Heat and Thermodynamics

Given,

Volume of glass vessel and liquid $=V_o$

Change in temperature $=\Delta T$

Linear expansion of glass $=\alpha_g$

Volume expansion of liquid $=\gamma_l$

Let final volume of glass and liquid be $V_g$ and $V_l$, then

$\begin{aligned} & V_g=V_o\left(1+3 \alpha_g \Delta T\right) \\ & V_l=V_o\left(1+\gamma_l \Delta T\right) \end{aligned}$

Volume of liquid flowing out,

$V_l-V_g=V_0\left(1+\gamma_l \Delta T\right)-V_0\left(1+3 \alpha_g \Delta T\right)=V_o \Delta T\left(\gamma_l-3 \alpha_g\right)$

Given, sink temperature, $T_2=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

Case I Efficiency, $\eta_1=40 \%$

Case II Efficiency, $\eta_2=60 \%$

Let temperature of source in two cases be $T_1$ and $T_1^{\prime}$.

As we know that,

$\begin{aligned} \eta_1 & =1-\frac{T_2}{T_1} \\ \Rightarrow \quad \frac{40}{100} & =1-\frac{300}{T_1} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad T_1=500 \mathrm{~K} \\ & \text { and } \quad \eta_2=1-\frac{T_2}{T_1^{\prime}} \\ & \Rightarrow \quad \frac{60}{100}=1-\frac{300}{T_1^{\prime}} \Rightarrow \frac{300}{T_1^{\prime}}=\frac{40}{100} \\ & \Rightarrow \quad T_1^{\prime}=750 \mathrm{~K} \\ & \therefore \text { Change is temperature, } \Delta T=T_1^{\prime}-T_1 \\ & =750-500=250 \mathrm{~K} \\ & \end{aligned}$

Given, nature of gas = Diatomic (at constant pressure)

As we know that,

Internal energy, $U=n C_V \Delta T$ ...... (i)

Heat energy, $Q=n C_p \Delta T$ ...... (ii)

where, $n=$ number of mole

$C_V=$ specific heat at constant volume

$C_p=$ specific heat at constant pressure

and $C_p=C_V+R$

If $f=$ degree of freedom, then

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

Now, dividing Eq. (i) by Eq. (ii), we get

$\Rightarrow \quad \frac{U}{Q}=\frac{n C_V \Delta T}{n C_p \Delta T}=\frac{C_V}{C_p}=\frac{5 R / 2}{7 R / 2}=\frac{5}{7}$

According to given $p$ - $V$ diagram,

As we know that work is equal to area under $p$-$V$ graph.

$\therefore \quad W_A > W_B > W_C > W_D$

and by using 1st law of thermodynamics,

$\Delta Q=\Delta U+\Delta W$

Here, internal energy, $\Delta U=0$

$\because$ Initial and final positions are same.

$\therefore$ Heat $(\Delta Q)$ sequence will be

$\Delta Q_A > \Delta Q_B > \Delta Q_C > \Delta Q_D $

Hence, $Q_A-Q_D=W_A-W_D$

The gas has its temperature increased from 27$^\circ$C to 127$^\circ$C.

First, change the temperatures from degrees Celsius to Kelvin:

$T_1 = 27 + 273 = 300\,K$

$T_2 = 127 + 273 = 400\,K$

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

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

Here, $R$ is the gas constant and $M$ is the molar mass. Because $R$ and $M$ stay the same, $v_{rms}$ depends only on temperature. We can say:

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

So, the ratio of the rms speeds at $T_1$ and $T_2$ is:

$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \frac{2}{\sqrt{3}} $

To find the percentage increase in $v_{rms}$, use:

$ \text{Percentage increase} = \left(\frac{v_2 - v_1}{v_1}\right) \times 100 $

Substitute the values:

$ \left(\frac{v_2}{v_1} - 1\right) \times 100 = \left(\frac{2}{\sqrt{3}} - 1\right) \times 100 $

This gives about 15.5% increase in $v_{rms}$.

Specific heat is the amount of heat required to raise the temperature of water by $1^{\circ} \mathrm{C}$.

For boiling of water to turn into steam, $\Delta T=0$

Since, $\Delta Q=n C_p \Delta T$

If $\Delta T=0, C_p=\frac{\Delta Q}{\Delta T}=\frac{\Delta Q}{0}=\infty$

Given,

Volume change, $\frac{\Delta V}{V}=\frac{0.12}{100}=12 \times 10^{-4}$

Change in temperature, $\Delta T=20 \Upsilon C$

Since, $\Delta V / V=3 \alpha \Delta T$

where, $\alpha$ is coefficient of linear expansion.

