Heat and Thermodynamics

Given,

Number of moles, $n=2$

$ \gamma=\frac{4}{3} \quad \text { (for triatomic gas) } $

$C_V$ for triatomic gas $=3 R$

Initial volume $=V_i$

Final volume $=8 V_i=V_f$

In adiabatic process work done is given by

$ \Delta U=n C_V \Delta T $

We have to find $\Delta T$ first, for that $p V^{\gamma-1}=$ constant (adiabatic condition)

$ \begin{aligned} & \frac{T_2}{T_1}=\left(\frac{V_f}{V_i}\right)^{\gamma-1} \\ & \quad\left[T_1=327^{\circ} \mathrm{C}=600 \mathrm{~K} \text { given }\right] \\ & T_2=T_1\left(\frac{1}{8}\right)^{\frac{4}{3}-1} \Rightarrow T_2=\frac{600}{2}=300 \mathrm{~K} \end{aligned} $

Now work done is $\Delta U=n C_V \Delta T$

$ =2 \times 3 R \times 300=1800 R $

Work done in isochoric process is zero because volume remains constant after above process.

Hence, work done is $1800 R$.

Given,

Initial temperature $T_1=27^{\circ} \mathrm{C}$

Final temperature $T_2=159^{\circ} \mathrm{C}$

RMS speed of gas is given by

$ \begin{gathered} v_{\text {rms }}=\sqrt{\frac{3 K T}{m}} \\ v_{\text {rms }} \propto \sqrt{T} \\ \text { for } T=27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ v_{\text {rms }} \propto \sqrt{300}=17.32 \mathrm{~m} / \mathrm{s} \\ v_{\text {rms }} \propto \sqrt{432}=20.78 \mathrm{~m} / \mathrm{s} \end{gathered} $

Percentage increase

$ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \times 100 \Rightarrow \frac{20.78-17.32}{17.32} \times 100=20 \% $

Hence, option (d) is the correct option.

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: $3:2$

To determine the pressure of the helium (He) gas in the identical vessel, we can use the ideal gas law:

$PV = nRT$

Given that both vessels are identical, they have the same volume, $V$. Also, each vessel contains one mole of gas, so $n = 1$ for both cases. Let's first express the pressure $p$ of the oxygen (O₂) gas in terms of the given conditions:

For the oxygen gas:

$pV = nRT$

Substituting the given values:

$pV = (1 \text{ mole}) \cdot R \cdot T$

Therefore,

$pV = RT$

Now, consider the helium gas in the other vessel at temperature $4T$. According to the ideal gas law:

$P_{He}V = nRT_{He}$

Here, ${T_{He} = 4T}$. Substituting the known values:

$P_{He}V = (1 \text{ mole}) \cdot R \cdot (4T)$

Therefore,

$P_{He}V = 4RT$

Since $pV = RT$ from our earlier equation for the oxygen gas, we can substitute $RT$ with $pV$ in the helium pressure equation:

$P_{He}V = 4(pV)$

Dividing both sides by $V$ to isolate $P_{He}$:

$P_{He} = 4p$

Therefore, the pressure of the helium gas at temperature $4T$ is $4p$.

Thus, the correct answer is:

Option C: 4p

A Carnot heat engine operates between two temperatures: the hot reservoir temperature ($ T_h $) and the cold reservoir temperature ($ T_c $). To find the efficiency ($ \eta $) of a Carnot engine, we use the formula:

$ \eta = 1 - \frac{T_c}{T_h} $

where both temperatures ($ T_h $ and $ T_c $) must be in absolute temperature, i.e., Kelvin (K).

The given temperatures are $ 527^{\circ} \mathrm{C} $ and $ 127^{\circ} \mathrm{C} $. First, we convert these temperatures to Kelvin:

$ T_h = 527^{\circ} \mathrm{C} + 273 = 800 \, \mathrm{K} $

$ T_c = 127^{\circ} \mathrm{C} + 273 = 400 \, \mathrm{K} $

Next, we calculate the efficiency of the Carnot engine:

$ \eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5 $

The efficiency ($ \eta $) is equal to the fraction of heat converted to work. Therefore, the fraction of heat used to do mechanical work per second is:

$ \eta = \frac{W}{Q_h} $

where $ W $ is the work done per second (which is $ 1.0 \, \text{kW} $ or $ 1000 \, \text{W} $), and $ Q_h $ is the heat consumed per second.

