{p V M}{R T} \text { in Eq. (i) } \\ & \qquad d=\frac{p V M}{R T V} \Rightarrow d=\frac{p M}{R T} \\ & \Rightarrow \quad d \propto \frac{1}{T} \end{aligned} $

$ \begin{gathered} \text { } \Delta L_{\text {steel }}=L_S \alpha_S \Delta t \\ \Delta L_{\text {copper }}=L_C \alpha_C \Delta t \end{gathered} $

Given, $L_S-L_C=4 \mathrm{~cm}$ for any temperature

$ \begin{array}{ll} \therefore & L_S \alpha_S \Delta t=L_C \alpha_C \Delta t \\ & \frac{L_S}{L_C}=\frac{\alpha_C}{\alpha_S}=\frac{1.7 \times 10^{-5}}{1.1 \times 10^{-5}}=\frac{1.7}{1.1}=\frac{17}{11} \end{array} $

By Newton's law of cooling,

$ \frac{-d T}{d t}=b\left[T-T_0\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $\frac{-d T}{d t}=$ Cooling rate,

$T=$ Body's temperature,

$T_0=$ Surrounding temperature and $b=$ Constant.

∴ From Eq. (i),

$ \begin{aligned} & \int_{T_i}^T \frac{d T}{T-T_0}=-b \int_0^t d t \\ \Rightarrow \quad & T=T_0+\left(T_i-T_0\right) e^{-b t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From the first condition of question,

$ \begin{aligned} T & =40^{\circ} \mathrm{C} \\ T_0 & =10^{\circ} \mathrm{C} \\ T_i & =100^{\circ} \mathrm{C} \\ t & =10 \mathrm{~min} \end{aligned} $

$ \text { Putting these value in Eq. (ii), } $

$ \begin{array}{ll} & 40=10+(100-10) e^{-b \times 10} \\ \Rightarrow & \frac{30}{90}=e^{-10 b} \\ \Rightarrow & \frac{1}{3}=e^{-10 b} \\ \Rightarrow & \ln \left(\frac{1}{3}\right)=-10 b \ln m \\ \Rightarrow & \ln 1-\ln 3=-10 b \\ \Rightarrow & -\ln 3=-10 b \\ \Rightarrow & b=\frac{\ln 3}{10}=\frac{11}{10} \end{array} $

Now, by next condition of question,

$ \begin{aligned} & T=20^{\circ} \mathrm{C} \\ & T_i=70^{\circ} \mathrm{C} \\ & T_a=10^{\circ} \mathrm{C} \end{aligned} $

From Eq. (ii), $20=10+(70-10) e^{-b t}$

$ \begin{gathered} \frac{10}{60}=e^{\frac{1.1}{10} t} \\ \Rightarrow \quad \ln \left(\frac{1}{6}\right)=-\frac{1.1}{6} t \ln e \\ \Rightarrow \ln 1-\ln 6=-\frac{1.1}{10} t \Rightarrow-\ln 6=\frac{-11}{100} t \\ \Rightarrow \quad t=\frac{100 \times \ln 6}{11}=\frac{100 \times 1.8}{11}=16.3 \mathrm{~min} \end{gathered} $

We know, from first law of thermodynamics,

$ \Delta Q=\Delta U+W \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Given, volume of liquid, $V_l=10^{-3} \mathrm{~m}^3$

And volume of gas or volume of steam,

$ V_g=2001 \mathrm{~m}^3 $

Now, $V_g \gg V_l$

$ \begin{aligned} & \therefore \quad\left(V_g-V_l\right) \approx V_g \\ & W=p V_g=\left(10^5 \times 2.001\right) \mathrm{J}=2.001 \times 10^2 \mathrm{~kJ} \\ & =200 \mathrm{~kJ} \end{aligned} $

Now, heat required for phase change to convert the liquid into steam,

$ \Delta Q=m L=1 \times 2000 \mathrm{~kJ} $

Put in Eq (i), $\Delta U=\Delta Q-W$

$ \Rightarrow \quad \Delta U=(2000-200) \mathrm{kJ} $

[from Eq. (i)]

$ \Rightarrow \quad \Delta U=1800 \mathrm{~kJ} $

Given, $W=100 \mathrm{~J}$

From ideal gas equation,

$ \begin{aligned} & & p \Delta V & =n R \Delta T\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \because & W & =p \Delta V \end{aligned} $

∴ Eq. (i) becomes,

$ W=n R \Delta T \Rightarrow \Delta T=\frac{W}{n R}=\frac{100}{n R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, we know the relation for change in internal energy,

$ \begin{aligned} & \Delta U=n C_V \Delta T \\ & \Rightarrow \quad \Delta U=n \frac{R}{\gamma-1} \Delta T \quad \ldots(\text { iii }) \\ &\left(\because C_V=\frac{R}{\gamma-1}\right) \end{aligned} $

and since it is a monoatomic gas,

$ \therefore \quad \gamma=\frac{5}{3} $

From Eqs. (ii) and (iii)

$ \begin{array}{lc} \Rightarrow & \Delta U=150 \mathrm{~J} \\ \Rightarrow & \Delta U=\frac{n R \Delta T}{\frac{5}{3}-1}=\frac{3 n R}{2} \times \frac{100}{n R}=150 \mathrm{~J} \end{array} $

Now, the heat given to the gas in the process is

$ \begin{aligned} & & \Delta Q & =\Delta U+W \\ \Rightarrow & & \Delta Q & =150+100=250 \mathrm{~J} \\ \Rightarrow & & \Delta Q & =250 \mathrm{~J} \end{aligned} $