Heat and Thermodynamics

Given a metal block composed of various metals:

Aluminium mass: $ m_{\text{Al}} = 2.4 \, \text{kg} $

Brass mass: $ m_{\text{Brass}} = 1.6 \, \text{kg} $

Copper mass: $ m_{\text{Cu}} = 0.8 \, \text{kg} $

The specific heats of the components are:

Aluminium: $ C_{\text{Al}} = 0.216 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

Brass: $ C_{\text{Brass}} = 0.0917 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

Copper: $ C_{\text{Cu}} = 0.0931 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

The heat supplied to the block is 44.4 cal. To find the final temperature of the block, we use the formula for heat energy:

$ Q = m C \Delta T $

The total heat equation is:

$ Q = (m_{\text{Al}} C_{\text{Al}} + m_{\text{Brass}} C_{\text{Brass}} + m_{\text{Cu}} C_{\text{Cu}}) \Delta T $

Substitute the given values:

$ 44.4 = (2.4 \times 0.216 + 1.6 \times 0.0917 + 0.8 \times 0.0931) \Delta T $

Calculate the product inside the brackets:

$ 44.4 = [0.5184 + 0.14672 + 0.07448] \Delta T $

Simplifying further:

$ 44.4 = 0.7396 \Delta T $

Solving for $ \Delta T $:

$ \Delta T = \frac{44.4}{0.7396} $

$ \Delta T = 60 $

The initial temperature $ T_i = 20^{\circ} \text{C} $. Therefore, the final temperature $ T_f $ is:

$ T_f = \Delta T + T_i $

$ T_f = 60 + 20 $

$ T_f = 80^{\circ} \text{C} $

During an adiabatic process, the given relationship is $ p V^{\frac{3}{2}} = \text{constant} $. In general, for an adiabatic process, the equation is $ p V^\gamma = \text{constant} $. By comparing these equations, we deduce that $ \gamma = \frac{3}{2} $.

Consider the initial volume $ V_1 = V_0 $ and the final volume $ V_2 = \frac{V_0}{4} $.

The relation between temperature and volume in an adiabatic process is given by:

$ T V^{\gamma-1} = \text{constant} $

Thus:

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

Substituting the known values, we have:

$ T \left(V_0\right)^{\gamma-1} = T_2 \left[\frac{V_0}{4}\right]^{\gamma-1} $

This simplifies to:

$ T_2 = T (4)^{\frac{3}{2}-1} $

Therefore, the final temperature is:

$ T_2 = T \sqrt{2} $

The condition $ dW = dQ $ is valid in an isothermal process.

According to the first law of thermodynamics:

$ dQ = dW + dU $

In an isothermal process, the temperature remains constant, which implies $ dU = 0 $. Therefore:

$ \therefore \quad dQ = dW $

To find the temperature of the sink ($T_2$), we begin by analyzing the efficiency increase of a Carnot engine.

Initial Conditions

Initial efficiency: $\eta = 25\%$ or 0.25.

Efficiency formula:

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

Substitute the given efficiency:

$ 0.25 = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \quad \Rightarrow \quad \frac{T_{\text{sink}}}{T_{\text{source}}} = 0.75 $

Thus, the temperature of the sink is:

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

After Increasing the Source Temperature

The source temperature is increased by 100 K:

$ T_{\text{source}}^{\prime} = T_{\text{source}} + 100 $

New efficiency: $\eta^{\prime} = 40\%$ or 0.40. Using the efficiency formula again:

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

Substitute the known values:

$ 0.40 = 1 - \frac{0.75 \times T_{\text{source}}}{T_{\text{source}} + 100} $

Solving for $T_{\text{source}}$, we find:

$ T_{\text{source}} = 400 \text{ K} $

Calculating the Sink Temperature

Substitute $T_{\text{source}}$ in the sink temperature equation:

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

Thus, the temperature of the sink ($T_2$) is 300 K.

For a deeper understanding of how the heat capacity ratio $\gamma = \frac{C_p}{C_v}$ varies with different types of gases based on their degrees of freedom, consider the following explanations:

Monoatomic Gases: The ratio $\gamma$ is typically $\frac{5}{3}$. These gases have three translational degrees of freedom, which leads to this specific ratio.

Diatomic Gases (Rigid): For diatomic gases that behave rigidly (i.e., where vibrational modes are not considered), $\gamma$ is $\frac{7}{5}$. The degrees of freedom include three translational and two rotational.

Diatomic Gases (Non-Rigid): For non-rigid diatomic gases, where vibrational modes are considered, $\gamma$ is $\frac{9}{7}$. Additional vibrational degrees of freedom affect this ratio.

