Heat and Thermodynamics

For process $\mathrm{A}$

$\begin{aligned} & \log P=\gamma \log \mathrm{V} \Rightarrow \mathrm{P}=\mathrm{V}^\gamma,(\gamma>1) \\ & P V^{-\gamma}=\text { Constant } \end{aligned}$

$C_A=C_V+\frac{R}{1+\gamma}$ ..... (i)

Likewise for process $\mathrm{B} \rightarrow P V^{-1}=$ Constant

$\begin{aligned} & C_B=C_v+\frac{R}{1+1} \\ & C_B=C_v+\frac{R}{2} \quad \text{..... (ii)} \\ & C_P=C_v+R \quad \text{..... (iii)} \end{aligned}$

By (i), (ii) & (iii)

$C_P>C_B>C_A>C_v$ [No answer matching]

$\begin{aligned} & \sqrt{\frac{3 R T}{2}}=\sqrt{\frac{3 R(320)}{32}} \\ & T=\frac{320}{16}=20 \mathrm{~K} \end{aligned}$

$\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT} \\ & \mathrm{N}=\text { Total no. of molecules } \\ & \mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT} \\ & 1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T} \\ & 1.01 \times 10^5=2 \times 10^2 \times \mathrm{T} \\ & \mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K} \end{aligned}$

$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}$

For diatomic gas $\mathrm{f}_{\mathrm{eq}}=5$

$\begin{aligned} & 5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2} \\ & 5 \mathrm{~N}+10=6 \mathrm{~N}+6 \\ & \mathrm{~N}=4 \end{aligned}$

Work done $\mathrm{AB}=\frac{1}{2}(8000+4000)$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3=\left(6000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Work done $\mathrm{BC}=-\left(4000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Total work done $=2000$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3$

$\begin{aligned} =2 \times 10^3 & \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}} R T_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}} \mathrm{RT}_{\mathrm{B}}} \\ & \text { Given } \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \\ & \text { And } \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{1 / 2}{1 / 32}=16 \end{aligned}$

For an adiabatic process, the following relation holds:

$P V^{\gamma} = \text{constant}$

where P is the pressure, V is the volume, and $\gamma = \frac{C_p}{C_v}$.

We are given that the pressure is proportional to the cube of the absolute temperature:

$P \propto T^3$.

Using the ideal gas law, $PV = nRT$, we can rewrite this as:

$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$.

Substituting this into the adiabatic relation, we get:

$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$.

Simplifying, we have:

$P^{1-\gamma} T^{2\gamma} = \text{constant}$.

Since P is proportional to $T^3$, we can write:

$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$.

This simplifies to:

$T^{3-3\gamma + 2\gamma} = \text{constant}$.

For this equation to hold, the exponent of T must be zero. Therefore:

$3 - 3\gamma + 2\gamma = 0$.

Solving for $\gamma$, we get:

$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$.

Therefore, the correct answer is Option B.

$\begin{aligned} & {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \\ & \text { And }[\mathrm{V}]=[\mathrm{b}] \\ & \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}] \end{aligned}$

The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom. For a diatomic molecule such as oxygen ($O_2$), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of $\frac{1}{2} kT$ per molecule, where $k$ is the Boltzmann constant and $T$ is the temperature in kelvins.

However, since we are dealing with moles, we'll use the universal gas constant $R$ instead of the Boltzmann constant $k$, because $R = k \cdot N_A$ where $N_A$ is the Avogadro constant (the number of molecules in a mole). Therefore, the average energy per mole for each degree of freedom is $\frac{1}{2}RT$.

To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:

For diatomic oxygen:

Given that temperature $ T $ is $ 27^{\circ} \mathrm{C} $, we first convert it to kelvins:

Now we plug in the values for $ R $ and $ T_{\text{K}} $:

When we calculate this, we find:

So the total kinetic energy of 1 mole of oxygen at $27^{\circ} \mathrm{C}$ is approximately 6232.5 J.

To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:

$\Delta U = m c_v \Delta T$

Where:

• $\Delta U$ is the change in internal energy,
• $m$ is the mass of the substance (in this case, air),
• $c_v$ is the specific heat at constant volume,
• $\Delta T$ is the change in temperature.

Given that:

• The mass of air, $m = 0.08 \, \text{kg},$
• The specific heat of air at constant volume, $c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C},$
• The change in temperature, $\Delta T = 5^{\circ} \text{C},$
• And the conversion factor from calories to Joules, $1 \, \text{cal} = 4.18 \, \text{J}.$

First, convert the specific heat from kcal to Joules:

$c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C} \times 1000 \, \text{cal/kcal} \times 4.18 \, \text{J/cal} = 710.6 \, \text{J/kg}^{\circ}\text{C}$

Now, substitute the values into the formula:

$\Delta U = 0.08 \times 710.6 \times 5$

$\Delta U = 284 \, \text{J}$

Thus, the change in internal energy is approximately 284 Joules.

To find the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$, we use the equation for the average kinetic energy of a molecule in terms of temperature:

$K_{\text{avg}} = \frac{3}{2} k_B T$

Where:

• $K_{\text{avg}}$ is the average kinetic energy
• $k_B$ is the Boltzmann constant, given as $1.38 \times 10^{-23} \, \text{J/K}$
• $T$ is the temperature in Kelvin

First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, so:

$K_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Substituting $K_{\text{avg}}$ and $k_B$ into the equation:

$\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Solving for $T$:

$T = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}$

After performing the calculations:

$T \approx 3200 \, \text{K}$

Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$ is approximately 3200 K.

Steeper curve (B) is adiabatic

Adiabatic $\Rightarrow \mathrm{PV}^v=$ const.

Or $\mathrm{P}\left(\frac{\mathrm{T}}{\mathrm{P}}\right)^v=$ const.

$\frac{\mathrm{T}^v}{\mathrm{P}^{v-1}}=\text { const. }$

Curve (A) is isothermal

$\mathrm{T}=$ const.

$\mathrm{PV}=$ const.

