Heat and Thermodynamics

In isothermal expansion,

$ \begin{aligned} & p_1 V_1=p_2 V_2 \\ \Rightarrow & p_0 V_1=p_2 .8 V_1 \\ \Rightarrow & p_2=p_0 / 8 \end{aligned} \quad\left[\because V_2=8 V_1\right] $

Now, gas is slowly and adiabatically compressed back to its initial volume. Hence, for this adiabatic compression,

$ \begin{aligned} & p_1=p_0 / 8, V_1=8 \mathrm{~V} \\ & p_2=?, V_2=V \end{aligned} $

∴ Using adiabatic relation,

$ \begin{aligned} & & p_1 V_1^\gamma & =p_2 V_2^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot(8 V)^\gamma & =p_2(V)^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma \cdot V^\gamma & =p_2 V^{\prime} \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma & =p_2 \\ \Rightarrow & & p_2 & =\frac{p_0}{8} \cdot 8^\gamma \\ & & & =\frac{p_0}{8} \cdot(8)^{\frac{4}{3}} \left[\because \gamma=\frac{4}{3}\right]\\ & & & =\frac{p_0}{8} \cdot 16 \\ \Rightarrow & & p_2 & =2 p_0 \end{aligned} $

$ \begin{aligned} &\text { ∴ Average kinetic energy per molecule, }\\ &\begin{array}{ll} & K=\frac{3}{2} k T=\frac{3}{2} \frac{R}{N} T=\frac{3}{2} \frac{p V}{N} \,\,\,\,\,\,\,[\because p V=R T]\\ \Rightarrow & \quad K \propto p \\ \because & \frac{K_2}{K_1}=\frac{p_2}{p_1} \\ \Rightarrow & \frac{K_2}{K_1}=\frac{2 p_0}{p_0} \\ \Rightarrow & \frac{K_2}{K_1}=2 \end{array} \end{aligned} $

Given, atomic mass of Ne,

$ M_1=20.2 \mathrm{u} $

Atomic mass of He ,

$ \begin{aligned} M_2 & =4.0 \mathrm{u} \\ T_2 & =-33^{\circ} \mathrm{C} \\ & =(-33+273) \mathrm{K} \\ & =240 \mathrm{~K} \end{aligned} $

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

$ \begin{array}{ll} \text { Since, }\left(V_{\text {rms }}\right) \mathrm{He} \\ \Rightarrow \sqrt{\frac{3 R T_2}{M_2}}=\sqrt{\frac{3 R T_1}{M_1}} \\ \Rightarrow \frac{T_2}{M_2}=\frac{T_1}{M_1} \\ \Rightarrow \frac{240}{4}=\frac{T_1}{20.2} \\ \Rightarrow T_1=\frac{20.2 \times 240}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, =1212 \mathrm{~K} \end{array} $

The volume of metal at $32^{\circ} \mathrm{C}$.

$ \begin{aligned} V_{32 \mathrm{C}} & =\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\ & =\frac{(49-39) g}{1.2 \times g}=8.33 \mathrm{~cm}^3 \end{aligned} $

Similarly, volume of metal at $42^{\circ} \mathrm{C}$.

$ V_{42 \mathrm{C}}=\frac{(49-40) \mathrm{g}}{1 \mathrm{~g}}=9 \mathrm{~cm}^3 $

Now, $V_{42 \mathrm{C}}=V_{32 \mathrm{C}}\left[1+\gamma\left(t_2-t_1\right)\right]$

$ \begin{aligned} \Rightarrow \quad \gamma & =\frac{V_{42{ }^{\circ} \mathrm{C}}-V_{322^{\circ} \mathrm{C}}}{V_{322^{\circ} \mathrm{C}}\left(t_2-t_1\right)}=\frac{9-8.33}{8.33(42-32)} \\ & =\frac{0.66}{83.3}=0.0079 /{ }^{\circ} \mathrm{C} \\ \alpha & =\frac{\gamma}{3}=\frac{0.0079}{3}=\frac{7.9}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \\ & \simeq \frac{8}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \end{aligned} $

