Heat and Thermodynamics

In a P-V diagram (Pressure vs. Volume), the total work done during a cyclic process is equivalent to the area enclosed by the cycle.

The cycle $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ is moving in a clockwise direction.

In a clockwise cycle on a P-V diagram, the work done by the system is positive.

The area enclosed by the triangle ABC is calculated using the formula for the area of a right-angled triangle:

$ \begin{aligned} & \text { Work Done }=\text { Area }=\frac{1}{2} \times \text { Base × Height } \\ & \text { Base }=V_C-V_A=5 \mathrm{~m}^3-2 \mathrm{~m}^3=3 \mathrm{~m}^3 \\ & \text { Height }=P_B-P_C=300 \mathrm{~Pa}-100 \mathrm{~Pa}=200 \mathrm{~Pa} \end{aligned} $

Substituting the values into the area formula:

$ \text { Work Done }=\frac{1}{2} \times 3 \mathrm{~m}^3 \times 200 \mathrm{~Pa}=300 \mathrm{~J} $

Therefore, the total work done by the system during the cycle ABC is 300 J . Hence, the correct answer is $\mathbf{3 0 0}$.

For a constant volume process, the heat supplied ( Q ) is given by the formula :

$ \mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{~T} $

Where :

• 1. $\mathrm{n}=$ number of moles of the gas.

• 2. $ \mathrm{C}_{\mathrm{v}}=$ molar specific heat capacity at constant volume.

• 3. $\Delta \mathrm{T}=$ change in temperature $50^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=30^{\circ} \mathrm{C}=30 \mathrm{~K}$

We are given the molar specific heat at constant pressure, $C_p=\frac{7}{2} R$. Using Mayer's Relation $\left(C_p-C_v=R\right)$ :

$ C_v=C_p-R $

$\Rightarrow $ $ C_v=\frac{7}{2} R-R=\frac{5}{2} R $

Using the formula of heat,

$ 300=\mathrm{n} \times\left(\frac{5}{2} \times 8.314\right) \times 30 $

$ \mathrm{n}=\frac{300 \times 2}{5 \times 8.314 \times 30} \approx 0.481 \mathrm{moles} $

So, the amount of the gas is 0.481 moles. As the molar mass of the gas is not given in the question, we can't determine the mass of the gas directly.

The problem states the gas is filled in a closed cylinder. That means the volume of the gas remains constant during the heating process ( $\mathrm{V}=$ constant).

For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature.

$ \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} $

Initial pressure is $\mathrm{P}_1=3.23 \mathrm{kPa}=3.23 \times 1000 \mathrm{~Pa}=3230 \mathrm{~Pa}$

Initial temperature is $\mathrm{T}_1=50^{\circ} \mathrm{C}=(50+273) \mathrm{K}=323 \mathrm{~K}$

The final temperature is $\mathrm{T}_2=2 \times \mathrm{T}_1=2 \times 323 \mathrm{~K}=646 \mathrm{~K}$

Using the relation $\frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}$ :

$ \mathrm{P}_2=\mathrm{P}_1 \times \frac{\mathrm{T}_2}{\mathrm{~T}_1} $

$\Rightarrow $ $P_2=3230 \times \frac{646}{323}$

$ \mathrm{P}_2=3230 \times 2=6460 \mathrm{~Pa} $

Therefore, the modified pressure is 6460 Pa . Hence, the correct answer is 6460 .

Other Solution :

$\begin{aligned} & \mathrm{V}=\text { constant } \\ & \text { so } \mathrm{P} \propto \mathrm{T} \\ & \mathrm{T}_{\mathrm{i}}=50^{\circ} \mathrm{C}=323 \mathrm{~K} \\ & \mathrm{~T}_{\mathrm{f}}=100^{\circ} \mathrm{C}=373 \mathrm{~K} \\ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{P}_{\mathrm{i}}}=\frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \\ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{3.23 \mathrm{kPa}}=\frac{373}{323} \\ & \Rightarrow \mathrm{P}_{\mathrm{f}}=3730 \mathrm{~Pa}\end{aligned}$

Initial state of the gas,

Initial Volume, $V_1=60 \mathrm{~cm}^3$

Initial Temperature, $\mathrm{T}_1=27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{~K}$

Initial Pressure, $\mathrm{P}_1=2 \mathrm{~atm}$

Final state of the gas is,

Final Volume, $V_2=20 \mathrm{~cm}^3$

Final Temperature, $\mathrm{T}_2=77^{\circ} \mathrm{C}=(77+273) \mathrm{K}=350 \mathrm{~K}$

We need to calculate the pressure of the gas at final state.

Using Ideal gas equation, $\mathrm{PV}=\mathrm{nRT} \Rightarrow \frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}=$ constant

$ \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} $

$\Rightarrow \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_2}{\mathrm{~T}_1 \mathrm{~V}_2}$

$\Rightarrow $ $P_2=\frac{2 \times 60 \times 350}{300 \times 20} \mathrm{~atm}$

$\Rightarrow $ $\mathrm{P}_2=7 \mathrm{~atm}$

An isobaric process occurs at constant pressure (P). For isobaric process :

The work done is $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$

The change in internal energy is $\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}$

Heat added to the system is, $\mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

The ratio of specific heats is defined as $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{V}}}$.

Also, from Mayer's relation $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$.