Rearranging the formula, we get:

$ Q_h = \frac{W}{\eta} $

Substituting the given values:

$ Q_h = \frac{1000 \, \mathrm{W}}{0.5} = 2000 \, \mathrm{W} $

To convert the heat consumed per second from Watts to kcal/s, we use the conversion factor $ 1 \, \mathrm{kcal/s} = 4184 \, \mathrm{W} $:

$ Q_h = \frac{2000 \, \mathrm{W}}{4184 \, \mathrm{W/kcal}} \approx 0.477 \, \mathrm{kcal/s} $

Therefore, the amount of heat the Carnot engine should consume per second to deliver mechanical work at the rate of $ 1.0 \, \text{kW} $ is:

Option A: 0.477 kcal/s

n = 0.5

T = 300

For v_rms to be doubled T' = 4 $\times$ 300 = 1200

$\Rightarrow$ Heat transferred

$ = (0.5)\left( {{5 \over 2}} \right)(8.32)(900)$

$ = 9360$ J

$1 - {{{T_L}} \over {{T_H}}} = 0.5$ ...... (1)

$1 - {{{T_L} - 40} \over {{T_H}}} = 0.65$ .... (2)

$ \Rightarrow {T_H} = {{800} \over 3}\,K \simeq 266.7\,K$

Average momentum $ = \left\langle {\overrightarrow P } \right\rangle = 0$

${v_{rms}} = \sqrt {{{3RT} \over M}} $

If temperature is doubled and oxygen atoms are used then

$v{'_{rms}} = \sqrt {{{3R(2T)} \over {M/2}}} = 4\,{v_{rms}}$

${\eta _{net}} = {{{W_1} + {W_2}} \over {{Q_1}}}$

${\eta _{net}} = {{{W_1}} \over {{Q_1}}} + {{{W_2}} \over {{Q_1}}}$

${\eta _{net}} = {\eta _1} + {{{W_2}} \over {{Q_2}}} \times {{{Q_2}} \over {{Q_1}}}$

${\eta _{net}} = {\eta _1} + ({\eta _2})(1 - {\eta _1})$

$1 - {\eta _1} < 1$

$ \Rightarrow {\eta _{net}} < {\eta _1} + {\eta _2}$

Statement A is incorrect, statement B is correct by equipartition of energy. Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH₄ is a polyatomic gas so it has 6 degree of freedom. So only B and D are correct.

$450 - T = {{dQ} \over {dt}} \times {{{l_1}} \over {{K_1}{A_1}}}$

$T - 0 = {{dQ} \over {dt}} \times {{{l_2}} \over {{K_2}{A_2}}}$

So, ${{450 - T} \over T} = {{{K_2}{A_2}{l_1}} \over {{K_1}{A_1}{l_2}}} = 9 \times {1 \over 2} \times 2 = 9$

$450 - T = 9T$

$ \Rightarrow T = 45^\circ C$

From Newton's cooling law ${{dQ} \over {dt}} = - k(T - {T_s})$ the statement A is correct.

For B

$U = \sigma eA{T^4}$

So, ${{{U_1}} \over {{U_2}}} = {\left( {{{283} \over {293}}} \right)^4} \simeq {1 \over {1.15}}$

Statement B is correct

For C

$\eta = 1 - {{{T_1}} \over {{T_2}}} = 1 - {{100} \over {400}} = {3 \over 4}$

So, efficiency is 75% C is correct

For D

From Newton's law of cooling ${{dQ} \over {dt}} = - k(T - {T_s})$

The statement is wrong.

${v_{rms}} = \sqrt {{{3RT} \over {{M_0}}}} $ because T is same

v_rms will be same so, A is correct D is incorrect

${{{P_1}} \over {{P_2}}} = {{{n_1}R{T_1}/{V_1}} \over {{n_2}R{T_2}/{V_2}}} = {{{n_1}} \over {{n_2}}} = {1 \over 4}$

B is correct

C is incorrect

${{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$

$ \Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$

$ \Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}$

$ \simeq 61.1 \times {10^{ - 5}}$ kg/s

${C_V} = {{nR} \over 2}$

And ${C_P} = {{nR} \over 2} + R$

$ \Rightarrow {{{C_V}} \over {{C_P}}} = {{{{nR} \over 2}} \over {{{nR} \over 2} + R}} = {n \over {n + 2}}$

$\Delta U = n{C_v}\Delta T$

$ = 7 \times {{3R} \over 2} \times 40$

$ = 3486\,J$

$P{V^\gamma }=$ constant

$ \Rightarrow P{V^\gamma } = (P'){\left( {{v \over 8}} \right)^\gamma }$ where $\gamma = 5/3$