Given, base area of cooking pot,

$ A=0.2 \mathrm{~m}^2 $

Thickness, $l=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$

Boiling rate, $R=\frac{m}{t}=3 \mathrm{~kg} / \mathrm{min}=\frac{1}{20} \mathrm{~kg} / \mathrm{s}$ Thermal conductivity,

$ K=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1} $

Heat of evaporation, $L=2 \times 10^6 \mathrm{~J} / \mathrm{kg}$

The amount of heat flowing into water through the base of cooking pot,

$ Q=\frac{K A\left(T_1-T_2\right) t}{l} $

where, $T_1=$ Temperature of the cooking pot and

$ T_2=\text { Boiling point of water } . $

Heat required to boil water, $Q^{\prime}=m L$

$ \begin{aligned} & \frac{K A\left(T_1-T_2\right)}{l} t=m L \\ & \begin{aligned} \Rightarrow T_1-T_2 & =\frac{m L l}{K A t}=\left(\frac{m}{t}\right) \frac{L l}{K A} \\ & =\frac{1}{20} \times \frac{2 \times 10^6 \times 2 \times 10^{-2}}{120 \times 0.2}=83^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

In given process

Pressure $\propto$ Volume $\Rightarrow p=k V$ work done in process,

$ \begin{aligned} \Delta W & =\int_{V_0}^{3 V_0} \rho d V=\int_{V_0}^{3 V_0} k V \cdot d V \\ & =k\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0} \\ & =\frac{k}{2}\left\{9 V_0^2-V_0^2\right\} \\ & =k \cdot 4 V_0^2=\frac{p_0}{V_0} \cdot 4 V_0^2 \end{aligned} $

$or\,\, \Delta W=4 p_0 V_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)$

As gas is monoatomic, $C_V=\frac{3}{2} R$

Change in internal energy of gas,

$ \Delta U=n C_v\left(T_f-T_i\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(ii)$

Now, from $p V=n R T$;

$ T_i=\frac{p_0 V_0}{n R} $

and $\quad T_f=\frac{9 p_0 V_0}{n R}$

So, from Eq. (ii)

$ \Delta U=n \cdot \frac{3}{2} R \cdot \frac{8 p_0 V_0}{n R}=12 p_0 V_0 $

By $\Delta Q=\Delta U+\Delta W$, we have Heat absorbed

$ \Delta Q=12 p_0 V_0+4 p_0 V_0=16 p_0 V_0 $

Given, $p_1=p, V_1=V, T_1=T$

$ p_2=p, V_2=4 V, T_2=\frac{T}{2} $

For the adiabatic relation of perfect gas,

$ \begin{aligned} & & T V^{\gamma-1} & =\text { constant } \\ & \Rightarrow & T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ & \Rightarrow & T V^{\gamma-1} & =\frac{T}{2}(4 V)^{\gamma-1} \end{aligned} $

$ \begin{aligned} \Rightarrow & & V^{\gamma-1} & =\frac{1}{2} \cdot(4)^{\gamma-1} \cdot V^{\gamma-1} \\ \Rightarrow & & 1 & =\frac{1}{2}(4)^{\gamma-1} \Rightarrow 2=(4)^{\gamma-1} \\ \Rightarrow & & 2 & =2^{2(\gamma-1)} \Rightarrow 2(\gamma-1)=1 \\ \Rightarrow & & \gamma-1 & =1 / 2 \\ \Rightarrow & & \gamma & =3 / 2 \end{aligned} $

$ \begin{aligned} &\text { Work done in adiabatic expansion, }\\ &\begin{aligned} W & =\frac{n R}{\gamma-1}\left[T_1-T_2\right] \\ & =\frac{n R}{\frac{3}{2}-1}[T-T / 2] \\ & =2 n R \times \frac{T}{2}=n R T \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad W=p V=\alpha p V \quad & \text { (given) } \\ & {[\because p V=n R T] } \end{aligned} $