$\therefore \quad \alpha=\frac{\Delta V}{V} \times \frac{1}{3 \Delta T} \Rightarrow \alpha=\frac{12 \times 10^{-4}}{3 \times 20}=2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$

As we know that,

In case of isothermal condition, T = constant

By ideal gas law,

$pV = nRT$

$\Rightarrow \quad pV=$ constant

$\Rightarrow \quad p\propto \frac{1}{V}$

$\therefore p$-$V$ graph is given as:

According to given graph,

As we know that,

$C=\frac{R}{\gamma-1}-\frac{R}{m-1}$ ..... (i)

where, $C_V$ is heat coefficient at constant volume, R is universal gas constant = 8.314 J K$^{-1}$, $\gamma$ is adiabatic constant and m is polytropic constant.

For path $A B, p V^2=$ constant

$\therefore \quad m=2$

and $\gamma_{\text {monoatomic }}=5 / 2$

Put these values in Eq (i), we get

$C=\frac{R}{\frac{5}{3}-1}-\frac{R}{2-1}=\frac{3 R}{2}-R=\frac{R}{2}$

For path $B C$, density is fixed

$\therefore$ Volume will remain unchanged.

Since, $U=n C_V \Delta T$

where, $U$ is internal energy and $C_V$ (monoatomic gas) $=\frac{3}{2} R$.

Isochoric process, $\Delta V=0$

But temperature is decreasing and $C_V=\frac{3}{2} R$

$\therefore$ Option (c) is incorrect.

For path $C A$ (isothermal)

$W=n R T \ln \frac{p_2}{p_1}=n R T \ln \frac{4 p_0}{p_0}=n R T \ln 4$

and $\quad U_0=\frac{3}{2} n R T$

$\therefore \quad W=\frac{2}{3} U_0(\ln 4)$

As we know that,

Kinetic energy, per unit volume, $E=\frac{1}{2} \rho V^2$ ..... (i)

and pressure per unit volume, $p=\frac{1}{3} \rho V^2$ ..... (ii)

From Eqs. (i) and (ii), we get

$\therefore \quad p=\frac{1}{3} 2 E \Rightarrow p=\frac{2}{3} E$

Hence, $p \propto E$.

As we know that, during expansion due to heating, change in volume $\Delta V=V_0 \gamma \Delta T$

Here, $V_0, \gamma$ and $\Delta T$ are initial volume, volumetric expansion coefficient and change in temperature.

Since, density $(\rho)=\frac{\operatorname{mass}(m)}{\text { volume }(V)}$

As volume increases by $\Delta V$.

$\therefore$ Density decreases.

As we know that,

(A) Since, time period $(T)=2 \pi \sqrt{l / g}$ where, $l$ is length and $g$ is gravity.

$\therefore \quad \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}$ ..... (i)

and by using concept of linear expansion,

$\frac{\Delta l}{l}=\alpha \Delta T$ ..... (ii)

Substituting in Eq. (i), we get

$\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta T$

$(\mathrm{A} \rightarrow 4)$

(B) Now, from Eq. (ii)

$\begin{aligned} \Delta l & =l \alpha \Delta T \\ \Rightarrow \quad l_0-l & =l \alpha \Delta T \Rightarrow l_0=l+l \alpha \Delta T \\ \Rightarrow \quad l_0 / l & =l+\alpha \Delta T \end{aligned}$

(B) $\rightarrow(3)$

$\begin{array}{ll} \text {(C)} \because p \Delta V=n R T \quad .....\text{(i)}\\ \text { and } \Delta V=V_0 \gamma \Delta T \\ \Rightarrow \Delta V \propto \gamma \\ \Rightarrow \gamma \propto T\quad \text {[Using Eq. (i)]}\end{array}$

(C) $\rightarrow(2)$

(D) Since, $Y=\frac{F l}{A \Delta l} \Rightarrow \frac{\Delta l}{l}=\frac{F}{A Y}=\alpha \Delta T$

(D) $\rightarrow$ (1)

$\therefore \quad \mathrm{A} \rightarrow 4, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 2, \mathrm{D} \rightarrow \mathrm{l}$ is correct.

As we know that,

Irreversible process is the phenomenon which can’t be reversed back to initial condition.

Since, conduction is the phenomenon in which heat flows from higher temperature to lower temperature.