$\begin{aligned} & \mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right) \\ & K\left(P_0, 3 V_0, 3 T_0\right) \\ & \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \\ & \mathrm{L}\left(2 \mathrm{P}_{\circ}, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \\ & \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \\ & \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \\ & \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \\ & \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ell \mathrm{n}\left(\frac{1}{2}\right)=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\ & \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\ & \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT}_0 \\ & \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT}_0 \ln 2 ; \Delta \mathrm{U}=0 \\ & \mathrm{WD}_{\text {nes }}=-2 \mathrm{nRT}_0 \ln 2 \\ & \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2 \\ & \end{aligned}$

Temperature at which the clock loses time, $ T=38^{\circ} \mathrm{C} $

Temperature at which the clock gains

time, $ T_{2}=18^{\circ} \mathrm{C} $

$ \begin{aligned} \Delta T & =T_{1}-T_{2} \\\\ & =(38-18)^{\circ} \mathrm{C}=20^{\circ} \mathrm{C} \end{aligned} $

Time lost per day, $ \Delta t_{1}=-10.8 \mathrm{~s} $

Time gained per day, $ \Delta t_{2}=10.8 \mathrm{~s} $

The total change in time per day,

$ \Delta t=\Delta t_{1}-\Delta t_{2}=-10.8-10.8=-21.6 \mathrm{~s} $

Time period of a pendulum,

$ T=2 \pi \sqrt{\frac{L}{g}} $

The change in length $ \Delta L $ due to a temperature change $ \Delta T $ is given by

$ \Delta L=L \alpha \Delta T $

where, $ \alpha= $ coefficient of linear expansion

So, $ \quad \frac{\Delta L}{L}=\alpha \Delta T $

The fractional change in the time period is approximately half the fractional change in length.

$ \begin{array}{l} \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta L}{L}=\frac{1}{2} \alpha \Delta T \\\\ \frac{\Delta t_{\text {day }}}{T_{\text {day }}}=\frac{-21.6}{86400} \end{array} $

Putting value in Eq. (ii) from Eq. (i) and Eq. (iii), we get

$ \begin{aligned} \frac{-21.6}{86400} & =\frac{1}{2} \times \alpha \times 20 \\\\ \alpha & =-2.5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

Negative sign indicated the clock runs slower at higher temperature.

Newton's law of cooling says that,

$ \begin{array}{l} \frac{\theta_{1}-\theta_{2}}{t} \propto\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right) \\\\ \frac{\theta_{1}-\theta_{2}}{t}=k\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right) \end{array} $

where, $ \theta_{1}= $ Initial temperature, $ \theta_{2}= $ Final temperature,

$ \theta_{0}= $ Room temperature, $ t= $ time and $ k= $ constant that depends on the characteristics of the liquid and the environment.

For first cooling period

$ \begin{array}{l} \frac{368-358}{22}=k\left(\frac{368+358}{2}-\theta_{0}\right) \\\\ \frac{10}{22}=k\left(363-\theta_{0}\right) \Rightarrow k=\frac{10}{22\left(363-\theta_{0}\right)} \ldots(\mathrm{i}) \\\\ \text { For second cooling period } \\\\ \frac{358-353}{12.5}=k\left(\frac{358+353}{2}-\theta_{0}\right) \\\\ \frac{5}{12.5}=k\left(355.5-\theta_{0}\right) \Rightarrow k=\frac{5}{12.5\left(355.5-\theta_{0}\right)} \end{array} $

On comparing Eq. (i) with Eq. (ii), we get

$ \begin{aligned} \frac{10}{22\left(363-\theta_{0}\right)} & =\frac{5}{12.5\left(355.5-\theta_{0}\right)} \\\\ 25\left(355.5-\theta_{0}\right) & =22\left(363-\theta_{0}\right) \\\\ 901.5 & =3 \theta_{0} \Rightarrow \theta_{0}=300.5 \mathrm{~K} \\\\ \Rightarrow \quad \theta_{0} & =300.5-273 \\\\ \theta_{0} & =27.5^{\circ} \mathrm{C} \end{aligned} $

Given, $ U=3+1.5 \mathrm{pV} $

At constant volume, the change in internal energy is given by the change in temperature

$ d U=C_{V} d T \Rightarrow \frac{d U}{d T}=C_{V} $

We know that, $ p V=n R T $

So, $ U=3+1.5 n R T $

Putting value in Eq. (ii), we get

$ \begin{array}{l} C_{V}=\frac{d}{d T}(3+1.5 n R T) \\\\ C_{V}=1.5 n R \end{array} $

For $ C_{p} $, we use the relation

$ \begin{array}{l} C_{p}=C_{V}+n R \\\\ C_{p}=1.5 n R+n R=2.5 n R \end{array} $

The ratio of the specific heat capacities of the gas at constant volume and constant pressure is

$ \begin{array}{l} \frac{C_{V}}{C_{p}}=\frac{1.5 n R}{2.5 n R} \\\\ \frac{C_{V}}{C_{p}}=\frac{15}{25} \\\\ \frac{C_{V}}{C_{p}}=\frac{3}{5} \end{array} $

We know that,

$ p V=n R T $

Given, Initial pressure, temperature and mass of a gas are $ p, 127^{\circ} \mathrm{C} $ and 21 g respectively.

$ T_{1}=127^{\circ} \mathrm{C}=400 \mathrm{~K} $

Let $ M $ be the molar mass of the gas, then

$ p V=\left(\frac{21}{m}\right) \times R \times 400 $

After the gas leakes out,

New pressure $ =\frac{2 p}{3} $, New temperature

$ =T_{2} $

Mass of gas remaining $ =(21-5) \mathrm{g}=16 \mathrm{~g} $

Now using ideal gas equation,

$ \frac{2 p V}{3}=\left(\frac{16}{m}\right) \times R \times T_{2} $

On solving Eq. (i) and Eq. (ii), we get

$ \begin{array}{l} \frac{21}{m} \times R \times 400=\frac{16}{m} \times R \times T_{2} \times \frac{3}{2} \\\\ \frac{21 \times 400 \times 2}{16 \times 3}=T_{2} \\\\ T_{2}=350 \mathrm{~K} \\\\ T_{2}=350-273 \\\\ T_{2}=77^{\circ} \mathrm{C} \end{array} $