Polyatomic Gases: These gases can have more complex molecular structures, resulting in more degrees of freedom. Here, for any polyatomic gas, the heat capacities are given by $C_p = 4 + f$ and $C_V = 3 + f$, where $f$ represents the additional degrees of freedom available to the gas. Therefore, the ratio $\gamma$ is expressed as:

$ \gamma = \frac{C_p}{C_V} = \frac{4 + f}{3 + f} $

Using this understanding, the correct pairing is as follows:

A (Monoatomic): Corresponds to II ($\frac{5}{3}$)

B (Diatomic Rigid): Corresponds to III ($\frac{7}{5}$)

C (Diatomic Non-Rigid): Corresponds to IV ($\frac{9}{7}$)

D (Polyatomic): Corresponds to I ($\frac{4+f}{3+f}$)

Consequently, the correct match is A-II, B-III, C-IV, D-I.

Given,

Area of thickness is same for both plates.

Free face of brass $=0^{\circ} \mathrm{C}$

Free face of copper $=100^{\circ} \mathrm{C}$

$ \frac{K_B}{K_C}=\frac{1}{4} \Rightarrow K_C=4 K_B $

$ \begin{aligned} &\text { Let, the temperature of interface }=T^{\circ} \mathrm{C}\\ &100^{\circ} \mathrm{C} \quad T^{\circ} \mathrm{C} \quad 0^{\circ} \mathrm{C} \end{aligned} $

$ \begin{array}{|l|l|} \hline \text { Copper } & \text { Brass } \\ \hline \end{array} $

Heat transfer by copper = Heat gained by brass

$ \begin{aligned} & \left(\frac{Q}{t}\right)_{\mathrm{Cu}}=\left(\frac{Q}{t}\right)_{\text {brass }} \\ & \Rightarrow \quad \frac{K_C A\left(100^{\circ}-T\right)}{l}=\frac{K_B A\left(T-0^{\circ}\right)}{l} \\ & \Rightarrow \quad 4 K_B\left(100^{\circ}-T\right)=K_B(T-0) \\ & \Rightarrow \quad 400-4 T=T \Rightarrow 400=5 T \\ & \Rightarrow \quad T=\frac{400}{5}=80^{\circ} \mathrm{C} \end{aligned} $

For an ideal monatomic gas, the internal energy depends only on the temperature, not on the process by which the temperature change occurs. Here’s a quick breakdown:

The internal energy of a monatomic ideal gas is given by:

$U = \frac{3}{2} nRT$

where:

$n$ is the number of moles,

$R$ is the gas constant,

$T$ is the temperature.

The change in internal energy when the gas is heated from $T_1$ to $T_2$ is therefore:

$\Delta U = \frac{3}{2} nR (T_2 - T_1)$

This expression is valid regardless of whether the heating occurs at constant volume or constant pressure, because the internal energy is a state function—it depends only on the initial and final temperatures.

Thus, the change in internal energy is the same in both cases.

Correct Answer: Option C.

When the temperatures are $ T_1 = 200^{\circ} \mathrm{C} = 473 \mathrm{~K} $ and $ T_2 = 0^{\circ} \mathrm{C} = 273 \mathrm{~K} $, the efficiency $\eta_1$ of the Carnot engine is calculated as:

$ \eta_1 = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} = 1 - 0.577 = 0.423 $

When the temperatures are $ T_1 = 0^{\circ} \mathrm{C} = 273 \mathrm{~K} $ and $ T_2 = -200^{\circ} \mathrm{C} = 73 \mathrm{~K} $, the efficiency $\eta_2$ is calculated as:

$ \eta_2 = 1 - \frac{T_2}{T_1} = 1 - \frac{73}{273} = 1 - 0.267 = 0.733 $

Therefore, the ratio of the efficiencies $\frac{\eta_1}{\eta_2}$ is given by:

$ \frac{\eta_1}{\eta_2} = \frac{0.423}{0.733} \approx 0.577 $

Thus, $\frac{\eta_1}{\eta_2} \approx 0.58$.

We know that,

In a cyclic process, $\Delta U=0$

According to first law of thermodynamics,

$ \begin{aligned} & \Delta Q=\Delta U+W \\ & \Delta Q=0+\text { Area under } p-V \text { curve } \end{aligned} $

$ \begin{aligned} & \Delta Q=\pi \times\left(10 \times 10^3\right) \times\left(10 \times 10^{-3}\right) \\ & \Delta Q=\pi \times 10 \times 10 \\ & \Delta Q=10^2 \pi \mathrm{~J} \end{aligned} $

According to the law of equipartition of energy, the energy associated with each degree of freedom of a molecule is $\frac{1}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin.

For a polyatomic gas with $n$ degrees of freedom, the total mean kinetic energy per molecule can be calculated as:

$ \text{Mean kinetic energy per molecule} = \frac{1}{2} n k_B T $

This formula reflects that each degree of freedom contributes $\frac{1}{2} k_B T$ to the total energy, and the number of degrees of freedom, $n$, multiplies this contribution to give the total mean kinetic energy of each molecule in the gas.

A perfect black body is defined as an object that absorbs all incident electromagnetic radiation. This means the fraction of radiation it absorbs, known as the absorption coefficient or absorptivity, is exactly $1$.