$\therefore$ Conduction is irreversible process.

Given, internal energy = U

Temperature = T

Since, $U=\frac{3}{2}KT$

$\therefore \quad U \propto T$

Given, coefficient of performance $(\mathrm{COP})=0.25$

Heat coming out, $Q_c=250 \mathrm{~J}$

Since,

$\begin{aligned} & \operatorname{COP}=\frac{Q_L}{\operatorname{Work}(W)} \\ & \Rightarrow \quad \mathrm{COP}=\frac{Q_L}{Q_H-Q_L} \Rightarrow \frac{Q_H-Q_L}{Q_L}=\frac{100}{25}=4 \\ & \Rightarrow \quad \frac{Q_H}{Q_L}-1=4 \Rightarrow \frac{Q_H}{Q_L}=5 \\ & \Rightarrow \quad Q_L=\frac{250}{5}=50 \\ & \therefore \quad W=Q_H-Q_L \\ & =250-50=200 \mathrm{~J} \\ \end{aligned}$

Given, mass of oxygen, $m_i=6 \mathrm{~g}$

Initial pressure $p_i=p$

Initial temperature $T_i=400 \mathrm{~K}$

Mass of oxygen leak $=m^{\prime}$

Final pressure, $p_f=\frac{p}{2}$

Final temperature, $T_f=300 \mathrm{~K}$

Since, $p V=\frac{m}{M} R T$

where, $m$ is given mass

$M$ is molar mass and $R$ is gas constant

Now, $\frac{p_i}{T_i m_i}=\frac{p_f}{T_f m_f}$

$\Rightarrow m_f=\left(\frac{p_f}{T_f}\right) \cdot\left(\frac{T_i m_i}{p_i}\right)=\left(\frac{p}{2 \times 300} \times \frac{400 \times 6}{p}\right)=4 \mathrm{~g}$

$\therefore$ Mass of oxygen leakes out $=6-4=2 \mathrm{~g}$

Given, initial and final length of rod is $l_1$ and $l_2$ i.e., $20 \mathrm{~cm}$ and $9 \mathrm{~cm}$ from $A$, and temperature at $A$ and $B$ is $T_A$ and $T_B$ i.e., $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$.

Now, since, heat flow rate $=\frac{d Q}{d t}=\frac{K A \Delta \theta}{\lambda}$

As the flow is steady,

$\begin{aligned} & \therefore & \frac{K A(100-\theta)}{9} & =\frac{K A(\theta-0)}{11} \\\\ & \Rightarrow & 1100-11 \theta & =9 \theta \\\\ &\Rightarrow & 1100 =20 \theta \\\\ & \therefore & \theta =55^{\circ} \mathrm{C} \end{aligned}$

Given, length of rods are $L$ and $2 L$.

Coefficient of linear expansion $\alpha$ and $2 \alpha$.

As, we know that,

$\Delta l=l_0 \alpha \Delta T$

where, $\Delta l$ is change in length,

$T$ is temperature

and $l_0$ is initial length

let final heat coefficient be $\alpha^{\prime}$

$\begin{array}{llrl} & \therefore & (L+2 L) \alpha^{\prime} \Delta T & =L \alpha \Delta T+2 L \cdot 2 \alpha \cdot \Delta T \\ \Rightarrow & & 3 \alpha^{\prime} & =\alpha+4 \alpha=5 \alpha \\ \Rightarrow & & \alpha^{\prime} & =5 \alpha / 3 \end{array}$

Given, heat absorbed and work done by two systems be $Q_1, Q_2$ and $W_1, W_2$ and as we know that, By using Ist law of thermodynamics,

$\Delta Q=\Delta U+\Delta W \Rightarrow \Delta U=\Delta Q-\Delta W$

where, $\Delta Q$ is heat change, $\Delta U$ is internal energy change i.e. state function and $\Delta W$ is work done.

$\begin{array}{lc} \therefore & \Delta Q_1-\Delta W_1=\Delta Q_2-\Delta W_2 \\ \text { or } & Q_1-W_1=Q_2-W_2 \end{array}$

Given, adiabatic constant $\gamma=1.5=3 / 2$

Let initial volume, $V_1=V$

$\therefore$ Final volume, $V_2=V / 4$

Initial and final pressure be $p_1$ and $p_2$.