To determine the ratio of the masses of steam and water in equilibrium, we begin with the given data:

Mass of steam, $ m_1 = 60 \, \text{g} $

Temperature of steam, $ \theta_1 = 100^{\circ} \text{C} $

Mass of water, $ m_2 = 360 \, \text{g} $

Temperature of water, $ \theta_2 = 40^{\circ} \text{C} $

Latent heat of steam, $ L = 540 \, \text{cal/g} $

Specific heat of water = $ 1 \, \text{cal/g}^{\circ} \text{C} $

According to the principle of calorimetry, the heat lost by steam equals the heat gained by water:

Heat gain by water:

Using the formula $ Q = m s \Delta T $, where $ m $ is the mass, $ s $ is the specific heat, and $ \Delta T $ is the change in temperature:

$ Q = 360 \times 1 \times (100 - 40) = 21600 \, \text{cal} $

Heat lost by steam:

First, consider the latent heat: $ Q = mL = 60 \times 540 = 32400 \, \text{cal} $

Calculate the mass of steam $ x $ that would release 21600 calories of energy:

$ \begin{align*} 1 \, \text{g of steam} & = 540 \, \text{cal} \\ x \, \text{g of steam} & = 21600 \, \text{cal} \\ x & = \frac{21600}{540} = 40 \, \text{g} \end{align*} $

Thus, 40 grams of steam converted to water, releasing 21600 calories of energy.

Calculate the total mass of water after the steam condenses:

Total amount of water = initial water + condensed steam:

$ 360 \, \text{g} + 40 \, \text{g} = 400 \, \text{g} $

Calculate the remaining steam:

Steam left = initial steam - condensed steam:

$ 60 \, \text{g} - 40 \, \text{g} = 20 \, \text{g} $

Determine the ratio of steam to water:

Ratio of steam to water:

$ \frac{20}{400} = \frac{1}{20} = 1:20 $

Thus, the ratio of the masses of steam to water in equilibrium is $ 1:20 $.

Given, ratio of flow of heat (ratio) $ =\frac{5}{9} $

Ratio of radii of cylinder $ A $ and $ B=\frac{1}{2} $ We know that rate flow of heat is given as

$ \frac{d Q}{d t}=-k A \frac{d \theta}{l_{1}} $

[ $ l= $ length of conductor]

[ $ k= $ conductivity constant]

[ $ d \theta= $ temperature difference]

$ \left(\frac{d Q}{d t}\right)_{A}=-k d \theta \frac{A_{1}}{1} $

$ \begin{aligned} \frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}} & =\frac{A_{1}}{l_{1}} \times \frac{l_{2}}{A_{2}} \times \frac{\Delta \theta_{1}}{\Delta \theta_{2}}[k, \text { is constant }] \\\\ & =\frac{\pi r_{1}^{2}}{l_{1}} \times \frac{l_{2}}{\pi r_{2}^{2}} \times \frac{\Delta \theta_{1}}{\Delta \theta_{2}} \\\\ \frac{5}{9} & =\left(\frac{r_{1}}{r_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}} \times \frac{2}{3} \\\\ \frac{5}{9} & =\frac{1}{4} \times \frac{2}{3} \times \frac{l_{2}}{l_{1}} \\\\ \frac{l_{1}}{l_{2}} & =\frac{3}{10} \\\\ l_{1}: l_{2} & =3: 10 \end{aligned} $

According to the first law of thermodynamics:

$ dQ = dU + dW $

where $ dQ $ is the heat supplied, $ dU $ is the change in internal energy, and $ dW $ is the work done by the system.

For a Monoatomic Gas:

The heat supplied $ Q_1 $ can be expressed as:

$ Q_1 = n C_V dT + W $

Here, the change in internal energy $ dU $ is given by:

$ dU = n C_V dT = n \left(\frac{3}{2}\right) R dT $

Substituting this into the equation for $ Q_1 $:

$ Q_1 = \frac{3}{2} n R dT + W $

Since $ n R dT = W $ (since $ PV = nRT $, therefore work done $ W = PV $):

$ Q_1 = \frac{3}{2} W + W $

Thus:

$ Q_1 = \frac{5W}{2} $

For a Diatomic Gas:

The heat supplied $ Q_2 $ is:

$ Q_2 = n \left(\frac{5}{2}\right) R dT + 2W $

Substituting for $ n R dT = 2W $:

$ Q_2 = \frac{5}{2} \times 2W + 2W $

This simplifies to:

$ Q_2 = 7W $

Ratio of $ Q_1 $ to $ Q_2 $:

To find the ratio $ Q_1 : Q_2 $:

$ \frac{Q_1}{Q_2} = \frac{\frac{5W}{2}}{7W} = \frac{5}{14} $

Thus, the ratio $ Q_1 : Q_2 $ is:

$ 5 : 14 $

To solve this problem, we need to find the temperature at which the root mean square (rms) speed of oxygen molecules is 75% of the rms speed of nitrogen molecules at a given temperature of $287^{\circ} \mathrm{C}$.

Given:

The rms speed of oxygen is 75% of the rms speed of nitrogen.

Temperature of nitrogen ($T_{\text{nitrogen}}$) = $287^{\circ} \mathrm{C}$.

First, convert this nitrogen temperature from Celsius to Kelvin:

$ T_{\text{nitrogen}} = 287 + 273 = 560\, \text{K} $

Formula for rms speed:

The rms speed ($v_{\text{rms}}$) is given by:

$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $

Where:

$R$ is the gas constant,

$T$ is the temperature in Kelvin,

$M$ is the molar mass of the gas.