$ \therefore \quad \alpha=1 $

$ \text { } \begin{aligned} \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\sqrt{\frac{3 R T}{M_{\mathrm{H}_2}}}=\sqrt{\frac{3 R \times 20}{2}} \\ & =\sqrt{30 R} \\ \left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} & =\sqrt{\frac{3 R T^{\prime}}{M_{\mathrm{O}_2}}}=\sqrt{\frac{3 R T^{\prime}}{32}} \end{aligned} $

According to question,

$ \begin{aligned} & & \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} \\ \Rightarrow & & \sqrt{30 R} & =\sqrt{\frac{3 R T^{\prime}}{32}} \Rightarrow 30 R=\frac{3 R T^{\prime}}{32} \\ \Rightarrow & & 30 \times 32 & =3 T^{\prime} \Rightarrow T^{\prime}=320 \mathrm{~K} \end{aligned} $

$ \begin{aligned} &\alpha=2 \times 10^{-5} \mathrm{~K}^{-1} \Rightarrow T=230^{\circ} \mathrm{C}\\ &\text { We know, } L=L_0[1+\alpha \Delta T]\\ &\begin{aligned} & =5\left[1+2 \times 10^{-5} \times(230-30)\right] \\ L & =5\left[1+\left(2 \times 10^{-5} \times 200\right)\right]=5.02 \mathrm{~cm} \end{aligned} \end{aligned} $

We know, $Q=m c \Delta T$

$ \begin{aligned} 2100 & =m \times 900 \times 2 \\ \Rightarrow \quad m & =\frac{2100}{1800}=1167 \mathrm{~kg} \end{aligned} $

which is nearest to option (a).

The given situation is shown below.

Work done in a cyclic process $=$ Area inside the cycle $A B C$

∴ Work done $=$ Area of $\triangle A B C$

$ \begin{aligned} & =\frac{1}{2} \times\left(2 V_1-V_1\right)\left(3 p_1-p_1\right) \\ & =\frac{1}{2} \times V_1 \times 2 p_1=p_1 V_1 \end{aligned} $

From 1st law of thermodynamics,

$ \begin{aligned} & \Delta Q=p \Delta V+n C_V \Delta T \\ = & p \Delta V+n \frac{R}{\gamma-1} \Delta T=p \Delta V\left[1+\frac{1}{\gamma-1}\right] \\ \Rightarrow & \Delta Q=200\left[1+\frac{1}{\gamma-1}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

We know, $\frac{C_p}{C_V}=\gamma$

$ \begin{aligned} & \Rightarrow \quad \frac{C_p}{C_p-R}=\gamma \quad\left(\because C_p-C_V=R\right) \\ & \Rightarrow \quad \gamma=\frac{\frac{7}{2} R}{\frac{7}{2} R-R}=\frac{\frac{7}{5}}{\frac{5}{2}}=\frac{7}{5} \end{aligned} $

Put $\gamma=\frac{7}{5}$ is Eq. (i)

$ \begin{array}{ll} \Delta Q & =200\left[1+\frac{1}{\left(\frac{7}{5}-1\right)}\right] \\ \Rightarrow & \Delta Q=200\left[1+\frac{5}{2}\right] \\ \Rightarrow & \Delta Q=200 \times \frac{7}{2}=700 \mathrm{~J} \end{array} $

Statement I is false because gas thermometers are more sensitive than liquid thermometers. This is because gas expands more than a liquid.

Statement II is true because $R / N_A=k_B$ where, $k_B=$ Boltzmann Constant

$R=$ Universal gas constant

$N_A=$ Avogadro number

Statement III is true because

$ \begin{array}{r} \text { Density }=\frac{\text { mass }}{\text { volume }} \\   d=m / V ...(i)\end{array} $

Now, by ideal gas equation

$ p V=n R T $

$p V=\frac{m}{M} R T$, where $M$ is the molar mass of gas.