The ratio of Heat to Work is:

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

Substituting $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ :

$ \frac{Q}{W}=\frac{C_p}{C_p-C_v} $

$\Rightarrow $ $ \frac{\mathrm{Q}}{\mathrm{~W}}=\frac{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}}{\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)-1}=\frac{\gamma}{\gamma-1} $

The work done is given as, $\mathrm{W}=100 \mathrm{~J}$ and the ratio of specific heats $\gamma=1.4$ (for a diatomic gas) So,

$ \frac{\mathrm{Q}}{100}=\frac{1.4}{1.4-1} $

$\Rightarrow $ $\frac{Q}{100}=\frac{1.4}{0.4}$

$\Rightarrow $ $\frac{Q}{100}=3.5$

$\Rightarrow $ $ Q=350 \mathrm{~J} $

Therefore, the heat given to the gas is 350 J .

Hence, the correct option is $\mathbf{3 5 0}$.

For an ideal gas, according to Mayer's relation

$ C_p-C_v=R $

Where R is the universal gas constant.

$ C_v=C_p-R $

Given:

• $\mathrm{C}_{\mathrm{p}}=7 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$

• $\mathrm{R}=2 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$

So, the molecular specific heat of oxygen at constant volume is,

$ \mathrm{C}_{\mathrm{v}}=7-2=5 \frac{\mathrm{cal}}{\mathrm{~mol}^{\circ} \mathrm{C}} $

Internal energy change is a state function that depends only on the temperature change. For n moles of an ideal gas, it is given by the formula:

$ \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} $

The given values are,

• Number of moles $\mathrm{n}=10 \mathrm{~mol}$

• Initial temperature, $\mathrm{T}_1=30^{\circ} \mathrm{C}$

• Final temperature $\mathrm{T}_2=40^{\circ} \mathrm{C}$

So, the change in temperature is, $\Delta \mathrm{T}=\mathrm{T}_2-\mathrm{T}_1=40-30=10^{\circ} \mathrm{C}$

So, the change in internal energy is,

$ \Delta U=10 \times 5 \times 10=500 \mathrm{cal} $

Therefore, when 10 moles of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ internal energy increase by 500 cal.

Hence, the correct answer is 500 .

Let the mass of water cooled from $50^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$ be $x$ gram.

Then the mass of water evaporated at $50^\circ\mathrm{C}$ is $(1000-x)$ gram.

According to the question, the heat released by cooling $x$ gram of water from $50^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$ is equal to the heat required to evaporate $(1000-x)$ gram of water at $50^\circ\mathrm{C}$.

We use

$ Q = ms\Delta T $

and for evaporation,

$ Q = mL $

Convert gram into kg:

$ x \text{ gram} = \frac{x}{1000}\text{ kg} $

and

$ (1000-x)\text{ gram} = \frac{1000-x}{1000}\text{ kg} $

Now, heat released by cooling $x$ gram water:

$ Q_1 = \frac{x}{1000}\times 4200 \times (50-0) $

$ Q_1 = \frac{x}{1000}\times 4200 \times 50 $

Heat required to evaporate $(1000-x)$ gram water:

$ Q_2 = \frac{1000-x}{1000}\times 2256\times 10^3 $

Since $Q_1 = Q_2$,

$ \frac{x}{1000}\times 4200\times 50 = \frac{1000-x}{1000}\times 2256\times 10^3 $

Cancel $\frac{1}{1000}$ from both sides:

$ x \times 4200 \times 50 = (1000-x)\times 2256\times 10^3 $

$ 210000x = 2256000(1000-x) $

$ 210000x = 2256000000 - 2256000x $

$ 210000x + 2256000x = 2256000000 $

$ 2466000x = 2256000000 $

$ x = \frac{2256000000}{2466000} $

$ x \approx 914.84 $

So, the closest integer is

$ \boxed{915} $

Each engine has the exact same individual thermal efficiency, $\eta$.

The efficiency of any heat engine is defined as the ratio of work done $\left(\mathrm{W}_{\mathrm{i}}\right)$ to the heat absorbed $\left(\mathrm{Q}_{\mathrm{in}, \mathrm{i}}\right)$ :

$ \eta=\frac{W_i}{Q_{i n, i}}=1-\frac{Q_{o u t, i}}{Q_{i n, i}} $

So, the ratio of heat rejected to heat absorbed for any single engine in the series is :

$ \frac{\mathrm{Q}_{o u t, \mathrm{i}}}{\mathrm{Q}_{\mathrm{in}, \mathrm{i}}}=1-\eta $

For engine $\mathrm{E}_1$

Absorbed heat is $Q_0$ from the first reservoir.

Rejected heat is $Q_1=Q_0(1-\eta)$

Work done is $\mathrm{W}_1=\eta \mathrm{Q}_0$.

For engine $\mathrm{E}_2$ :

Absorbs heat rejected by $E_1$, so $Q_{\text {absorbed }, 2}=Q_1=Q_0(1-\eta)$.

Rejected heat is $Q_2=Q_1(1-\eta)=Q_0(1-\eta)^2$

Work done is $\mathrm{W}_2=\eta \mathrm{Q}_1=\eta \mathrm{Q}_0(1-\eta)$

For engine $\mathrm{E}_3$ :

Absorbed heat is $Q_2$.

Rejected heat is $Q_3=Q_2(1-\eta)=Q_0(1-\eta)^3$.