For oxygen ($M_{\text{oxygen}} = 32 \, \text{g/mol}$) and nitrogen ($M_{\text{nitrogen}} = 28 \, \text{g/mol}$), we establish the following relationship by squaring both sides and canceling $R$:

$ \sqrt{\frac{3RT_{\text{oxygen}}}{32}} = \frac{3}{4} \sqrt{\frac{3RT_{\text{nitrogen}}}{28}} $

Simplifying:

Square both sides to eliminate the square roots:

$ \frac{3T_{\text{oxygen}}}{32} = \frac{9}{16} \times \frac{3T_{\text{nitrogen}}}{28} $

Cancel out the constants and solve for $T_{\text{oxygen}}$:

$ T_{\text{oxygen}} = \frac{9}{16} \times \frac{560 \times 32}{28} $

Calculate $T_{\text{oxygen}}$:

$ T_{\text{oxygen}} = 360 \, \text{K} $

Convert back to Celsius:

$ T_{\text{oxygen}} = 360 - 273 = 87^{\circ} \mathrm{C} $

Thus, the temperature at which the rms speed of oxygen achieves 75% of the rms speed of nitrogen molecules at $287^{\circ} \mathrm{C}$ is $87^{\circ} \mathrm{C}$.

When a large liquid drop splits into $n$ similar smaller drops under isothermal conditions, the following process occurs:

Volume Consistency: The total volume remains the same before and after the split. This is due to the conservation of mass principle, where the sum of the volumes of all the small drops equals the initial volume of the large drop.

Surface Area Increase: The surface area increases. The surface area $A$ of a sphere is given by $A = 4\pi r^2$, where $r$ is the radius. If a large drop with radius $R$ has a volume $V$ given by $V = \frac{4}{3}\pi R^3$, splitting into $n$ smaller drops each with radius $r$ retains the total volume as:

$ n \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 $

Solving this gives:

$ n r^3 = R^3 \Rightarrow r = R \left(\frac{1}{n}\right)^{\frac{1}{3}} $

Each small drop has a surface area $4\pi r^2$, so the total surface area for $n$ small drops is:

$ n \times 4\pi r^2 = n \times 4\pi \left(R \left(\frac{1}{n}\right)^{\frac{1}{3}}\right)^2 = 4\pi R^2 \times n^{\frac{1}{3}} $

Since $n^{\frac{1}{3}} > 1$ for $n > 1$, the total surface area indeed increases when a large drop splits into smaller drops.

Energy Changes: The increase in surface area results in an increase in surface energy. According to thermodynamic principles, forming a new surface requires energy due to surface tension. Therefore, energy must be absorbed to create the additional surfaces of the new drops.

Consequently:

Option A: Incorrect, since the total volume remains constant.

Option B: Incorrect, since the total surface area increases.

Option C: Correct, energy is absorbed to increase the total surface area.

Option D: Incorrect, energy is not liberated; it is absorbed.

Thus, the correct answer is Option C: energy is absorbed.

water mass $ m_{1}=74 \mathrm{~g} $

$ \theta_{1}=70^{\circ} \mathrm{C} $

Ice mass, $ m_{2}=37 \mathrm{~g} $

$ \theta_{2}=0^{\circ} \mathrm{C} $

Ice latent heat of fusion, $ L=80 \mathrm{cal} / \mathrm{g} $

Specific heat capacity of water

$ s=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C} $

Now using principle of calorimetry

Heat lost by water is equal to heat gain by ice. Let $ \theta $ be the common temperature.

$ \begin{aligned} m_{1} s\left(\theta_{1}-\theta\right) & =m_{2} L+m_{2} s\left(\theta-\theta_{2}\right) \\\\ 74 \times 1 \times[70-\theta] & =37[80+1(\theta-0)] \\\\ 2[70-\theta] & =[80+\theta] \\\\ 140-2 \theta & =80+\theta \\\\ 140-80 & =3 \theta \Rightarrow 60=3 \theta \\\\ \theta & =20^{\circ} \mathrm{C} \end{aligned} $

Given,

Metal plate thickness,

$ d x=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} $

Area of plate, $ A=5 \mathrm{~cm}^{2}=5 \times 10^{-4} \mathrm{~m}^{2} $

Temperature difference,

$ d T=14^{\circ} \mathrm{C}=14 \mathrm{~K} $

Rate of thermal flow $ =42 \mathrm{~J} $

We know that equation for rate of thermal flow is

$ \begin{aligned} \frac{d Q}{d t} & =-K A \frac{d T}{d x} \\\\ \frac{d Q}{d T} & =-42 \mathrm{~J} \quad[\text { heat flows out }] \\\\ -42 & =\frac{-K \times 5 \times 10^{-4} \times 14}{5 \times 10^{-3}} \\\\ 42 & =1.4 K \\\\ K & =30 \mathrm{~W} / \mathrm{m} / \mathrm{K} \end{aligned} $

Given, ratio of $ \frac{C_{p}}{C_{V}} $ or $ \gamma=1.5 $

For adiabatic process,

Let initial pressure $ p_{a} $, Final pressure $ =p_{1} $ Initial volume $ =V $, Final volume $ =2 V $

Using adiabatic equation

$ \begin{aligned} p V^{\gamma} & =\text { constant } \\\\ p_{a} V_{1}^{\gamma} & =p_{1} V_{2}^{\gamma} \\\\ \text { (Initial) } & \text { (Final) } \\\\ \Rightarrow \quad & p_{a} \end{aligned}=p_{1}\left[\frac{2 V}{V}\right]^{\gamma} \text {, } $

For isothermal process, final pressure $ =p_{2} $ Using isothermal equation

$ \begin{aligned} p V & =\text { constant } \\\\ p_{\text {iso }} V_{1} & =p_{2} V_{2} \\\\ \text { (Initial) } & =(\text { Final) } \end{aligned} \quad \begin{aligned} p_{\text {iso }} & =p_{2}\left[\frac{2 V}{V}\right] \\\\ \Rightarrow \quad p_{2} & =\frac{p_{\text {iso }}}{2} \end{aligned} $

According to question,

$ \begin{aligned} p_{1} & =p_{2} \\\\ \frac{p_{a}}{[2]^{1.5}} & =\frac{p_{\text {iso }}}{2} \\\\ \frac{p_{a}}{p_{\text {iso }}} & =\frac{2 \sqrt{2}}{2} \quad\left[\text { as } a^{1.5}=a \cdot \sqrt{a}\right] \\\\ p_{a}: p_{\text {iso }} & =\sqrt{2}: 1 \end{aligned} $

The average translational kinetic energy (per molecule) of any ideal gas (diatomic or monoatomic or polyatomic) is always $ \frac{3}{2} k_{B} T $.