$\begin{aligned} & \text { Since, } \quad p V^\gamma=\text { constant } \\ & \therefore \quad p_1 V_1^\gamma=p_2 V_2^\gamma \\ & \Rightarrow \quad p_1 / p_2=\left(V_2 / V_1\right)^\gamma=\left(\frac{V}{4 \times V}\right)^{3 / 2} \\ \end{aligned}$

$\Rightarrow p_1/p_2=\frac{1}{8}$

Hence, $p_2:p_1=8:1$

Given, initial temperature $T_1 =30^{\circ} \mathrm{C} =303 \mathrm{~K}$

Change in radius, $\Delta r=0.08 \mathrm{~mm}$

Internal diameter of cylinder $d_c=15 \mathrm{~cm}$

$\therefore$ Internal radius of cylinder, $r_c=7.5 \times 10^{-2} \mathrm{~m}$

Radius of piston, $r_p=(7.5-0.008) \times 10^{-2} \mathrm{~m}$

$=7.492 \times 10^{-2} \mathrm{~m}$

Linear expansion of cylinder $\alpha_C$ and piston $\alpha_p$ are $1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ and $1.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}$.

Let, $\Delta r_C$ and $\Delta r_p$ is increase in radius of cylinder and piston respectively, for fully filled piston in cylinder.

By using equation of linear expansion,

$\begin{aligned} & r=r_0(1+\alpha \Delta T) \\ \therefore \quad & \Delta r_C=r_C \alpha_C \Delta T, \text { for cylinder } \end{aligned}$

New radius will be $r_C^{\prime}=r_C+\Delta r_C$

$\begin{aligned} & =r_C+r_C \alpha_C \Delta T=r_C\left(1+\alpha_C \Delta T\right) \\ & =7.5 \times 10^{-2}\left[1+1.2 \times 10^{-5}(T-303)\right] \end{aligned}$

Similarly, new radius of piston,

$r_p{ }^{\prime}=7.492 \times 10^{-2}\left[1+1.6 \times 10^{-5}(T-303)\right]$

Now, for fully fitted piston, $r_C=r_p$

$\begin{aligned} 7.5 \times 10^{-2}[1 & \left.+1.2 \times 10^{-5}(T-303)\right] \\ & =7.492 \times 10^{-2}\left[1+1.6 \times 10^{-5}(T-303)\right] \\ \Rightarrow \quad & 1.0011+1.2013 \times 10^{-5}(T-303) \\ & =1+1.6 \times 10^{-5}(T-303) \\ \Rightarrow 1.1 \times 10^{-3} & =(T-303) \times 10^{-5}(1.6-1.2013) \\ & =(T-303) \times 10^{-5} \times 0.3987 \\ \Rightarrow T-303 & =275 \\ \Rightarrow \quad T & =578 \mathrm{~K}=305^{\circ} \mathrm{C} \end{aligned}$

Given, initial volume, temperature and pressure be $V_1,T_1$ and $p_1$ are 1500 m$^3$, 27$^\circ$C and 4 atm.

Final volume, temperature and pressure are $V_2,T_2$ and $p_2$ are $V_2,-3^\circ$C and 2 atm

By using ideal gas equation,

$\begin{aligned} & & p V & =n R T \\ \Rightarrow & & \frac{p V}{T} & =\text { constant } \\ \Rightarrow & & \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow & & V_2 & =\frac{p_1 V_1 T_2}{T_1 p_2} \\ & \therefore & V_2 & =\frac{4 \times 1500 \times 270}{300 \times 2}=2700 \mathrm{~m}^3 \end{aligned}$

Work done in the cyclic process

= Area of the loop ABCD

= (2p $-$ p) $\times$ (2V $-$ V) = pV

Conductivity ($\sigma$) of a metallic conductor is given by

$\sigma = {1 \over \rho } = {{n{e^2}\tau } \over m}$

where, $\rho$ = resistivity of conductor,

$\tau$ = relaxation time,

n = number of free electrons

and m = mass.

For metals, as the temperature increases, the relaxation time $\tau$ decreases because collisions become more frequent. Hence, the conductivity of a metallic conductor decreases with the increases of temperature.

As we know,

${C_v} = {n \over 2}R$ ...... (i)

${C_p} = \left( {{n \over 2} + 1} \right)R$ ..... (ii)

On dividing Eq. (ii) by Eq. (i), we get

${{{C_p}} \over {{C_v}}} = \gamma = {{\left( {{n \over 2} + 1} \right)} \over {{n \over 2}}} = 1 + {2 \over n}$

$ \Rightarrow {{{C_p}} \over {{C_v}}} = 1 + {2 \over n}$