Work done $\mathrm{W}_3=\eta \mathrm{Q}_2=\eta \mathrm{Q}_0(1-\eta)^2$

For engine $\mathrm{E}_4$ :

Absorbed heat is $Q_3$.

Rejects heat is $Q_4=Q_3(1-\eta)=Q_0(1-\eta)^4$.

Work done is $\mathrm{W}_4=\eta \mathrm{Q}_3=\eta \mathrm{Q}_0(1-\eta)^3$.

For engine $\mathrm{E}_5$ :

Absorbed heat is $Q_4$.

Rejected heat is $Q_5=Q_4(1-\eta)=Q_0(1-\eta)^5$.

Work done is $\mathrm{W}_5=\eta \mathrm{Q}_4=\eta \mathrm{Q}_0(1-\eta)^4$

The total work W performed by all 5 engines combined per cycle is the sum of their individual works:

$ \mathrm{W}=\mathrm{W}_1+\mathrm{W}_2+\mathrm{W}_3+\mathrm{W}_4+\mathrm{W}_5 $

By the principle of energy conservation across the entire system, the total work done must equal the initial heat input to the first engine minus the final heat rejected by the last engine :

$ \mathrm{W}=\mathrm{Q}_0-\mathrm{Q}_5 $

$ W=Q_0-Q_0(1-\eta)^5=Q_0\left[1-(1-\eta)^5\right] $

The net efficiency $\eta_{\text {net }}$ of the combined system is given by:

$ \eta_{\text {net }}=\frac{W}{Q_0}=1-(1-\eta)^5 $

It is given that $\eta_{\text {net }}=\frac{211}{243}$.

$ 1-(1-\eta)^5=\frac{211}{243} $

$ \begin{gathered} (1-\eta)^5=1-\frac{211}{243} \\ (1-\eta)^5=\frac{243-211}{243} \\ (1-\eta)^5=\frac{32}{243} \\ 1-\eta=\left(\frac{32}{243}\right)^{1 / 5} \\ 1-\eta=\frac{2}{3} \\ \eta=1-\frac{2}{3}=\frac{1}{3} \end{gathered} $

Therefore, the value of $\eta$ is $\frac{1}{3}$

The two compartments, $S_1$ and $S_2$, separated by a rigid, conducting partition $P_1$.

Section $S_1$ is bounded by the outer insulated container walls and the immovable, rigid partition $P_1$.

Because the boundaries cannot move, its volume remains completely constant $\mathrm{V}=$ constant. Therefore, the gas in $\mathrm{S}_1$ undergoes an isochoric (constant volume) process.

Section $\mathrm{S}_2$ is bounded on one side by a freely movable, frictionless, insulated piston $\mathrm{P}_2$.

Since it moves freely to balance the external atmospheric pressure, the pressure of the gas inside $S_2$ remains constant $\mathrm{P}=$ constant $=\mathrm{P}_{\mathrm{atm}}$.

Therefore, the gas in $\mathrm{S}_2$ undergoes an isobaric (constant pressure) process.

Both compartments contain $\mathrm{n}_{\mathrm{m}}=1$ mole of an ideal monoatomic gas.

For a monoatomic ideal gas, the molar heat capacity at constant volume $\left(\mathrm{C}_{\mathrm{v}}\right)$ is $\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}$

The molar heat capacity at constant pressure $(C_p)$ is given by $C_p = C_v + R$

$C_p = \frac{3}{2}R + R = \frac{5}{2}R$

Let $T_1$ be the instantaneous temperature of the gas in $S_1$ and $T_2$ be the instantaneous temperature of the gas in $S_2$ at any time $t$.

Let us assume that $S_1$ is initially at a higher temperature than $S_2$, so $T_1 > T_2$.

As time progresses, heat $dQ$ flows across the conducting partition $P_1$ from the hotter section $S_1$ to the cooler section $S_2$.

For section $S_1$ the gas undergoes isochoric cooling.

The heat lost by $S_1$ causes its temperature to decrease by $dT_1$:

$dQ = -n_m C_v\, dT_1 = -(1)\left(\frac{3}{2}R\right)dT_1 \rightarrow dT_1 = -\frac{2}{3R}dQ$

For section $S_2$ where the gas undergoes isobaric heating.

The same amount of heat $dQ$ enters $S_2$, causing its temperature to increase by $dT_2$:

$dQ = n_m C_p\, dT_2 = (1)\left(\frac{5}{2}R\right)dT_2 \rightarrow dT_2 = \frac{2}{5R}dQ$

$d(\Delta T) = dT_1 - dT_2$

$d(\Delta T) = -\frac{2}{3R}dQ - \frac{2}{5R}dQ = -\left(\frac{2}{3R} + \frac{2}{5R}\right)dQ$

$d(\Delta T) = -\left(\frac{10 + 6}{15R}\right)dQ = -\frac{16}{15R}dQ$

According to the rate of heat flow $\frac{dQ}{dt}$ through a conducting barrier of thermal conductivity $K$, area $A$, and thickness $x$ under a temperature difference $(T_1 - T_2)$ is given by:

$\frac{dQ}{dt} = \frac{KA}{x}(T_1 - T_2) = \frac{KA}{x}\,\Delta T$

Therefore, the infinitesimal heat transferred in a time interval $dt$ is:

$dQ = \frac{KA}{x}\Delta T\, dt$

Substituting the expression for $dQ$,

$d(\Delta T) = -\frac{16}{15R}\left(\frac{KA}{x}\Delta T\right)$

$\frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR}dt$

At $\mathrm{t}=0$, the initial temperature difference is $\Delta \mathrm{T}=\Delta \mathrm{T}_0$.