It depends only on temperature and is independent of nature of gas.

Hence, the ratio of two gases is $ 1: 1 $.

Given, mass of ice at $-20^{\circ} \mathrm{C}$,

$ m_{\mathrm{ice}}=54 \mathrm{~g} $

Mass of steam at $100^{\circ} \mathrm{C}, m_{\text {meam }}=25 \mathrm{~g}$

Specific heat capacity of ice,

$ s_{\text {iee }}=0.5 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C} $

Heat required to bring ice at $0^{\circ} \mathrm{C}$

$ \begin{aligned} & =m_{\text {ce }} s_{\text {cle }} \Delta T \\\\ q_1 & =54 \times 0.5 \times(20)=540 \mathrm{cal} \end{aligned} $

Heat required to melt ice at $0^{\circ} \mathrm{C}$

$ \begin{aligned} & =m_{\mathrm{ice}} \times L_{\mathrm{ice}} \\\\ & q_2=54 \times 80=4320 \mathrm{cal} \end{aligned} $

Condense steam at $100^{\circ} \mathrm{C}$ to water at $100^{\circ} \mathrm{C}$.

$ \begin{aligned} q_3 & =m_{\text {steam }} \times L_{\text {steam }}=25 \times 540 \\\\ & =13500 \mathrm{cal} \end{aligned} $

To cold water from $100^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$,

$ \begin{aligned} q_4 & =m_{\text {water }} \times s_{\text {water }} \times \Delta T \\\\ & =25 \times 1 \times 100=2500 \mathrm{cal} \end{aligned} $

$\therefore$ Total heat gained by ice,

$ Q_{\text {gain }}=q_1+q_2=4860 \mathrm{cal} $

Heat lost by steam, $Q_{\text {lott }}=q_3=13500 \mathrm{cal}$ Heat gained by ice to melt will be provided by steam i.e, $\frac{4860}{540}=9 \mathrm{~g}$

Thus, only 9 g steam is needed to completely melt the ice,

Hence, $(54+9) \mathrm{g}$ of water will be present along with $(25-9) \mathrm{g}$ of steam at $100^{\circ} \mathrm{C}$.

Lets calculate the temperature of mixture.

Remaining heat of steam;

$ \begin{aligned} Q & =16 \times 540 \\\\ & =8640 \mathrm{cal} \end{aligned} $

Heat required to raise the tempature of 63 g upto $T^{\circ}$ will be

$ \begin{array}{rlrl} Q & =m s_{\text {water }} \times(\Delta T) \\\\ 8640 & =63 \times 1(\Delta T) \\\\ \Rightarrow \quad \Delta T & >100^{\circ} \mathrm{C} \end{array} $

Hence, not all steam will be converted into water. The steam required to raise 63 g water upto $100^{\circ} \mathrm{C}$ will be,

$ m=\frac{6300}{540}=11.66 \mathrm{~g} $

$\therefore$ Total steam used $=11.66+9$

$ \approx 20 \mathrm{~g} $

Hence 20 g of steam will be used to bring 54 g of ice at $-20^{\circ} \mathrm{C}$ upto $100^{\circ} \mathrm{C}$ water.

Total water $=54+20 \approx 74 \mathrm{~g}$

and steam $=5 \mathrm{~g}$

Which is close to option (b).

Total surface area of solid sphere

$ A_1=4 \pi r^2 $

Total surface area of solid hemisphere,

$ A_2=3 \pi r^2 $

As, we know that, $E \propto A$

$ \therefore \quad \frac{E_2}{E_1}=\frac{A_2}{A_1}=\frac{3}{4} $

For diatomic gas, $C_p=\frac{7}{2} R$

$ \begin{aligned} C_v & =\frac{5}{2} R \\\\ \therefore \quad d U & =n C_v d T \end{aligned} $

$ \begin{aligned} d Q & =n C_p d T \\\\ d W & =p \times d V \\\\ & =n R d T \\\\ \therefore d W: d U: d Q & =n R d T: n \frac{5}{2} R d T: n \frac{7}{2} R d T \\\\ & =1: \frac{5}{2}: \frac{7}{2} \\\\ & =2: 5: 7 \end{aligned} $

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

According to question, $v_1=\sqrt{\frac{3 R(300)}{M}}$

$ v_2=\sqrt{\frac{3 R(432)}{M}} $

Thus, $\frac{v_2-v_1}{v_1}=\frac{\sqrt{432}-\sqrt{300}}{\sqrt{300}}=\frac{3.46}{17.32}$

$ \begin{aligned} \% \text { increase } & =\frac{v_2-v_1}{v_1} \times 100 \% \\\\ & =\frac{3.46}{17.32} \times 100 \% \\\\ & =19.99 \% \\\\ & \approx 20 \% \end{aligned} $

The relationship between the two scales,

$ F=\frac{9}{5} C+32 \quad \ldots \text { (i) } $

If the temperature is twice the celsius temperature, we can write

$ F=2 C \quad \ldots \text { (ii) } $

Now, $\frac{9}{5} \times C+32=2 C$

$ \begin{aligned} & \frac{9}{5} \times C+32=\frac{10}{5} C \\\\ & 32=\frac{10}{5} C-\frac{9}{5} C \Rightarrow 32=\frac{1}{5} C \\\\ & C=160 \quad \ldots \text { (iii) } \end{aligned} $

Fahrenheit temperature,

$ \begin{aligned} F & =\frac{9}{5} \times 160+32 \\\\ \quad F & =288+32 \\\\ \Rightarrow \quad F & =320 \end{aligned} $

Therefore, the temperature on the fahrenheit scale that is twice the temperature of $160^{\circ} \mathrm{C}$ on the celsius scale is $320^{\circ} \mathrm{F}$.