Here's a brief explanation:

The absorption coefficient represents the proportion of radiation absorbed.

For a perfect black body, by definition, no radiation is reflected or transmitted.

Therefore, the absorption coefficient is $1$, meaning it absorbs 100% of the incident radiation.

So, the correct answer is:

Option D: $1$.

Given that the initial temperature $ T_1 = 300 \, \text{K} $, the ratio of specific heats $ \gamma = 1.50 $, and the final volume $ V_2 = 2V_1 $, we can calculate the resultant fall in temperature during adiabatic expansion using the following process:

For an adiabatic expansion, the relation is given by:

$ TV^{\gamma-1} = \text{constant} $

This implies:

$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} $

Substitute the given values:

$ T_2 = T_1 \left(\frac{1}{2}\right)^{1.50-1} = 300 \left(\frac{1}{2}\right)^{0.5} $

Calculating $ T_2 $, we find:

$ T_2 = 213.13 \, \text{K} $

Thus, the resultant fall in temperature is:

$ = T_1 - T_2 \approx 88 \, \text{K} $

To solve this problem, we need to find how much we must increase the temperature of the source to double the efficiency of the Carnot engine from 25% to 50%.

Given:

Sink temperature, $ T_L = 300 \, \text{K} $

Initial efficiency, $ \eta = 25\% = 0.25 $

Step 1: Calculate the Original Source Temperature

Using the efficiency formula for a Carnot engine:

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

Substitute the known values:

$ 0.25 = 1 - \frac{300}{T_H} $

Solving for $ T_H $:

$ T_H = \frac{300}{1 - 0.25} = \frac{300}{0.75} = 400 \, \text{K} \quad \ldots \text{ (i) } $

Step 2: Calculate the New Source Temperature for $ 50\% $ Efficiency

New efficiency, $ \eta^{\prime} = 50\% = 0.50 $

Using the efficiency formula again:

$ \eta^{\prime} = 1 - \frac{T_L}{T_H^{\prime}} $

Substitute the values:

$ 0.50 = 1 - \frac{300}{T_H^{\prime}} $

Solving for $ T_H^{\prime} $:

$ T_H^{\prime} = \frac{300}{1 - 0.50} = \frac{300}{0.50} = 600 \, \text{K} \quad \ldots \text{ (ii) } $

Step 3: Calculate the Increase in Temperature

The increase in temperature required is:

$ = T_H^{\prime} - T_H = 600 \, \text{K} - 400 \, \text{K} = 200 \, \text{K} $

Thus, the required increase in the source temperature is 200 K.

Given the following:

Heat supplied, $ Q = 100\, \text{J} $

Increase in internal energy, $ \Delta U = 60\, \text{J} $

Using the first law of thermodynamics, we have:

$ Q = W + \Delta U $

Substitute the given values:

$ 100 = W + 60 $

$ W = 40\, \text{J} \quad \ldots \text{(i)} $

For a monoatomic gas, the heat capacity ratio $ \gamma = \frac{5}{3} $.

The ratio of work done ($ \Delta W $) to the heat supplied ($ \Delta Q $) for a monoatomic gas is:

$ \frac{\Delta W}{\Delta Q} = \frac{\gamma - 1}{\gamma} = \frac{\frac{5}{3} - 1}{\frac{5}{3}} = 0.4 \text{ or } 40\% \quad \ldots \text{(ii)} $

This calculation indicates that 40% of the supplied heat is converted into work done in a monoatomic gas.

Equation (i) confirms this analysis.

Therefore, the gas is monoatomic in nature.

Given:

Volume, $ V = 3 \, \text{m}^3 $

Pressure, $ p = 3 \times 10^5 \, \text{Pa} $

The internal energy of an ideal gas can be calculated using the formula:

$ E = \frac{3}{2} p V $

Substituting the given values:

$ E = \frac{3}{2} \times 3 \times 10^5 \times 3 $

Calculating the above expression gives:

$ E = 13.5 \times 10^5 \, \text{J} $

To determine the ratio of the lengths of two cylindrical rods, let's review the thermal conductivity formula:

$ K = \frac{Q \cdot L}{A \cdot \Delta T} $

Where:

$ Q $ is the rate of heat flow.

$ L $ is the length of the rod.

$ A $ is the cross-sectional area.

$ \Delta T $ is the temperature difference across the rod.