$ \Rightarrow P' = 32P$

Molar mass $M - {{2 \times 4 + n \times 1} \over {2 + n}}$ ..... (i)

Also, $\gamma = {{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}} \over {{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}} = {{2 \times 5R + n \times 7R} \over {2 \times 3R + n \times 5R}}$

$ \Rightarrow \gamma = {{10 + 7n} \over {6 + 5n}}$ ...... (ii)

Given that ${V_{rms}} = \sqrt 2 \,{V_{sound}}$

$ \Rightarrow \sqrt {{{3RT} \over M}} = \sqrt 2 \sqrt {{{\gamma RT} \over M}} $

$ \Rightarrow \gamma = {3 \over 2}$

$ \Rightarrow n = 2$

${\eta _1} = 1 - {{420} \over {720}} = {{300} \over {720}}$

And ${\eta _2} = 1 - {{320} \over {1220}} = {{900} \over {1220}}$

$ \Rightarrow {{{\eta _1}} \over {{\eta _2}}} = {{300} \over {720}} \times {{1220} \over {900}}$

$ \simeq 0.56$

Let AB is isothermal process and BC is adiabatic process then for AB process

P_AV_A = P_BV_B

$\Rightarrow$ P_B = 10⁷ Nm^$-$2

For process BC

P_BV$_B^r$ = P_C V$_C^r$

P_C = 3.536 x $\times$ 10⁶ Pa

Because KE $\propto$ T so A is correct, B is incorrect, statement C cannot be said, statement D is contradicting itself, statement E is incorrect (Isothermal process)

So no answer correct (Bonus)

If the statement of D would have been.

"Pressure of gas increases with increase in temperature at constant volume, "then statement D would have been correct, so in that case answer would have been 'A'.

Here volume is constant.

$\therefore$ ${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}$

$ \Rightarrow {{100} \over {273}} = {{180} \over {{T_2}}}$

$ \Rightarrow {T_2} = {{180} \over {100}} \times 273$

$ = 1.8 \times 273$

$ = 491\,K$

For monoatomic gas degree of freedom f = 3 and $\gamma$ = ${5 \over 3}$

Here for gas,

Initial pressure (P₁) = 75 kPa

Initial volume (V₁) = 1200 cm³

Final volume (V₂) = 150 cm³

Final pressure (P₂) = ?

For adiabatic process,

${P_1}V_1^\gamma = {P_2}V_2^\gamma $

$ \Rightarrow 75 \times {(1200)^\gamma } = {P_2}{(150)^\gamma }$

$ \Rightarrow {P_2} = 75 \times {\left( {{{1200} \over {150}}} \right)^{{5 \over 3}}}$

$ = 75 \times {(8)^{{5 \over 3}}}$

$ = 75 \times 32$ kPa

= 2400 kPa

Work done in adiabatic process,

$W = {{{P_2}{V_2} - {P_1}{V_1}} \over {1 - \gamma }}$

$ = {{2400 \times {{10}^3} \times 150 \times {{10}^{ - 6}} - 75 \times {{10}^3} \times 1200 \times {{10}^{ - 6}}} \over {1 - {5 \over 3}}}$

$ = {{(2400 \times 150 - 75 \times 1200) \times {{10}^{ - 3}}} \over { - {2 \over 3}}}$

$ = {{ - 810000 \times {{10}^{ - 3}}} \over 2}$

$ = - 405$ kJ

$\Delta D = D\alpha \Delta T$

$\Delta T = {{0.011} \over {6.230 \times 1.4 \times {{10}^{ - 5}}}}$

= 126.11$^\circ$C

$\Rightarrow$ T_f = T + $\Delta$T

= (27 + 126.11)$^\circ$C

= 153.11$^\circ$C

Comparing the area under the PV graph

A₃ > A₁ > A₂

$\Rightarrow$ W₃ > W₁ > W₂

Total number of moles are

$n = {n_{{H_2}}} + {n_{{O_2}}}$

$ = {{16} \over 2} + {{128} \over {32}}$

= 12 moles

Using $PV = nRT$

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

$ = {{12 \times 8.31 \times 273.15} \over {{{10}^5}}}$ m³

= 0.27 m³ = 27 $\times$ 10⁴ cm³

$\Delta Q = n{C_v}\Delta T$ (Isochoric process)

$ = 2 \times {{3R} \over 2} \times 20$

$ = 498$ J

From the equation

$[a] \equiv [P{V^2}]$

$[b] \equiv [V]$

$ \Rightarrow \left[ {{a \over b}} \right] \equiv [PV]$