For liquid $A$ and $B$ mixed to reach a final temperature of $20^{\circ} \mathrm{C}$, the equation based on the heat exchanged is

$ m \times C_B \times(24-20)=m \times C_A \times(20-15) $

Simplifying, we get

$ \begin{aligned} 4 m \times C_B & =5 m \times C_A \\\\ \frac{C_B}{C_A} & =\frac{5}{4} \end{aligned} $

For liquid $B$ and $C$ mixed to reach a final temperature of $26^{\circ} \mathrm{C}$,

$ m \times C_C \times(30-26)=m \times C_B \times(26-24) $

we get,

$ \begin{aligned} 4 m \times C_C & =2 m \times C_B \\\\ \frac{C_C}{C_B} & =\frac{1}{2} \Rightarrow \frac{C_B}{C_A} \times \frac{C_C}{C_B}=\frac{5}{4} \times \frac{1}{2}=\frac{5}{8} \end{aligned} $

Therefore,

$ \begin{aligned} & C_A: C_B: C_C=1: \frac{5}{4}: \frac{5}{8} \\\\ & C_A: C_B: C_C=8: 10: 5 \end{aligned} $

The efficiency of a reversible heat engine is given by the ratio of work output to heat input, which can be also be expressed in terms of the temperatures of the hot $\left(T_H\right)$ and cold $\left(T_C\right)$ reservoirs,

$ \eta=1-\frac{T_C}{T_H} $

Given, efficiency is $\mathbf{5 0 \%}$

$ 0.5=1-\frac{T_C}{T_H} $

Solving for $\frac{T_C}{T_H}$, we get

$ \frac{T_C}{T_H}=0.5 $

Coefficient of performance (COP),

$ \mathrm{COP}=\frac{Q_C}{W}=\frac{T_C}{T_B-T_C} $

For a reversible process, the (COP) can also be related to the temperatures of the hot and cold reservoirs,

$ \begin{aligned} \mathrm{COP} & =\frac{0.5 \times T_H}{T_H-0.5 \times T_H} \\\\ & =\frac{0.5}{1-0.5}=\frac{0.5}{0.5}=1 \end{aligned} $

Therefore, the coefficient of performance of the refrigerator working between the same two temperature but reverse direction is 1.

Given, $n=4$ moles

Temperature,

$ T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} $

We know that,

$ U=\frac{5}{2} \times n \times R \times T \quad \ldots \text { (i) } $

Putting above values in Eq. (i),

$ U=\frac{5}{2} \times 4 \times 8.314 \times 300 \mathrm{~K} $

$\therefore$ Universal gas constant,

$ \begin{aligned} R & =8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\\\ U & =5 \times 2 \times 8.31 \times 300 \\\\ & =10 \times 8.31 \times 300 \\\\ & =3000 \times 8.31 \\\\ U & =24930 \mathrm{~J}=24.930 \mathrm{~J} \end{aligned} $

Thus, $U=24.93 \mathrm{~kJ}$

Using first law of thermodynamies

$ \Delta U=\Delta Q-W $

For adiabatic process, $\Delta Q=0$

$ \begin{aligned} \therefore \quad W & =-\Delta U=-n C_V \Delta T \\ & =-n C_V\left[T_1-T_2\right] \\ \Rightarrow \quad W & =W_1=2 \times \frac{3}{2} R \times(80-50) \\ \Rightarrow \quad W & =90 R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ W_2 & =3 \times \frac{5}{2} R \times(50-20)=225R \\ & =225 \times \frac{W}{90} \,\,\,\,\,\,\,\,\,\,\,\,[form \,\,eq.ii]\\ & =2.5 \mathrm{~W} \end{aligned} $

By using the first law of thermodynamics,

$ \Delta U=Q-W $

Here, internal energy =?

Work done on the gas, $W=-12 \mathrm{~J}$

Heat absorbed, $Q=18 \mathrm{~J}$

From Eq. (i), $\Delta U=18-(-12)=30 \mathrm{~J}$

$ \Delta U=18+12=30 \mathrm{~J} $

So, the change in internal energy of the gas is 30 J .

The efficiency of a Carnot engine is calculated using the formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Initially, the efficiency $\eta_1$ is:

$ \eta_1 = 1 - \frac{2}{3} = \frac{1}{3} $

After the change in the temperature ratio, the efficiency $\eta_2$ becomes:

$ \eta_2 = 1 - \frac{3}{4} = \frac{1}{4} $

To determine the change in efficiency, we calculate $\Delta \eta$:

$ \Delta \eta = \eta_2 - \eta_1 = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12} $

The percentage change in efficiency is calculated as follows:

$ \frac{\Delta \eta}{\eta_1} \times 100\% = \frac{-\frac{1}{12}}{\frac{1}{3}} \times 100 = -\frac{1}{4} \times 100\% = -25\% $

The negative sign indicates that there is a 25% decrease in efficiency.

The molar specific heat capacities of gases at constant pressure vary depending on whether the gas is monoatomic or diatomic due to differences in their degrees of freedom.

For a monoatomic gas, the molar specific heat capacity at constant pressure, denoted as $ C_{p_m} $, is given by:

$ C_{p_m} = \frac{5}{2} R $

For a diatomic gas, the molar specific heat capacity at constant pressure, denoted as $ C_{p_d} $, is:

$ C_{p_d} = \frac{7}{2} R $

The ratio of the molar specific heat capacities for monoatomic and diatomic gases is therefore:

$ \frac{C_{p_m}}{C_{p_d}} = \frac{\left(\frac{5}{2} R\right)}{\left(\frac{7}{2} R\right)} = \frac{5 \times 2 \times R}{2 \times 7 \times R} = \frac{5}{7} $

Thus, the ratio is $ 5:7 $.

To find the mass of water $ m $ that results in a final mixture temperature of $ 6^\circ \mathrm{C} $, let's break down the heat exchanges that occur during the process.