Since the rods $ A $ and $ B $ are made of the same material, their thermal conductivity, $ K $, is equal. This gives us:

$ K_A = \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} \quad \text{and} \quad K_B = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Given that the rods have the same mass and material, their volumes are equal:

$ V_A = V_B $

Thus, the equation for equal volumes becomes:

$ A_A \cdot L_A = A_B \cdot L_B $

From this, we derive:

$ \frac{L_A}{L_B} = \frac{A_B}{A_A} $

Since $ K_A = K_B $, we equate the expressions:

$ \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Substituting the ratio of volumes, we get:

$ \frac{L_A}{L_B} = \frac{Q_B}{Q_A} \times \frac{A_A}{A_B} \times \frac{\Delta T_A}{\Delta T_B} $

Using the given values:

$ \frac{L_A}{L_B} = \frac{8}{3} \times \frac{L_B}{L_A} \times \frac{40}{60} $

Solving the equation:

$ \frac{L_A^2}{L_B^2} = \frac{8}{3} \times \frac{4}{6} = \frac{32}{18} = \frac{16}{9} $

Taking the square root gives:

$ \frac{L_A}{L_B} = \sqrt{\frac{16}{9}} = \frac{4}{3} $

Thus, the ratio of the lengths of rods $ A $ and $ B $ is $ \frac{4}{3} $.

Given the efficiency of the Carnot cycle, $\eta = \frac{1}{6}$, and a scenario where the sink's temperature is lowered by 65 K, resulting in an increased efficiency of $\eta_2 = \frac{1}{3}$.

Efficiency of the Carnot Cycle

The efficiency $\eta$ of a Carnot cycle is expressed by:

$ \eta_1 = 1 - \frac{T_2}{T_1} $

where $T_2$ is the temperature of the sink, and $T_1$ is the temperature of the source.

Given:

$ \frac{1}{6} = 1 - \frac{T_2}{T_1} $

Now, when the sink's temperature is reduced by 65 K, the efficiency becomes:

$ \eta_2 = 1 - \frac{(T_2 - 65)}{T_1} $

Calculation

Substitute the expression for $\frac{T_2}{T_1}$ from the first situation into the new efficiency equation:

$ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} $

So, for the new condition:

$ \frac{1}{3} = 1 - \left( \frac{T_2}{T_1} - \frac{65}{T_1} \right) $

Substitute the value from the previous step:

$ \frac{65}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{3}{18} = \frac{1}{6} $

Solve for $T_1$:

$ T_1 = 65 \times 6 = 390 \, \text{K} $

Initial Temperature of Sink

Calculate $T_2$ using the ratio determined earlier:

$ \frac{T_2}{T_1} = \frac{5}{6} $

Thus:

$ T_2 = \frac{5}{6} \times 390 = 325 \, \text{K} $

Final Temperature of Sink

When the sink's temperature is lowered by 65 K:

$ T_2 - 65 = 325 - 65 = 260 \, \text{K} $

The initial temperature of the sink is 325 K, and the final temperature of the sink is 260 K.

To determine the minimum power output of the motor needed to prevent ice melting in a refrigerator, we must use the concept of the Coefficient of Performance (COP) for a refrigeration system.

The COP is defined as:

$ \omega = \frac{T_2}{T_1 - T_2} $

Where:

$ T_2 = 273 $ K (temperature inside the refrigerator, equivalent to 0°C)

$ T_1 = 293 $ K (external temperature, equivalent to 20°C)

Plugging in the values:

$ \omega = \frac{273}{293 - 273} = \frac{273}{20} $

Next, calculate the rate of ice melting. Given that ice melts at the rate of 2 kg per hour, convert this to grams per second:

$ \text{Rate of mass of ice melting per second} = \frac{2 \times 1000}{3600} = \frac{5}{9} \text{ g/s} $

To prevent the ice from melting, the heat that needs to be expelled per second, $ Q_2 $, is determined using the latent heat of fusion for ice, $ 80 \, \text{cal/g} $:

$ Q_2 = \left(\frac{5}{9}\right) \times 80 = \frac{400}{9} = 44.44 \, \text{cal/s} $

Convert $ Q_2 $ from calories to Joules (since $ 1 \, \text{cal} = 4.186 \, \text{J} $):

$ Q_2 = 44.44 \times 4.186 = 186 \, \text{J/s} $

Given this heat must be extracted to prevent melting, and using the COP:

$ \omega = \frac{Q_2}{W} \implies W = \frac{Q_2}{\omega} = \frac{186}{\left(\frac{273}{20}\right)} = 13.6 \, \text{J} $

Thus, in an ideal refrigerator, the minimum power output of the motor required is:

$ P_{\text{min}} = \frac{W}{1 \, \text{s}} = 13.6 \, \text{J/s} = 13.6 \, \text{W} $

Therefore, the minimum necessary power output is 13.6 W.

The efficiency of a Carnot engine is expressed as:

$ \eta = 1 - \frac{T_2}{T_1} $

where:

$ T_1 $ is the temperature of the source,

$ T_2 $ is the temperature of the sink, and

$ T_1 > T_2 $.