$ \begin{aligned} & \therefore \quad m=\frac{p V M}{R T} \\ & \text { Put } m=\frac{p V M}{R T} \text { in Eq. (i) } \\ & \qquad d=\frac{p V M}{R T V} \Rightarrow d=\frac{p M}{R T} \\ & \Rightarrow \quad d \propto \frac{1}{T} \end{aligned} $

$ \begin{gathered} \text { } \Delta L_{\text {steel }}=L_S \alpha_S \Delta t \\ \Delta L_{\text {copper }}=L_C \alpha_C \Delta t \end{gathered} $

Given, $L_S-L_C=4 \mathrm{~cm}$ for any temperature

$ \begin{array}{ll} \therefore & L_S \alpha_S \Delta t=L_C \alpha_C \Delta t \\ & \frac{L_S}{L_C}=\frac{\alpha_C}{\alpha_S}=\frac{1.7 \times 10^{-5}}{1.1 \times 10^{-5}}=\frac{1.7}{1.1}=\frac{17}{11} \end{array} $

By Newton's law of cooling,

$ \frac{-d T}{d t}=b\left[T-T_0\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $\frac{-d T}{d t}=$ Cooling rate,

$T=$ Body's temperature,

$T_0=$ Surrounding temperature and $b=$ Constant.

∴ From Eq. (i),

$ \begin{aligned} & \int_{T_i}^T \frac{d T}{T-T_0}=-b \int_0^t d t \\ \Rightarrow \quad & T=T_0+\left(T_i-T_0\right) e^{-b t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From the first condition of question,

$ \begin{aligned} T & =40^{\circ} \mathrm{C} \\ T_0 & =10^{\circ} \mathrm{C} \\ T_i & =100^{\circ} \mathrm{C} \\ t & =10 \mathrm{~min} \end{aligned} $

$ \text { Putting these value in Eq. (ii), } $

$ \begin{array}{ll} & 40=10+(100-10) e^{-b \times 10} \\ \Rightarrow & \frac{30}{90}=e^{-10 b} \\ \Rightarrow & \frac{1}{3}=e^{-10 b} \\ \Rightarrow & \ln \left(\frac{1}{3}\right)=-10 b \ln m \\ \Rightarrow & \ln 1-\ln 3=-10 b \\ \Rightarrow & -\ln 3=-10 b \\ \Rightarrow & b=\frac{\ln 3}{10}=\frac{11}{10} \end{array} $

Now, by next condition of question,

$ \begin{aligned} & T=20^{\circ} \mathrm{C} \\ & T_i=70^{\circ} \mathrm{C} \\ & T_a=10^{\circ} \mathrm{C} \end{aligned} $

From Eq. (ii), $20=10+(70-10) e^{-b t}$

$ \begin{gathered} \frac{10}{60}=e^{\frac{1.1}{10} t} \\ \Rightarrow \quad \ln \left(\frac{1}{6}\right)=-\frac{1.1}{6} t \ln e \\ \Rightarrow \ln 1-\ln 6=-\frac{1.1}{10} t \Rightarrow-\ln 6=\frac{-11}{100} t \\ \Rightarrow \quad t=\frac{100 \times \ln 6}{11}=\frac{100 \times 1.8}{11}=16.3 \mathrm{~min} \end{gathered} $

We know, from first law of thermodynamics,

$ \Delta Q=\Delta U+W \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Given, volume of liquid, $V_l=10^{-3} \mathrm{~m}^3$

And volume of gas or volume of steam,

$ V_g=2001 \mathrm{~m}^3 $

Now, $V_g \gg V_l$

$ \begin{aligned} & \therefore \quad\left(V_g-V_l\right) \approx V_g \\ & W=p V_g=\left(10^5 \times 2.001\right) \mathrm{J}=2.001 \times 10^2 \mathrm{~kJ} \\ & =200 \mathrm{~kJ} \end{aligned} $

Now, heat required for phase change to convert the liquid into steam,

$ \Delta Q=m L=1 \times 2000 \mathrm{~kJ} $

Put in Eq (i), $\Delta U=\Delta Q-W$

$ \Rightarrow \quad \Delta U=(2000-200) \mathrm{kJ} $

[from Eq. (i)]

$ \Rightarrow \quad \Delta U=1800 \mathrm{~kJ} $

Given, $W=100 \mathrm{~J}$

From ideal gas equation,

$ \begin{aligned} & & p \Delta V & =n R \Delta T\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \because & W & =p \Delta V \end{aligned} $