At $\mathrm{t}=\tau$, the final temperature difference becomes $\Delta \mathrm{T}=\frac{\Delta \mathrm{T}_0}{2}$.

$ \int\limits_{\Delta \mathrm{T}_0}^{\Delta \mathrm{T}_0 / 2} \frac{\mathrm{~d}(\Delta \mathrm{~T})}{\Delta \mathrm{T}}=-\frac{16 \mathrm{KA}}{15 \times \mathrm{R}} \int\limits_0^\tau \mathrm{dt} $

$\Rightarrow $ $[ n\left(\Delta T\right)]^{\Delta T_0/2}_{\Delta T_0} = -\frac{16KA}{15xR}\,\tau$

$\Rightarrow $ $\ln\left(\frac{\Delta T_0/2}{\Delta T_0}\right) = -\frac{16KA}{15xR}\,\tau$

$\Rightarrow $ $\ln \frac{1}{2} = -\frac{16KA}{15xR}\,\tau$

$\Rightarrow $ $-\ln 2 = -\frac{16KA}{xR}\,\tau \Rightarrow \ln 2 = \frac{16KA}{15xR}\,\tau$

$\Rightarrow $ $\tau = \frac{15}{16}\ln(2)\frac{xR}{KA}$

Using the given approximation $\ln 2 \approx 0.7$:

$\tau = \frac{15}{16} \times (0.7) \times \frac{xR}{KA} = \frac{10.5\,xR}{16\,KA} = 0.65625\,\frac{xR}{KA}$

$\Rightarrow n \approx 0.66$

Therefore, the correct answer is 0.66.

$\begin{aligned} & \mathrm{R}_1=\frac{\ell_1}{\mathrm{~K}_1 \mathrm{~A}_1}, \mathrm{R}_2=\frac{\ell_2}{\mathrm{~K}_2 \mathrm{~A}_2} \\ & \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\Delta \mathrm{T}}{\mathrm{R}} \\ & \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\left(\frac{\mathrm{dQ}}{\mu}\right)_2 \\ & \frac{400-\mathrm{T}}{\mathrm{R}_1}=\frac{\mathrm{T}-200}{\mathrm{R}_2} \\ & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\ell_1}{\ell_2}\right)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2 \times \frac{\mathrm{K}_2}{\mathrm{~K}_1} \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times 2 \\ & =\left(\frac{1}{4}\right) \end{aligned}$

$\begin{aligned} & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{1}{4} \\ & 1600-4 \mathrm{~T}=\mathrm{T}-200 \\ & 5 \mathrm{~T}=1800 \\ & \mathrm{~T}=360 \mathrm{~K} \end{aligned}$

To determine the emissivity ($\varepsilon$) of the wire used in the bulb, we start with the given parameters and the formula for power radiated by a black body:

Length of the wire ($L$): 10 cm = 0.1 m

Diameter of the wire ($d$): 0.5 mm = 0.0005 m

Temperature of the wire ($T$): $1727^{\circ} \mathrm{C} = 2000 \mathrm{K}$

Power radiated ($P$): 94.2 W

The power radiated by the wire can be expressed as:

$ P = \varepsilon \sigma A T^4 $

Where:

$\sigma$ is the Stefan-Boltzmann constant ($6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$)

$A$ is the surface area of the wire, calculated using the formula for the surface area of a cylinder: $A = \pi d L$

Substituting the given values into the formula:

$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times (3.14 \times 0.0005 \times 0.1) \times (2000)^4 $

Simplifying the expression inside the parentheses:

$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4 $

On solving for $\varepsilon$:

$ \varepsilon = \frac{94.2}{[(6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4]} $

Given that $\varepsilon$ is $\frac{x}{8}$, evaluating this expression provides the value of $x = 5$. Thus, the emissivity $\varepsilon$ is $\frac{5}{8}$.

$\begin{aligned} &\text { Area of circle, } W=\frac{\pi}{4} d_1 d_2\\ &\begin{aligned} & \mathrm{W}=\frac{\pi}{4}(500-300) \times 10^3(350-150) \times 10^{-6} \\ & \mathrm{~W}=31.4 \mathrm{Joule} \\ & \mathrm{~W}=314 \times 10^{-1} \mathrm{Joule} \end{aligned} \end{aligned}$

To determine the internal energy of the air in a room that measures $4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m}$ at 1 atmospheric pressure, we can follow these steps using the properties of a diatomic gas:

The volume of the room, $ V $, is calculated as:

$ V = 4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m} = 48 \, \mathrm{m}^3 $

For a diatomic gas like air, the internal energy $ U $ is given by the formula:

$ U = nC_v T = n \frac{5RT}{2} $

Where:

$ n $ is the number of moles,

$ C_v $ is the molar heat capacity at constant volume,

$ R $ is the ideal gas constant, and

$ T $ is the temperature in Kelvin.