Case 1:

For a Carnot engine operating between 800 K and 500 K, the efficiency is:

$ \eta_1 = 1 - \frac{500}{800} $

Case 2:

For another Carnot engine with unknown source temperature $ x $ and a sink at 600 K, the efficiency is:

$ \eta_2 = 1 - \frac{600}{x} $

Given that the efficiencies are the same in both cases:

$ \begin{aligned} &\eta_1 = \eta_2 \\ &1 - \frac{600}{x} = 1 - \frac{500}{800} \\ &\frac{600}{x} = \frac{5}{8} \end{aligned} $

Solving for $ x $:

$ \begin{aligned} x &= \frac{600 \times 8}{5} \\ x &= \frac{4800}{5} \\ x &= 960 \, \text{K} \end{aligned} $

The rms velocity (root mean square velocity) of gas molecules is given by the formula:

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

Where:

$ R $ is the Universal gas constant

$ T $ is the temperature of the gas in Kelvin

$ M $ is the molar mass of the gas

For a constant $ R $ and $ M $, the rms velocity is proportional to the square root of the temperature:

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

Given a change in temperature from $ T' = 27^{\circ} \mathrm{C} $ to $ T'' = 90^{\circ} \mathrm{C} $, we first convert these temperatures to Kelvin:

$ T' = 273 + 27 = 300 \, \text{K} $

$ T'' = 273 + 90 = 363 \, \text{K} $

The ratio of the rms velocities at these temperatures is:

$ \frac{v_{\text{rms,new}}}{v_{\text{rms,old}}} = \sqrt{\frac{T''}{T'}} = \sqrt{\frac{363}{300}} $

This calculation gives:

$ \frac{v_{\text{rms,new}}}{v_{\text{rms,old}}} = \sqrt{1.21} \approx 1.1 $

The percentage increase in the rms velocity of the gas molecules is calculated by:

$ \left( \frac{v_{\text{rms,new}} - v_{\text{rms,old}}}{v_{\text{rms,old}}} \right) \times 100 = (1.1 - 1) \times 100 = 0.1 \times 100 = 10\% $

Thus, the increase in the rms velocity of the gas molecules is 10%.

According to Stefan's law, the rate of cooling is proportional to the difference between the fourth powers of the object's temperature and the ambient temperature:

$ H \propto \left(T^4 - T_0^4\right) $

Given:

Ambient temperature, $ T_0 = 300 \, \text{K} $

Rate of cooling at $ 600 \, \text{K} $ is $ H $

We are asked to find the rate of cooling at $ 900 \, \text{K} $.

Using Stefan's law, compare the rates of cooling:

$ \frac{H'}{H} = \frac{(900)^4 - (300)^4}{(600)^4 - (300)^4} $

Calculate:

$ 900^4 - 300^4 $

$ 600^4 - 300^4 $

$ \Rightarrow \frac{H'}{H} = \frac{9^4 - 3^4}{6^4 - 3^4} $

Simplify the expression:

$ 9^4 = 6561 $

$ 3^4 = 81 $

$ 6^4 = 1296 $

Substitute these values back:

$ \frac{H'}{H} = \frac{6561 - 81}{1296 - 81} = \frac{6480}{1215} = \frac{16}{3} $

Thus, the rate of cooling at $ 900 \, \text{K} $ is:

$ H' = \frac{16}{3} H $

Explanation:

For a Carnot cycle, the efficiency ($\eta$) of the heat engine is defined as:

$ \eta = 1 - \frac{T_2}{T_1} $

where $ T_1 $ is the absolute temperature of the heat source (in Kelvin), and $ T_2 $ is the absolute temperature of the heat sink (in Kelvin).

Convert the given temperatures from Celsius to Kelvin:

$ T_1 = 127^\circ \mathrm{C} + 273 = 400 \, \mathrm{K} $

$ T_2 = 27^\circ \mathrm{C} + 273 = 300 \, \mathrm{K} $

Plug these values into the efficiency formula:

$ \eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25 $

This means the efficiency is $ \frac{1}{4} $.

The efficiency also relates to the work output ($ W_{\text{out}} $) and the heat absorbed ($ W_{\text{absorb}} $) as follows:

$ \eta = \frac{W_{\text{out}}}{W_{\text{absorb}}} $

Given that the heat absorbed ($ W_{\text{absorb}} $) is $ 5 \times 10^4 $ cal, we can calculate the work output:

$ W_{\text{out}} = \frac{1}{4} \times 5 \times 10^4 = 1.25 \times 10^4 \, \mathrm{cal} $

Thus, the amount of heat converted to work is $ 1.25 \times 10^4 $ cal.

To find the value of $\gamma$ for the mixture of two gases, we use the formula:

$ \frac{n_1 + n_2}{\gamma_{\text{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} $

For our specific case:

$n_1 = 1$, $\gamma_1 = \frac{7}{5}$

$n_2 = 1$, $\gamma_2 = \frac{4}{3}$

Substituting these into the formula gives:

$ \frac{1 + 1}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{7}{5} - 1} + \frac{1}{\frac{4}{3} - 1} $

This simplifies to:

$ \frac{2}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{3}} $

Calculating further:

$ \frac{2}{\gamma_{\text{mix}} - 1} = \frac{5}{2} + 3 = \frac{5}{2} + \frac{9}{3} = \frac{5}{2} + \frac{6}{2} = \frac{11}{2} $

Now, solving for $\gamma_{\text{mix}}$:

$ \gamma_{\text{mix}} - 1 = \frac{2}{\frac{11}{2}} = \frac{4}{11} $

So, $\gamma_{\text{mix}}$ is:

$ \gamma_{\text{mix}} = 1 + \frac{4}{11} = \frac{15}{11} $

A Carnot heat engine with an efficiency of 10% can be transformed into a refrigerator. To find the coefficient of performance (COP) of this refrigerator, we use the relationships between efficiency and temperature in a Carnot cycle.