Step 1: Heating Ice from $-20^\circ \mathrm{C}$ to $0^\circ \mathrm{C}$

The heat required is given by:

$ Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T = 5\, \text{g} \times 0.5\, \text{cal/g}{}^\circ\mathrm{C} \times 20^\circ\mathrm{C} = 50\, \text{cal} $

Step 2: Melting Ice at $0^\circ \mathrm{C}$ to Water

The heat required for melting the ice is:

$ Q_2 = m_{\text{ice}} \times L_f = 5\, \text{g} \times 80\, \text{cal/g} = 400\, \text{cal} $

Step 3: Heating Water from Ice from $0^\circ \mathrm{C}$ to $6^\circ \mathrm{C}$

The heat required to raise the temperature of the resulting water is:

$ Q_3 = m_{\text{water from ice}} \cdot c_{\text{water}} \cdot \Delta T = 5\, \text{g} \times 1\, \text{cal/g}{}^\circ\mathrm{C} \times 6^\circ\mathrm{C} = 30\, \text{cal} $

Total Heat Required

Summing these components, the total heat absorbed by the ice and converted water is:

$ Q_{\text{total gained}} = Q_1 + Q_2 + Q_3 = 50 + 400 + 30 = 480\, \text{cal} $

Step 4: Heat Lost by Water Initially at $30^\circ \mathrm{C}$ Cooling to $6^\circ \mathrm{C}$

The heat lost is calculated by:

$ Q_{\text{lost}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T = m \times 1\, \text{cal/g}{}^\circ\mathrm{C} \times 24^\circ\mathrm{C} = 24m\, \text{cal} $

Equilibrium Condition

The heat gained must equal the heat lost:

$ Q_{\text{total gained}} = Q_{\text{lost}} $

$ 480 = 24m $

Solving for $ m $:

$ m = \frac{480}{24} = 20\, \text{g} $

Therefore, the mass of water required is $ 20\, \text{g} $.

Work Done During Isothermal Expansion:

For an ideal gas expanding isothermally, the work done $ W $ is given by:

$ W = nRT \ln\left(\frac{V_f}{V_i}\right) $

where $ n $ is the number of moles, $ R $ is the ideal gas constant, $ T $ is the temperature, $ V_f $ is the final volume, and $ V_i $ is the initial volume.

Gas A Expansion:

The pressure of gas $ A $ decreases by 50%, thus:

$ p_f = 0.5 p_i $

Using the ideal gas law $ pV = nRT $, we have:

$ p_f V_f = p_i V_i $

$ 0.5 p_i V_f = p_i V_i \Rightarrow V_f = 2 V_i, \text{ thus } \frac{V_f}{V_i} = 2 $

Therefore, the work done by gas $ A $ is:

$ W_A = nRT_1 \ln(2) $

Gas B Expansion:

The pressure of gas $ B $ decreases by 75%, thus:

$ p_f = 0.25 p_i $

Using the ideal gas law again:

$ p_f V_f = p_i V_i $

$ 0.25 p_i V_f = p_i V_i \Rightarrow V_f = 4 V_i, \text{ thus } \frac{V_f}{V_i} = 4 $

The work done by gas $ B $ is:

$ W_B = nRT_2 \ln(4) $

Determining the Temperature Ratio

Given that the work done by both gases is the same, $ W_A = W_B $, we equate the expressions for work:

$ nRT_1 \ln(2) = nRT_2 \ln(4) $

Solving for the temperature ratio:

$ T_1 \ln(2) = T_2 \ln(4) $

$ T_1 \ln(2) = T_2 \cdot 2 \ln(2) $

$ \frac{T_1}{T_2} = \frac{2 \ln(2)}{\ln(2)} = 2 $

Thus, the ratio of temperatures $ T_1 : T_2 $ is:

$ 2:1 $

Given:

Heat absorbed, $ Q = 80 \, \text{J} $

Change in volume, $ \Delta V = 16 \times 10^{-5} \, \text{m}^3 $

First Law of Thermodynamics

$ Q = \Delta U + W $

For a monoatomic ideal gas, the change in internal energy ($ \Delta U $) can be expressed as:

$ \Delta U = \frac{3}{2} n R \Delta T $

The work done ($ W $) by the gas during expansion is:

$ W = p \Delta V $

Using the ideal gas law:

$ \begin{aligned} & p V = n R T \\ & p \Delta V = n R \Delta T \end{aligned} $

Substitute into the internal energy equation:

$ \Delta U = \frac{3}{2} p \Delta V $

Now substitute equations for $ W $ and $ \Delta U $ into the first law:

$ \begin{aligned} Q & = \frac{3}{2} p \Delta V + p \Delta V \\ 80 & = \frac{5}{2} \times p \times 16 \times 10^{-5} \end{aligned} $

Solve for pressure $ p $:

$ \begin{aligned} \frac{2 \times 80}{16 \times 10^{-5} \times 5} & = p \\ \frac{160}{80 \times 10^{-5}} & = p \\ 2 \times 10^5 \, \text{N/m}^2 & = p \end{aligned} $

Thus, the pressure of the gas is $ 2 \times 10^5 \, \text{N/m}^2 $.

The efficiency $ (\eta) $ of a Carnot engine is determined by the formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Given that the initial efficiency is $ 25\% $ and the source temperature is $ 127^\circ \mathrm{C} $, which converts to $ 400 \, \mathrm{K} $ in absolute temperature, we have:

$ 0.25 = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

This implies:

$ \frac{T_{\text{sink}}}{T_{\text{source}}} = 0.75 $

From which:

$ T_{\text{sink}} = 0.75 \times 400 = 300 \, \mathrm{K} $

If we decrease the sink temperature by $ 10\% $, the new sink temperature $ T_{\text{sink}}^{\prime} $ becomes:

$ T_{\text{sink}}^{\prime} = T_{\text{sink}} - \frac{T_{\text{sink}} \times 10}{100} = 300 - 30 = 270 \, \mathrm{K} $

The new efficiency $ \eta^{\prime} $ of the engine is calculated as follows:

$ \eta^{\prime} = 1 - \frac{T_{\text{sink}}^{\prime}}{T_{\text{source}}} \times 100\% $

Substituting the values, we get:

$ \eta^{\prime} = 1 - \frac{270}{400} \times 100\% $

$ = 1 - 0.675 \times 100\% $

$ \eta^{\prime} = 0.325 \times 100\% $

Therefore, the new efficiency is:

$ \eta^{\prime} = 32.5\% $

For a monoatomic gas:

Number of moles, $ n_m = 2 $.