∴ Eq. (i) becomes,

$ W=n R \Delta T \Rightarrow \Delta T=\frac{W}{n R}=\frac{100}{n R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, we know the relation for change in internal energy,

$ \begin{aligned} & \Delta U=n C_V \Delta T \\ & \Rightarrow \quad \Delta U=n \frac{R}{\gamma-1} \Delta T \quad \ldots(\text { iii }) \\ &\left(\because C_V=\frac{R}{\gamma-1}\right) \end{aligned} $

and since it is a monoatomic gas,

$ \therefore \quad \gamma=\frac{5}{3} $

From Eqs. (ii) and (iii)

$ \begin{array}{lc} \Rightarrow & \Delta U=150 \mathrm{~J} \\ \Rightarrow & \Delta U=\frac{n R \Delta T}{\frac{5}{3}-1}=\frac{3 n R}{2} \times \frac{100}{n R}=150 \mathrm{~J} \end{array} $

Now, the heat given to the gas in the process is

$ \begin{aligned} & & \Delta Q & =\Delta U+W \\ \Rightarrow & & \Delta Q & =150+100=250 \mathrm{~J} \\ \Rightarrow & & \Delta Q & =250 \mathrm{~J} \end{aligned} $

We know, $v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}$ (where $M=$ molar mass of gas)

For $\mathrm{N}_2$ :

$ \begin{aligned} \Rightarrow \quad & 100=\sqrt{\frac{3 R T \times 1000}{28}} \\ & \left(\text { for } N_2 \text { molar mass, } M=\frac{28}{1000} \mathrm{~kg}\right) \end{aligned} $

$ \Rightarrow \quad 100=\sqrt{\frac{3000 R T}{28}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Now, for He, $v_{\text {rms }}^{\prime}=\sqrt{\frac{3 R T \times 1000}{4}}$

$i.e\,\, v_{\mathrm{rms}}^{\prime}=\sqrt{\frac{3000 R T}{4}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (ii) by Eq. (i), we get

$ \begin{array}{ll} & \frac{v_{\mathrm{rms}}^{\prime}}{100}=\sqrt{\frac{\frac{3000 R T}{4}}{\frac{3000 R T}{28}}} \\ \Rightarrow & \frac{v_{\mathrm{rms}}^{\prime}}{100}=\sqrt{\frac{28}{4}} \\ \Rightarrow & v_{\mathrm{rms}}^{\prime}=100 \sqrt{7} \mathrm{~m} / \mathrm{s} \end{array} $

Given, $s_w=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$

$\begin{aligned} & s_{\text {lece }}=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1} \\ & L_f=80 \mathrm{cal} \mathrm{g}^{-1} \end{aligned}$

Mass of ice, $m_{\text {cee }}=5 \mathrm{~g}$

Mass of water, $m_w=20 \mathrm{~g}$

Let final temperature of the mixture be $T^{\circ} \mathrm{C}$.

According to principle of calorimetry,

Heat lost by water $=$ Heat gained by ice

$\begin{aligned} & m_w s_w(35-T)=m_{\text {cce }} s_{\text {ce }}[0-(-30)] \\ & +\mathrm{m}_{i c c} \times L_f+m_{\text {kec }} \times S_w \times T \\ & \Rightarrow 20 \times 1 \times(35-T)=5 \times 0.5 \times 30+5 \times 80+5 \times 0.5 \times T \\ \end{aligned}$

On solving, we get

$T=9^{\circ} \mathrm{C}$

Given :

• Diameter of the sphere, $D$

Initial surface area, $A$ is:

$ A = 4 \pi R^2 $

where $R = \frac{D}{2}$.

So,

$ A = 4 \pi \left(\frac{D}{2}\right)^2 = \pi D^2 \quad \text{(i)} $

After heating by temperature $\delta T$, the new surface area, $A'$, is:

$ A' = 4 \pi \left(\frac{D'}{2}\right)^2 = \pi \left(D'\right)^2 \quad \text{(ii)} $

Here, $D'$ is the new diameter.