The efficiency $\eta$ of a Carnot engine is given by:

$ \eta = 1 - \frac{T_2}{T_1} $

Rearranging this formula, we find:

$ \frac{T_1}{T_2} = \frac{1}{1 - \eta} \tag{i} $

For a refrigerator, the coefficient of performance $\omega$ is defined as:

$ \omega = \frac{T_2}{T_1 - T_2} $

Substituting equation (i) into this, we have:

$ \omega = \frac{1}{\left(\frac{T_1}{T_2}\right) - 1} = \frac{1}{\left[\frac{1}{1 - \eta}\right] - 1} = \frac{1 - \eta}{\eta} $

Plugging in the given efficiency ($\eta = 0.1$):

$ \omega = \frac{1 - 0.1}{0.1} = \frac{0.9}{0.1} = 9 $

Therefore, the coefficient of performance of the refrigerator is 9.

To determine the relationship between the root-mean-square (rms) velocity of gas molecules and their mass, we use the formula:

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

In this equation:

$ v_{\mathrm{rms}} $ is the rms velocity,

$ R $ is the gas constant,

$ T $ is the temperature,

$ m $ is the mass of the gas molecules.

From the equation, it's clear that:

$ v_{\mathrm{rms}} \propto \frac{1}{\sqrt{m}} $

This means the rms velocity of gas molecules is inversely proportional to the square root of their mass. As the mass $ m $ of the molecules increases, the rms velocity decreases, and vice versa.

The dimensions of various thermal properties are given and described as follows:

Thermal Conductivity (K):

Formula: $ K = \frac{Q \cdot d}{A \cdot \Delta T} $

Dimensional Analysis:

$ \left[ \frac{(ML^2T^{-3}) (L)}{(L^2)(K)} \right] = [MLT^{-3}K^{-1}] $

Boltzmann Constant (k):

Defined as the ratio of energy to temperature:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}] $

Latent Heat (L):

Defined as the ratio of energy to mass:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}] $

Specific Heat (C):

Defined as the ratio of energy to the product of mass and temperature:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[M][K]} = [M^0L^2T^{-2}K^{-1}] $

These analyses help in determining the correct dimension pairings for each property.

Given:

Length of the rod, $ x = 20 \, \text{cm} = 0.2 \, \text{m} $

Temperatures at the ends:

$ T_1 = 0^{\circ} \text{C} = 273 \, \text{K} $

$ T_2 = 100^{\circ} \text{C} = 373 \, \text{K} $

Cross-sectional area, $ A = 4 \times 10^{-4} \, \text{m}^2 $

Mass of ice melted, $ m = 5 \, \text{g} $

Time, $ t = 1 \, \text{min} = 60 \, \text{s} $

Latent heat of fusion, $ L_{\text{ice}} = 80 \, \text{cal/g} = 336 \, \text{J/g} $

Calculation:

Total Heat Required to Melt the Ice:

$ Q = m \times L_{\text{ice}} = 5 \times 336 = 1680 \, \text{J} $

Rate of Heat Transfer:

$ \frac{Q}{t} = \frac{1680}{60} = 28 \, \text{J/s} $

Thermal Conductivity ($ K $):

$ K = \frac{\Delta Q \cdot x}{A \cdot \Delta T} = \frac{28 \times 0.2}{4 \times 10^{-4} (373 - 273)} $

$ K = \frac{1.4 \times 10^4}{100} $

$ K = 1.4 \times 10^2 $

$ K = 140 \, \text{W/mK} $

Given:

Initial work done by the gas, $ W_1 = W $

Final volume for the first process, $ V_{f_1} = 2V $

Initial volume for the first process, $ V_{i_1} = V $

Temperature for the first process, $ T_1 = T $

Number of moles for the first process, $ n = 2 $

The work done by an ideal gas during an isothermal expansion is given by the equation:

$ W = n \cdot R \cdot T \ln \left(\frac{V_f}{V_i}\right) $

Substituting the given values for the first process:

$ W_1 = 2 \cdot R \cdot T \ln \left(\frac{2V}{V}\right) = 2 \cdot R \cdot T \ln 2 $

For the second process:

Final volume, $ V_{f_2} = 8V $

Initial volume, $ V_{i_2} = V $

Temperature, $ T_2 = \frac{T}{2} $

Number of moles, $ n_2 = 4 $

The work done for the second process is:

$ W_2 = n_2 \cdot R \cdot T_2 \ln \left(\frac{V_{f_2}}{V_{i_2}}\right) $

Substituting the values:

$ W_2 = 4 \cdot R \cdot \frac{T}{2} \ln \left(\frac{8V}{V}\right) $

$ W_2 = 2 \cdot R \cdot T \ln \left(2^3\right) $

$ W_2 = 3 \cdot 2 \cdot R \cdot T \ln 2 $

This simplifies to:

$ W_2 = 3 \cdot W $

Hence, the work done by the ideal gas in the second process is $ 3W $.