Using the ideal gas law, $ PV = nRT $, we can substitute for $ nRT $:

$ nRT = PV $

Therefore, the internal energy $ U $ becomes:

$ U = \frac{5}{2} PV $

Substituting the given values for atmospheric pressure $ P = 10^5 \, \mathrm{Pa} $ (1 atm) and volume $ V = 48 \, \mathrm{m}^3 $:

$ U = \frac{5}{2} \times 10^5 \, \mathrm{Pa} \times 48 \, \mathrm{m}^3 = 12 \times 10^6 \, \mathrm{J} $

Thus, the internal energy of the air in the room is $ 12 \times 10^6 \, \mathrm{J} $.

To determine the value of $ n $, we start by considering the specific heat ratios for gases A and B.

For gas A, which is monoatomic with 3 translational degrees of freedom:

The degrees of freedom, $ f_A $, is 3.

Therefore, $ \gamma_A = \frac{f_A + 2}{f_A} = \frac{3 + 2}{3} = \frac{5}{3} $.

For gas B, which is polyatomic with 3 translational, 3 rotational degrees of freedom, and 1 vibrational mode:

The total degrees of freedom, $ f_B $, is $ 3 + 3 + 2 \times 1 = 8 $ (since one vibrational mode contributes 2 degrees of freedom: 1 kinetic + 1 potential).

Therefore, $ \gamma_B = \frac{f_B + 2}{f_B} = \frac{8 + 2}{8} = \frac{10}{8} = \frac{5}{4} $.

Now, the relation given is:

$ \frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right) $

Substitute the values of $ \gamma_A $ and $ \gamma_B $:

$ \frac{\frac{5}{3}}{\frac{5}{4}} = \frac{5}{3} \times \frac{4}{5} = \frac{20}{15} = \frac{4}{3} $

Equating to the given form:

$ \frac{4}{3} = 1 + \frac{1}{n} $

Thus, we have:

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

$ \frac{1}{3} = \frac{1}{n} $

Solving for $ n $:

$ n = 3 $

Hence, the value of $ n $ is 3.

To solve this problem, we can use the combined gas law in terms of pressure ($P$) and temperature ($T$). For a gas with a fixed volume, the relationship is given as:

$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $

where:

$ P_1 $ is the initial pressure,

$ T_1 $ is the initial temperature in Kelvin,

$ P_2 $ is the final pressure,

$ T_2 $ is the final temperature in Kelvin.

Given that the initial temperature $ T_1 = 27°C $, we need to convert it to Kelvin:

$ T_1 = 27°C + 273.15 = 300.15 \, \text{K} $

The final pressure $ P_2 $ is twice the initial pressure ($ P_2 = 2P_1 $). Substituting these values into the gas law equation gives:

$ \frac{P_1}{300.15} = \frac{2P_1}{T_2} $

Solving for $ T_2 $:

Cancel $ P_1 $ from both sides:

$ \frac{1}{300.15} = \frac{2}{T_2} $

Rearrange to solve for $ T_2 $:

$ T_2 = 2 \times 300.15 = 600.3 \, \text{K} $

Convert the final temperature back to Celsius:

$ T_2 = 600.3 \, \text{K} - 273.15 = 327.15°C $

Therefore, to double the pressure of the gas, the temperature should be raised to approximately 327°C.

$ Q = n\,C_P\,\Delta T \quad \text{and} \quad \Delta U = n\,C_V\,\Delta T $

For a monoatomic ideal gas, the molar specific heats are given by:

$ C_V = \frac{3}{2}R \quad \text{and} \quad C_P = C_V + R = \frac{5}{2}R $

Thus, the ratio of the heat added to the change in internal energy is:

$ \frac{Q}{\Delta U} = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} $

Given that:

$ \frac{E_1}{E_2} = \frac{x}{9} $

We equate the two ratios:

$ \frac{5}{3} = \frac{x}{9} $

Multiplying both sides by 9:

$ x = \frac{5}{3} \times 9 = 15 $

Thus, the value of $x$ is 15.

We’re given an ideal gas initially at $T_i = 0^\circ \mathrm{C} = 273\,\mathrm{K}.$ The gas is suddenly compressed to one-fourth of its initial volume, and we are told that

$\gamma = \frac{c_p}{c_v} = \frac{3}{2}\,.$

For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be

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

Even though the compression is “sudden” (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states.

Here are the steps:

Write the adiabatic relation between the initial and final states:

$T_i\, V_i^{\gamma-1} = T_f\, V_f^{\gamma-1}\,.$

Since the gas is compressed to one-fourth of its volume, we have

$V_f = \frac{1}{4} V_i\,.$

Substitute this into the adiabatic relation:

$273\,V_i^{\gamma-1} = T_f \left(\frac{1}{4} V_i\right)^{\gamma-1}\,.$

With $\gamma = \frac{3}{2},$

$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}\,.$

So the relation becomes:

$273\,V_i^{1/2} = T_f \left(\frac{1}{4}\right)^{1/2} V_i^{1/2}\,.$

Cancel the common factor $V_i^{1/2}$ (provided it is nonzero) and simplify:

$273 = T_f \left(\frac{1}{2}\right)\,.$

Now solve for the final temperature $T_f$:

$T_f = 273 \times 2 = 546\,\mathrm{K}\,.$

The change in temperature is then:

$\Delta T = T_f - T_i = 546 - 273 = 273\,\mathrm{K}\,.$

So, the change in temperature due to the thermodynamic process is $\boxed{273\,\mathrm{K}}.$

${R_1} = {{2L} \over {{K_1}A}}$

(As ${l_1} = {l_2} = {l_3} = l$

${A_1} = {A_2} = A/2$

${A_3} = A$)