Using the equation for linear expansion:

$ D' = D(1 + \alpha \delta T) \quad \text{(iii)} $

Substituting the value of $D'$ from equation (iii) into equation (ii), we get:

$ \begin{aligned} A' & = \pi D^2 (1 + \alpha \delta T)^2 \\ & = \pi D^2 \left(1 + \alpha^2 \delta T^2 + 2 \alpha \delta T\right) \quad \text{(iv)} \end{aligned} $

Simplifying further:

$ \begin{aligned} A' & = \pi D^2 \left(1 + \alpha^2 \delta T^2 + 2 \alpha \delta T\right) - \pi D^2 \\ & = \pi D^2 \left[1 + \alpha^2 \delta T^2 + 2 \alpha \delta T - 1\right] \\ & = \pi D^2 \left[\alpha^2 \delta T^2 + 2 \alpha \delta T\right] \end{aligned} $

The change in surface area is:

$ \begin{aligned} \Delta A & = A' - A \\ & = \pi D'^{ 2} - \pi D^{ 2 } \\ & = \pi D^2 \left(\alpha \delta T (\alpha \delta T + 2)\right) \end{aligned} $

To find the energy exhausted by the Carnot engine, we need to understand the relationship between the work done, the heat absorbed, and the heat exhausted in a Carnot cycle.

The efficiency ($\eta$) of a Carnot engine operating between two temperatures $T_{h}$ (hot reservoir) and $T_{c}$ (cold reservoir) is given by the formula:

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

In this case, the temperatures are:

$T_{h} = 400 \, \text{K}$

$T_{c} = 300 \, \text{K}$

Substituting these values into the efficiency formula, we get:

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

$\eta = 1 - 0.75$

$\eta = 0.25$

The efficiency of the Carnot engine is 0.25 (or 25%). This means that 25% of the heat absorbed (Q_h) is converted into work (W), and the rest is exhausted as waste heat (Q_c).

We know that the work done by the engine (W) is 400 J. Using the efficiency, we can find the heat absorbed (Q_h):

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

$400 = 0.25 \cdot Q_{h}$

$Q_{h} = \frac{400}{0.25}$

$Q_{h} = 1600 \, \text{J}$

The heat absorbed by the engine is 1600 J. To find the energy exhausted (Q_c), we use the relationship:

$Q_{h} = W + Q_{c}$

Substituting the known values, we get:

$1600 = 400 + Q_{c}$

$Q_{c} = 1600 - 400$

$Q_{c} = 1200 \, \text{J}$

Therefore, the energy exhausted by the engine is 1200 J.

The correct answer is:

Option B: 1200 J

To solve this problem, we need to understand the relationship between the slopes of isothermal and adiabatic processes in a pressure-volume ($p-V$) diagram for a gas, as well as the heat capacity ratio (also known as the adiabatic index $\gamma$).

For an isothermal process (constant temperature), the pressure and volume relationship is given by Boyle's law:

$ pV = \text{constant} $

Differentiating the above with respect to volume, we get:

$ \frac{dp}{dV} = -\frac{p}{V} $

Thus, the slope of the isothermal $p-V$ graph, $S_I$, is:

$ S_I = -\frac{p}{V} $

For an adiabatic process (no heat exchange), the relationship between pressure and volume is given by:

$ pV^\gamma = \text{constant} $

where $\gamma$ is the heat capacity ratio. Differentiating the above with respect to volume, we get:

$ \frac{dp}{dV} = -\gamma \frac{p}{V} $

Thus, the slope of the adiabatic $p-V$ graph, $S_A$, is:

$ S_A = -\gamma \frac{p}{V} $

From the above, we can see that the ratio of $S_I$ to $S_A$ is:

$ \frac{S_I}{S_A} = \frac{-\frac{p}{V}}{-\gamma \frac{p}{V}} = \frac{1}{\gamma} $

Given that the heat capacity ratio $\gamma = \frac{3}{2}$, we substitute this value into the ratio:

$ \frac{S_I}{S_A} = \frac{1}{\frac{3}{2}} = \frac{2}{3} $

Thus, the correct answer is:

Option B: $\frac{2}{3}$

The number of rotational degrees of freedom of a diatomic molecule is an important concept in molecular physics, which helps in understanding how the molecule can rotate in space.