To analyze the increase in internal energy when heat is absorbed by a monoatomic gas:

First, we know that the heat absorbed, $ Q $, is given as 40 J.

For a monoatomic ideal gas, the specific heat capacities are:

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

$ C_V = \frac{3R}{2} $

The relationship for heat transfer in terms of $ C_p $ is:

$ Q = n C_p \Delta T $

Substituting in known values:

$ 40 = n \cdot \frac{5R}{2} \cdot \Delta T $

From this, we obtain:

$ nR \Delta T = 16 \quad (i) $

Next, we determine the increase in internal energy, $ \Delta U $, which is given by:

$ \Delta U = n C_V \Delta T $

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

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

Using Equation (i), we substitute for $ nR \Delta T $:

$ \Delta U = \frac{3}{2} \times 16 $

This simplifies to:

$ \Delta U = 24 \, \text{J} $

Thus, the increase in the internal energy of the gas is 24 J.

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

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

Given that the absolute temperature of the source ($T_{\text{source}}$) is 25% more than that of the sink ($T_{\text{sink}}$), we can express this relationship as:

$ T_{\text{source}} = T_{\text{sink}} + \frac{25 \times T_{\text{sink}}}{100} $

Simplifying, this becomes:

$ T_{\text{source}} = 1.25 \times T_{\text{sink}} $

Substituting into the efficiency equation, we have:

$ \eta = \left(1 - \frac{T_{\text{sink}}}{1.25 \times T_{\text{sink}}}\right) \times 100\% $

This simplifies to:

$ \eta = \left(1 - \frac{1}{1.25}\right) \times 100\% $

$ \eta = (1 - 0.8) \times 100\% $

$ \eta = 0.2 \times 100\% $

Thus, the efficiency of the Carnot engine is:

$ \eta = 20\% $

Given:

Molar heat capacity at constant pressure for a diatomic gas, $ C_p = C $

We need to find the molar heat capacity at constant volume for a monoatomic gas, $ C_V $.

Calculations:

For a Diatomic Gas:

For a diatomic gas, the molar heat capacity at constant volume is:

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

Therefore, the relation between $ C_p $ and $ C_V $ is:

$ C_p = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R $

Since $ C_p = C $, we have:

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

Solving for $ R $, we get:

$ R = \frac{2C}{7} \tag{i} $

For a Monoatomic Gas:

The molar heat capacity at constant volume for a monoatomic gas is:

$ C_V = \frac{3}{2} R $

Substituting the value of $ R $ from Equation (i):

$ C_V = \frac{3}{2} \cdot \frac{2C}{7} = \frac{3C}{7} $

Therefore, the molar specific heat capacity of a monoatomic gas at constant volume is $ \frac{3C}{7} $.

The relationship between the root-mean-square (rms) speed ($v_{rms}$) and the average speed ($v_{avg}$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:

Where:

• $R$ is the ideal gas constant
• $T$ is the temperature in Kelvin
• $M$ is the molar mass of the gas
• $\pi$ is the mathematical constant pi

In this problem, the rms speed of the oxygen molecule is given by:

Now, let's divide the expression for $v_{rms}$ by the expression for $v_{avg}$:

By simplifying the expression, we get:

Square both sides of the equation:

Now we will substitute the provided value of $\pi = \frac{22}{7}$:

By simplifying the expression, we get:

Now let's solve for $x$:

Multiplying both sides by $x$:

Finally, solving for $x$:

So, the value of $x$ is $\boxed{28}$.

To answer this question, we need to find the efficiency of a Carnot engine operating between the boiling and freezing points of water. The boiling point of water is 100°C (373 K) and the freezing point is 0°C (273 K). The efficiency of a Carnot engine is given by the formula:

Efficiency = $1 - \frac{T_{cold}}{T_{hot}}$

Plugging in the values for the boiling and freezing points of water:

Efficiency = $1 - \frac{273}{373} = 1 - 0.732 = 0.268$

So, the efficiency of a Carnot engine operating between these temperatures is approximately 26.8%.

Now, let's analyze the given options: A. efficiency more than 27% - This is incorrect, as the Carnot efficiency is 26.8% and no engine can be more efficient than a Carnot engine.

B. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures - This is correct, as real engines are less efficient than Carnot engines operating between the same temperatures.