${R_2} = {{2L} \over {{K_2}A}}$

${R_3} = {L \over {{K_3}A}}$

${I_1} + {I_2} = {I_3}$

$ \Rightarrow {{100 - \theta } \over {{{2L} \over {{K_1}A}}}} + {{100 - \theta } \over {{{2L} \over {{K_2}A}}}} = {{\theta - 0} \over {{L \over {{K_3}A}}}}$

$ \Rightarrow {{100 - \theta } \over 2}({K_1} + {K_2}) = {K_3}\theta $

$ \Rightarrow 50({K_1} + {K_2}) = \theta \left[ {{K_3} + {{{K_1} + {K_2}} \over 2}} \right]$

$ \Rightarrow 50(60 + 120) = \theta \left[ {135 + {{60 + 120} \over 2}} \right]$

$ \Rightarrow 50 \times 80 = \theta (135 + 90)$

$ \Rightarrow \theta = {{50 \times 180} \over {225}}$

$ \Rightarrow \theta = 40^\circ C$

$ \begin{aligned} & \mathrm{nRT}_{\mathrm{w}}=\mathrm{P}_{\mathrm{w}} \mathrm{~V}_{\mathrm{w}}=1 \mathrm{~J} \\ & \mathrm{P}_{\mathrm{W}}=\frac{1}{64} \times 10^6 \mathrm{~Pa} \end{aligned} $

For WX process

$ \begin{aligned} & P_X V_X^y=P_W V_W^y \\ & \Rightarrow P_X=P_W\left(\frac{V_W}{V_X}\right)^y \end{aligned} $

amount of heat absorbed in XY process

$ \begin{aligned} \mathrm{Q} & =\mathrm{nCP} \Delta \mathrm{~T}=\mathrm{n} \times \frac{5}{2} \mathrm{R} \times\left[\mathrm{T}_{\mathrm{Y}}-\mathrm{T}_{\mathrm{X}}\right] \quad\left[\text { For monoatomic gas } \mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2}\right] \\ \mathrm{Q} & =\frac{5}{2}\left[\mathrm{nRT}_{\mathrm{Y}}-\mathrm{nRT}_{\mathrm{X}}\right] \\ & =\frac{5}{2}\left[\mathrm{P}_{\mathrm{Y}} \mathrm{~V}_{\mathrm{Y}}-\mathrm{P}_{\mathrm{X}} \mathrm{~V}_{\mathrm{X}}\right] \\ & =\frac{5}{2} \mathrm{P}_{\mathrm{X}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \quad\left[\because \mathrm{P}_{\mathrm{X}}=\mathrm{P}_{\mathrm{Y}} ; \text { Isobaric process }\right] \\ & =\frac{5}{2} \times \mathrm{P}_{\mathrm{W}} \times\left[\frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{~V}_{\mathrm{X}}}\right]^{\mathrm{y}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \end{aligned} $

Putting values :

Q = 1.6 Joule

Extension in spring

$ \begin{aligned} & \mathrm{x}=0.5 \mathrm{~L}-0.4 \mathrm{~L} \\ & =0.1 \mathrm{~L} \end{aligned} $

FBD of piston

$ \begin{aligned} & \mathrm{kx}+\mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A} \\ & \mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A}-\mathrm{kx} \\ & \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{kL}}{\mathrm{~A}(10)} ........(i) \\ & \mathrm{P}_1 \mathrm{~V}=\mathrm{n}_1 \mathrm{RT} \\ & \mathrm{P}_2 \mathrm{~V}=\mathrm{n}_2 \mathrm{RT} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2} \end{aligned} $

$ \begin{aligned} & \mathrm{P}_1=\frac{3}{2} \mathrm{P}_2 ........(ii) \\ & \mathrm{P}_2=\frac{3}{2} \mathrm{P}_2-\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \frac{\mathrm{P}_2}{2}=\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \mathrm{P}_2=\frac{\mathrm{kL}}{5 \mathrm{~A}}=\frac{\mathrm{kL}}{\mathrm{~A}} \alpha \\ & \alpha=\frac{1}{5}=0.2 \end{aligned} $

Since the plates are blackbodies, they emit radiation according to the Stefan-Boltzmann law. The power emitted per unit area by a blackbody is given by.

$ P=\sigma T^4 . $

Where, $\sigma=$ Stefan - Boltzmann constant

$ \sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{k}^4 $

When two blackbodies face each other, the net power transfer per unit area from $P$ to $Q=$ power emitted by $P$ toward $Q$ - power emitted by $Q$ toward $P$

So,

$ W_0=\sigma T_p^4-\dot{\sigma} T_a^4=\sigma\left(T_p^4-T_a^4\right)\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

After introducing two additional plates: assuming temperature $T_1$ and $T_2$ for these plates.

In the steady state, the power transfer per unit area through each interface (From $P$ to plate $1,1 \rightarrow 2,2 \rightarrow Q$ ) must be same, because there is no accumulation of energy in the intermediate plates.