A diatomic molecule consists of two atoms connected by a bond. Due to this structure, the molecule can rotate around two axes perpendicular to the bond axis. However, rotation around the bond axis itself does not count as a rotational degree of freedom for rigid diatomic molecules because the moment of inertia around this axis is negligible.

Therefore, the diatomic molecule has:

• Two rotational degrees of freedom: rotation around two axes perpendicular to the bond axis.

Based on this explanation, the correct answer is:

Option C: 2

Given:

• Coefficient of linear expansion, $\alpha = 1 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1}$


• Initial temperature, $T_1 = 25^{\circ} \mathrm{C}$


• Final temperature, $T_2 = -15^{\circ} \mathrm{C}$

Firstly, calculate the change in temperature:

$ \Delta T = T_1 - T_2 = 25 - (-15) = 40^{\circ} \mathrm{C} $

Next, determine the change in length ($\Delta L$) with respect to the original length (L):

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

Thus, the fractional change in length is:

$ \frac{\Delta L}{L} = \alpha \Delta T $

$ \frac{\Delta L}{L} = 1 \times 10^{-5} \times 40 $

$ \frac{\Delta L}{L} = 4 \times 10^{-4} $

To express this as a percentage change in the measurement of length:

$ \frac{\Delta L}{L} \times 100 = 4 \times 10^{-4} \times 100 $

$ \frac{\Delta L}{L} \times 100 = 0.04\% $

Therefore, the percentage error in the measurement of length is 0.04%.

Hint : Work done by adiabatic expansion $=-$ (Change in internal energy) $=-(-30 \mathrm{~J})=30 \mathrm{~J}$

Initial temperature of the sink, $T_2=250 \mathrm{~K}$

Initial efficiency of carnot engine,

$\eta_1=25 \%=0.25$

If $T_1$ be the temperature of the sink, then

$\begin{aligned} \eta_1 & =1-\frac{T_2}{T_1} \\ 0.25 & =1-\frac{250}{T_1} \Rightarrow \frac{250}{T_1}=0.75 \\ \Rightarrow \quad T_1 & =\frac{1000}{3} \mathrm{~K} \end{aligned}$

When efficiency is increased by $50 \%$

i.e $\eta_2=50 \%=0.5$

Then if $T_2^{\prime}$ be temperature of sink, then

$\begin{array}{rlrl} & \eta_2 & =1-\frac{T_2^{\prime}}{T_1} \\ \Rightarrow & & 0.5 & =1-\frac{T_2^{\prime}}{1000 / 3} \\ \Rightarrow & & \frac{T_2^{\prime}}{1000 / 3} & =1-0.5=0.5=\frac{1}{2} \\ \Rightarrow & & T_2^{\prime} & =\frac{1}{2} \times \frac{1000}{3}=\frac{500}{3} \mathrm{~K} \end{array}$

Here we see that to increase the efficiency of carnot engine temperature of the sink should be decreased, but in question, asked about increasing temperature of sink to increase efficiency, which is not possible. Hence, no option is correct.

In non-rigid diatomic molecule with an additional vibrational mode, each molecule has additional energy equal to $2 \times \frac{1}{2} k_B T=k_B T$, because a vibrational frequency has both kinetic and potential energy modes.

$\begin{aligned} \therefore \quad U & =\left(\frac{5}{2} k_B T+k_B T\right) N_A=\frac{7}{2} k_B N_A T \\ & =\frac{7}{2} R T \quad\left[R=k_B N_A\right] \end{aligned}$

$\begin{aligned} \therefore \quad C_v & =\frac{d U}{d T}=\frac{d}{d T}\left(\frac{7}{2} R T\right)=\frac{7}{2} R \\ C_p & =C_V+R=\frac{7}{2} R+R=\frac{9}{2} R \\ \frac{C_p}{C_V} & =\frac{9 / 2 R}{7 / 2 R} \Rightarrow \frac{C_p}{C_V}=\frac{9}{7} \\ \Rightarrow \quad 49 C_p^2 & =81 C_V^2 \end{aligned}$