C. efficiency equal to $27 \%$ - This is incorrect, as the Carnot efficiency is 26.8%, not 27%.

D. efficiency less than $27 \%$ - This is correct, as the efficiency is less than 27% (26.8%).

Thus, the correct answer is "B and D only."

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$Q = m \times L$

where $m = 1 \mathrm{~kg}$ and $L = 2257 \mathrm{~kJ/kg}$:

$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$

Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:

$W = -P \Delta V$

where $P$ is the atmospheric pressure and $\Delta V$ is the change in volume. We are given that the atmospheric pressure is $P = 1 \times 10^5 \mathrm{~Pa}$, and the change in volume is

$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$.

Now, we can calculate the work done:

$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$

Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:

$\Delta U = Q + W$

Substitute the values of $Q$ and $W$:

$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$

The change in internal energy of the system during the process is +2090 kJ.

The root mean square speed ($v_{rms}$) of a gas is given by the formula:

$v_{rms} = \sqrt{\frac{3kT}{m}}$

where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molar mass of the gas molecules.

The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases. Their molar masses are:

1. Neon: $20.18\,\text{g/mol}$ (monoatomic)
2. Chlorine: $2 \times 35.45 = 70.90\,\text{g/mol}$ (diatomic)
3. Uranium hexafluoride: $238.03 + 6 \times 18.998 = 352.03\,\text{g/mol}$ (polyatomic)

Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:

$v_{rms} \propto \frac{1}{\sqrt{m}}$

The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:

$v{rms}(\text{mono}) > v{rms}(\text{dia}) > v_{rms}(\text{poly})$

The internal energy (U) of a gas depends on its degrees of freedom (f).

For a monatomic gas like neon, the degrees of freedom are f = 3 (translational). For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).

The internal energy for each component of the gas mixture can be calculated using the formula:

$U = \frac{f}{2}nRT$

where n is the number of moles, R is the universal gas constant, and T is the temperature.

For the 4 moles of neon (monatomic):

$U_{Ne} = \frac{3}{2} \cdot 4RT = 6RT$

For the 2 moles of oxygen (diatomic):

$U_{O_2} = \frac{5}{2} \cdot 2RT = 5RT$

Now, to find the total internal energy, we sum the internal energies of the individual components:

$U_{total} = U_{Ne} + U_{O_2} = 6RT + 5RT = 11RT$

Thus, the total internal energy of the system is 11RT.

An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings. This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.

Given the options:

Option A: There is no heat supplied to the system.

• This statement is TRUE. In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.

Option B: The temperature of the gas increases.

• This statement is TRUE. When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.

Option C: There is no change in the internal energy.

• This statement is NOT TRUE. In an adiabatic process, the change in internal energy of the system is equal to the work done on the system. When the gas is compressed, work is done on the gas, which increases its internal energy.

Option D: The change in the internal energy is equal to the work done on the gas.

• This statement is TRUE. As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

So, Option C is the statement that is not true for an adiabatic process.

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.

The final volume is $\frac{1}{8}$ of the initial volume, so $P_2 = P_1 \times \frac{V_1}{V_2} = P_1 \times 8 = 8P_1$.

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = \text{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.

Since $V_2 = \frac{V_1}{8}$, we have $P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) ^ \gamma = P_1 \times 8^\frac{5}{3} = P_1 \times 2^5 = 32P_1$.

The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1} = 4$.

Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).

Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).

Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (III) matches with (B).

Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom. So, option (IV) matches with (C).

The kinetic energy of an ideal gas is given by the equation:

$KE = \frac{3}{2} kT$

where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double. The original temperature is given as ($27^\circ C$), which is equal to (300 K) in absolute terms. Therefore, the final temperature ($T_f$) in kelvins is:

$T_f = 2 \cdot 300 K = 600 K$

Converting this back to degrees Celsius gives:

$T_f = 600K - 273 = 327 ^\circ C$

The efficiency of a Carnot engine is given by:

$\eta = 1 - \frac{T_{c}}{T_{h}}$

where:

• $\eta$ is the efficiency of the Carnot engine,
• $T_{c}$ is the temperature of the cold reservoir,
• $T_{h}$ is the temperature of the hot reservoir.

Note that the temperatures must be in Kelvin for this formula.

Here, the given temperatures are $127^{\circ}C$ and $27^{\circ}C$. Converting these to Kelvin gives:

$T_{h} = 127^{\circ}C + 273.15 = 400.15 K$ $T_{c} = 27^{\circ}C + 273.15 = 300.15 K$

So the efficiency of the Carnot engine is:

$\eta = 1 - \frac{300.15}{400.15} = 0.25$

The efficiency can also be defined as the work done divided by the heat supplied:

$\eta = \frac{W}{Q_{h}}$

where:

• $W$ is the work done by the engine,
• $Q_{h}$ is the heat transferred to the engine by the hot reservoir.

We can rearrange this equation to solve for $Q_{h}$:

$Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ$

Statement I: If heat is added to a system, its temperature must increase. This statement is not necessarily true. For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature. The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.