So

$ \begin{aligned} & \sigma\left(T_p^4-T_1^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \sigma\left(T_1^4-T_2^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

and $\sigma\left(T_2^4-T_a^4\right)=W_S\,\,\,\,...(iv)$

by adding $e q^n$ (2), (3) and (4),

$ \begin{aligned} & \sigma\left(T_p^4-T _1^4+T_1^4-T_2^4+T_2^4-T_Q^4\right)=3 W_S \\ \Rightarrow & \sigma\left(T_p^4-T_Q^4\right)=3 W_S \\ \Rightarrow & W_0=3 W_S \\ & \Rightarrow \frac{W_0}{W_S}=3 \end{aligned} $

Case - 1

$\begin{aligned} & P-P_0=\Delta P=\frac{4 T}{R} \\ & P=\left(P_0+\frac{4 T}{R}\right) \end{aligned}$

Case-2

$\begin{aligned} & \mathrm{P}_1-\frac{8 \mathrm{P}_0}{27}=\Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1} \\ & \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27} \end{aligned}$

Constant temperature process

$\begin{aligned} & \mathrm{PV}=\mathrm{P}_1 \mathrm{~V}_1 \\ & \left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}\right) \rightarrow \text { (Neglected) } \\ & \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \\ & \Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}=\frac{4 \mathrm{~T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{~Pa} \end{aligned}$

To solve this problem, we need to integrate the heat capacity over the given temperature range because the specific heat capacity is temperature dependent. We are given that the specific heat capacity $C$ is defined as $C = kT$, where $k$ is a constant, and $T$ is the absolute temperature.

The heat required to raise the temperature, $Q$, can be found using the following integral:

$ Q = \int_{T_1}^{T_2} C \, dT $

Given $C = kT$, the integral becomes:

$ Q = \int_{T_1}^{T_2} kT \, dT $

We need to convert the given temperatures from Celsius to Kelvin. The temperatures given are:

• Initial temperature: $-73^{\circ} \mathrm{C}$
• Final temperature: $27^{\circ} \mathrm{C}$

In Kelvin, these temperatures are:

• $T_1 = -73^{\circ} \mathrm{C} + 273 = 200 \, \mathrm{K}$
• $T_2 = 27^{\circ} \mathrm{C} + 273 = 300 \, \mathrm{K}$

Now we can evaluate the integral:

$ Q = k \int_{200}^{300} T \, dT $

Integrating, we get:

$ Q = k \left[ \frac{T^2}{2} \right]_{200}^{300} $

Substituting the limits of integration:

$ Q = k \left[ \frac{300^2}{2} - \frac{200^2}{2} \right] $

Solving the values inside the brackets:

$ Q = k \left[ \frac{90000}{2} - \frac{40000}{2} \right] $

$ Q = k \left[ 45000 - 20000 \right] $

$ Q = k \cdot 25000 $

We are given that this heat is equal to $nk$:

$ nk = k \cdot 25000 $

Dividing both sides by $k$, we get:

$ n = 25000 $

Thus, the value of $n$ is 25000.

The change in length of the rod when it is heated is given by the equation:

$\Delta L = L_0 \cdot \alpha \cdot \Delta T$

where

• $\Delta L$ is the change in length,
• $L_0$ is the original length,
• $\alpha$ is the coefficient of linear expansion, and
• $\Delta T$ is the change in temperature.

Substituting the given values:

$\Delta L = 1 \, \text{m} \cdot 10^{-5} \, \text{K}^{-1} \cdot 200 \, \text{K} = 0.002 \, \text{m}$

The rod is not allowed to extend or bend, so a stress is created in the rod. This stress can be calculated using Young's modulus (Y), which is the ratio of the stress (force per unit area, F/A) to the strain (change in length per unit length, $\Delta L / L_0$):

$Y = \frac{F/A}{\Delta L / L_0}$

Rearranging for F gives:

$F = Y \cdot A \cdot \frac{\Delta L}{L_0}$

Substituting the given values:

$F = 2 \times 10^{11} \, \text{N/m}^2 \cdot 10^{-4} \, \text{m}^2 \cdot \frac{0.002 \, \text{m}}{1 \, \text{m}} = 4 \times 10^{4} \, \text{N}$

So the compressive tension produced in the rod is $4 \times 10^{4} \, \text{N}$.

The amount of heat energy required to raise the temperature of a substance can be calculated as:

Q = m $ \times $ c $ \times $ ΔT

where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

The time required to heat a substance can be calculated as :

t = ${Q \over P}$

where t is the time required, and P is the power of the heating device.

The actual power output of the heating device can be calculated as:

P_actual = P_input $ \times $ efficiency

where P_input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.

Substituting the given values:

Q = 2 kg $ \times $ 4200 J/kg/K $ \times $ (60 - 10) = 2 kg $ \times $ 4200 J/kg/K $ \times $ 50 K = 4200 $ \times $ 50 $ \times $ 2 J = 420,000 J

P_input = 2000 W = 2000 J/s

P_actual = 2000 $ \times $ 0.7 = 1400 J/s

t = ${Q \over {{P_{actual}}}}$ = ${{420,000} \over {1400}}$ J/s = 300 s

So, the time required to heat 2 kg of water from 10°C to 60°C is approximately 300 s.