Given surface area of sphere, $A=4 \mathrm{~m}^2$

Temperature, $T=400 \mathrm{~K}$

Emissivity, $\varepsilon=0.5$

Environment temperature,

$T_0=200 \mathrm{~K}$

Stefan Boltzmann constant, $\sigma=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^4$

According to Stefan Boltzmann law, rate of energy exchange of the sphere

$\begin{aligned} E & =\varepsilon \sigma A\left(T^4-T_0^4\right) \\ & =0.5 \times 5.67 \times 10^{-8} \times 4\left(400^4-200^4\right) \\ & =27216 \times 10^3 \mathrm{~W}=2721.6 \mathrm{~W} \end{aligned}$

Efficiency of carnot engine,

$\eta=1-\frac{T_2}{T_1}\quad \text{... (i)}$

Where, $T_2=$ Temperature of sink and $T_1=$ temperature of source According to first condition,

$\begin{aligned} & T_2=27+273=300 \mathrm{~K} \\ & \eta=40 \%=0.4 \end{aligned}$

From Eq. (i), $0.4=1-\frac{300}{T_1}$

$\Rightarrow \quad \frac{300}{T_1}=0.6 \Rightarrow T_1=\frac{300}{0.6}=500 \mathrm{~K}$

$\Rightarrow \quad T_1=500 \mathrm{~K}$

Again, when, $\eta=50 \%=0.5$, then

$\begin{aligned} & \qquad 0.5=1-\frac{300}{T_1^{\prime}} \\ & \Rightarrow \quad \frac{300}{T_1^{\prime}}=0.5 \Rightarrow T_1^{\prime}=600 \mathrm{~K} \\ & \begin{aligned} & \therefore \text { Increase of source temperature }=T^{\prime}-T_1 \\ &=600-500=100 \mathrm{~K} \end{aligned} \end{aligned}$

given, power of engine, $P=20 \mathrm{~kW}$

$=2 \times 10^4 \mathrm{~W}$

Efficiency, $\eta=40 \%=0.4$

Time, $t=1$ hour $=60 \times 60 \mathrm{~s}$

Energy of car, $E=$ Power $\times$ Time

$=2 \times 10^4 \times 60 \times 60=7.2 \times 10^7 \mathrm{~J}$

$\begin{aligned} & \text { Energy generated by fuel Combustion }=\frac{E}{\eta} \\ & =\frac{7.2 \times 10^7}{0.4}=18 \times 10^7 \mathrm{~J} \\ & =18 \times 10^4 \times 10^3 \mathrm{~J}=18 \times 10^4 \mathrm{~kJ}=180000 \mathrm{~kJ} \end{aligned}$

A diatomic molecule has one normal mode of vibration, since it can only stretch or compress the single bond. Hence, it has only one vibrational degree of freedom.

Let the insulated container has n moles of the monoatomic gas. The loss of kinetic energy of the gas,

$\Delta {E_K} = {1 \over 2}(mn){v^2}$

If the change in temperature of the gas is $\Delta$T, then heat gained by the gas is,

$\Delta Q = {3 \over 2}nR\,\Delta T$

Now, according to question,

$\Delta Q = \Delta {E_K}$

${3 \over 2}nR\,\Delta T = {1 \over 2}mn{v^2}$

$ \Rightarrow \Delta T = {{m{v^2}} \over {3R}}$

In the curve, V $\propto$ $\frac{1}{T}$

$\Rightarrow$ V = m$\frac{1}{T}$ where, m = slope of curve.

$\Rightarrow$ VT = constant

$\therefore$ TdV + VdT = 0 ..... (i)

$\Rightarrow$ dV = $-\frac{V}{T}$dT

As, Q = $\Delta$U + W

CdT = C_VdT + pdV ..... (ii)

CdT = C_VdT $-$ $\frac{p}{T}$ VdT

$\Rightarrow$ C = C_V $-$ $\frac{pV}{T}$

= C_V $-$ R ($\because$ $\frac{pV}{T}$ = R)

For diatomic gas, C_V = $\frac{5}{2}$R

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