Temperature, $\Delta T_m = 27^{\circ} \mathrm{C} = 300\, \mathrm{K}$.

Internal energy, $ U_m = U $.

The total internal energy for a monoatomic gas is calculated as:

$ U_m = \frac{3}{2} n R \Delta T $

Substituting the values:

$ U = \frac{3}{2} \times 2 \times R \times 300 $

So,

$ U = 900R \quad \text{...(i)} $

For a diatomic gas:

Number of moles, $ n_d = 3 $.

Temperature, $\Delta T_d = 127^{\circ} \mathrm{C} = 400\, \mathrm{K}$.

The total internal energy for a diatomic gas is given by:

$ U_d = \frac{5}{2} n R \Delta T $

Substitute the known values:

$ U^{\prime} = \frac{5}{2} \times 3 \times R \times 400 $

Thus,

$ U^{\prime} = 3000R \quad \text{...(ii)} $

To find the relationship between $ U^{\prime} $ and $ U $, divide Eq. (ii) by Eq. (i):

$ \frac{U^{\prime}}{U} = \frac{3000R}{900R} $

Therefore,

$ U^{\prime} = \frac{10}{3} U $

The specific details are as follows:

Mass of water, $m_w = 200 \, \text{g}$

Specific heat capacity of water, $C_w = 1 \, \text{cal/g}^\circ \text{C}$

Change in temperature of the water, $\Delta T_w = 10^\circ \text{C}$ (from $80^\circ \text{C}$ to $70^\circ \text{C}$)

Mass of the metal, $m_m = 100 \, \text{g}$

Change in temperature of the metal, $\Delta T_m = 50^\circ \text{C}$ (from $20^\circ \text{C}$ to $70^\circ \text{C}$)

Calculating Heat Exchanged

Using the heat equation, we calculate:

Heat lost by the water:

$ Q_w = m_w \times C_w \times \Delta T_w $

Heat gained by the metal:

$ Q_m = m_m \times C_m \times \Delta T_m $

Setting Heat Lost Equal to Heat Gained

The principle of conservation of energy tells us that the heat lost by the water equals the heat gained by the metal:

$ m_w \times C_w \times \Delta T_w = m_m \times C_m \times \Delta T_m $

Substituting the values, we have:

$ 200 \times 1 \times 10 = 100 \times C_m \times 50 $

By solving for $C_m$:

$ C_m = \frac{200 \times 1 \times 10}{100 \times 50} = \frac{2}{5} \, \text{cal/g}^\circ \text{C} $

Calculating the Ratio

Finally, the ratio of the specific heat capacity of the metal to that of water is:

$ \frac{C_m}{C_w} = \frac{\frac{2}{5}}{1} = \frac{2}{5} $

To determine the efficiency of a heat engine operating between two specific temperatures, we use the formula:

$ \eta = 1 - \frac{T_C}{T_H} $

where:

$ T_C $ is the temperature of the cold reservoir.

$ T_H $ is the temperature of the hot reservoir.

From the temperature scale relationships:

The Celsius and Fahrenheit scales coincide at $-40^\circ$, which is equivalent to 233.15 Kelvin.

The Kelvin and Fahrenheit scales coincide at 574.59 Kelvin.

Using these temperatures, we calculate the efficiency:

$ \begin{aligned} \eta &= 1 - \frac{233.15}{574.59} \\ &\approx 1 - 0.4056 \\ &\approx 0.5944 \end{aligned} $

Therefore, the efficiency of the heat engine operating between these temperatures is approximately 59.44%.

Given the initial conditions of the problem:

Pressure, $ p_1 = 10^5 \, \mathrm{Nm}^{-2} $

Initial volume, $ V_1 = 16 \, \mathrm{L} $

Final volume, $ V_2 = 2 \, \mathrm{L} $

The molar specific heat at constant volume is given as $ \frac{3}{2} R $.

For an adiabatic process, the work done on the gas is calculated using the formula:

$ W = \frac{n R \Delta T}{\gamma-1} = \frac{p_2 V_2 - p_1 V_1}{\gamma - 1} $

First, we determine the final pressure $ p_2 $ using the adiabatic relation:

$ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^\gamma $

Substituting the given values:

$ p_2 = 10^5 \left(\frac{16}{2}\right)^{1.66} = 3.15 \times 10^6 \, \mathrm{N/m}^2 $

Next, we calculate the work done:

$ W = \frac{\left(3.15 \times 10^6 \times 2\right) - \left(10^5 \times 16\right)}{1.66 - 1} $

Simplifying further:

$ W = \frac{6.3 \times 10^6 - 16 \times 10^5}{0.66} $

$ W = \frac{63 \times 10^5 - 16 \times 10^5}{0.66} $

$ W = \frac{47 \times 10^5}{0.66} = 71.21 \times 10^5 $

Finally, converting the work done to kilojoules:

$ \text{Work done} = 71.21 \times 10^5 \times 10^{-3} \, \mathrm{J} = 7.1 \, \mathrm{kJ} $

Thus, the work done on the gas is 7.1 kJ.

The work done during a cycle is equal to the area under $p$ - $V$ diagram.

$ \begin{aligned} &\text { From the given } p \text { - } V \text { diagram, }\\ &\begin{aligned} W & =\frac{1}{2} \times A C \times B C \\ W & =\frac{1}{2} \times(3 V-V) \times(3 p-p) \\ & =\frac{1}{2} \times 2 V \times 2 p \\ W & =2 p V \end{aligned} \end{aligned} $

At NTP, the internal energy of a diatomic gas is primarily due to translational and rotational motions, which is given by $\frac{5}{2} k_B T$.

At high, temperature, vibrational modes becomes accessible and internal energy increases because these modes contribute additional degree of freedom. The internal energy increases because these modes contribute in temperature, $\frac{7}{2} k_B T$.

Therefore, the ratio of KE,

$\frac{\text { KE at high temperature }}{\text { KE at NTP }}=\frac{7 / 2 k_B T_{\text {high }}}{5 / 2 k_B T_{\text {NTP }}}$

This is a simplified model and assumes that all vibrational modes are fully excited, which may not be the case depending on the exact temperature. All options are incorrect.