Let the correct temperature be X$^\circ$C

$ \Rightarrow {{X - 0} \over {100 - 0}} = {{41 - 5} \over {95 - 5}} \Rightarrow X = 40$

$\Rightarrow$ Temperature is 273 + 40 K = 313 K

${P_1}V_1^\gamma = {P_2}V_2^2$

${{{P_1}} \over {d_1^\gamma }} = {{{P_2}} \over {d_2^\gamma }}$

${{{d_1}{T_1}} \over {d_1^\gamma }} = {{{d_2}{T_2}} \over {d_2^\gamma }}$

${T_2} = {\left( {{{{d_2}} \over {{d_1}}}} \right)^{\gamma - 1}}{T_1}$

$ = {(32)^{{2 \over 5}}}{T_1}$

${T_2} = 4\,{T_1}$

${C_V} = {f \over 2}R$

total degree of freedoms

$ = 1 \times 3 + 3 \times 5 = 18$

${{{\alpha ^2}} \over 4} = {{18} \over {2n}} = {{18} \over {2 \times 4}}$

$ \Rightarrow {\alpha ^2} = 9$

$\alpha = 3$

$f = 8$

$W = P\,dV = 150$

$Q = W + \Delta U$

$ = P\,dV + {f \over 2}\,PdV$

$Q = 5 \times 150 = 750\,J$

Heat lost by water = Heat gained by ice

$0.3 \times 4200 \times 25 = x \times 3.5 \times {10^5}$

$x = {{0.3 \times 4200 \times 25} \over {3.5 \times {{10}^5}}}$

$ = 90 \times 100 \times {10^5} \times {10^3}$ gram = 90 gm

$\Delta {L_1} = {\alpha _1}{L_1}\Delta T$

$\Delta {L_2} = {\alpha _2}{L_2}\Delta T$

${\alpha _1}{L_1} = {\alpha _2}{L_2}$

$1.2 \times {10^{ - 5}} \times {L_1} = 1.8 \times {10^{ - 5}} \times {L_2}$

${L_1} = {{1.8} \over {1.2}} \times 40 = 60$ cm

$H = {{{V^2}} \over R}\,.\,\Delta t$

$ \Rightarrow H = {{{V^2}} \over {{R_1}}}\,.\,20 = {{{V^2}} \over {{R_2}}}\,.\,60$ ..... (i)

Also, $H = {{{V^2}} \over {\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]}}\,.\,\Delta t$

$ = {4 \over 3}\,.\,{{{V^2}} \over {{R_1}}}\,.\,\Delta t$ [$\because$ ${R_2} = 3{R_1}$]

$ \Rightarrow \Delta t = 15$

${{{L_1}} \over {{K_1}{A_1}}} + {{{L_2}} \over {{K_2}{A_2}}} = {{{L_1} + {L_2}} \over {{K_{eff}}{A_{eff}}}}$

$ \Rightarrow {4 \over K} + {{2.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$

$ \Rightarrow {{10.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$

$ \Rightarrow {K_{eff}} = {{13K} \over {10.5}} = \left( {1 + {5 \over {21}}} \right)K$

$ \Rightarrow \alpha = 21$

$\eta = {W \over Q} = {{300 - 225} \over {300}}$

$ \Rightarrow {{75} \over {300}} = 1 - {{{T_L}} \over {{T_H}}}$

$ \Rightarrow {T_L} = {3 \over 4}\,{T_H} = {3 \over 4}(500) = 375\,K$

$ \Rightarrow {T_L} = 102^\circ C$

$U = 2\left( {{3 \over 2}R} \right)300$

$ = 3 \times 8.31 \times 300$

$ = 7479$ J

W = nR$\Delta$T = 400 J

$\therefore$ $\Delta$Q = nC_P$\Delta$T

$ = n \times {7 \over 2}R \times \Delta T = {7 \over 2} \times (400) = 1400$

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

$ = 1 - {{200} \over {800}} = {3 \over 4}$

$\therefore$ $\eta = {W \over {{Q_1}}}$

$ \Rightarrow {3 \over 4} = {{12000 \times {{10}^3}} \over {{Q_1}}}$

$ \Rightarrow {Q_1} = 16 \times {10^6}\,J$

$Q = ms\Delta T$

${{dQ} \over {dt}} = {\left( {{{dm} \over {dt}}} \right)_{water}}S\Delta T = {\left( {{{dm} \over {dt}}} \right)_{oil}}C$

$ \Rightarrow 2 \times 4.2 \times {10^3} \times 40 = {\left( {{{dm} \over {dt}}} \right)_{oil}} \times 8 \times {10^6}$

$ \Rightarrow {\left( {{{dm} \over {dt}}} \right)_{oil}} = {{8 \times 4.2 \times {{10}^4}} \over {8 \times {{10}^6}}}$ kg/minute

= 42 g/min

$\left( {1 - {{324} \over {{T_H}}}} \right) = {{300 - 180} \over {300}}$

$1 - {2 \over 5} = {{324} \over {{T_H}}}$

${T_H} = {{324 \times 5} \over 3} = 540$

$PV = nRT$

So ${{dP} \over P} \times 100 = {{dT} \over T} \times 100$

$0.4 = {1 \over T} \times 100$

$ \Rightarrow T = 250\,K$

$\Delta$Q_rej = 50 $\times$ 540 + 50 $\times$ 1 $\times$ (100 $-$ 20)

= 50 $\times$ [540 + 80]

= 50 $\times$ 620

= 31000 cal

= 31 $\times$ 10³ cal

By 1^st law,

$\Delta U = \Delta Q - {{\Delta Q} \over 4} = {3 \over 4}\Delta Q$

$ \Rightarrow n{C_v}\Delta T = {3 \over 4}nC\Delta T$

$ \Rightarrow C = {{4{C_v}} \over 3} = 2R$