Heat and Thermodynamics

Given, $t_1=20^{\circ} \mathrm{C}$ and $t_2=60^{\circ} \mathrm{C}$

Coefficient of linear expansion,

$ \alpha=10^{-5 \circ} \mathrm{C}^{-1} $

As we know that, coefficient of area expansion is

$ \begin{aligned} \beta & =\frac{\Delta A}{A \times \Delta T} \Rightarrow 2 \alpha=\frac{\Delta A}{A \times \Delta T}[\because \beta=2 \alpha] \\ \Delta A & =2 \alpha \times A \times \Delta T \\ & =2 \times 10^{-5} \times 40 \times 20 \times 10^{-4} \times(60-20) \\ & =0.64 \times 10^{-4} \mathrm{~m}^2=0.64 \mathrm{~cm}^2 \end{aligned} $

As, two bars are placed in same medium at a specified temperature. So, at steady state, the rate of heat flow is same in them.

As we know,

$ \begin{aligned} \frac{Q}{t} & =\frac{K A \Delta T}{L} \Rightarrow \frac{Q_A}{t}=\frac{Q_B}{t} \Rightarrow \frac{K_A}{L_A}=\frac{K_B}{L_B} \\ \therefore \frac{K_A}{K_B} & =\frac{L_A}{L_B} \end{aligned} $

Given, pressure, $p=4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$, $\Delta Q=2000 \mathrm{~J}$

Change in volume, $\Delta V=3 \times 10^{-3} \mathrm{~m}^3$

Now, the work done to change volume is

$ \begin{aligned} \Delta W & =p \Delta V=4 \times 10^5 \times 3 \times 10^{-3} \\ & =12 \times 10^2=1200 \mathrm{~J} \end{aligned} $

According to first law of thermodynamics,

$ \begin{aligned} &\text { So, change in internal energy, }\\ &\Delta U=\Delta Q-\Delta W=2000-1200=800 \mathrm{~J} \end{aligned} $

For isothermal process, $\Delta T=$ constant

As we know that, from first law of thermodynamics,

$ \begin{aligned} & & \Delta Q & =\Delta U+\Delta W=n C_V \Delta T+\Delta W \\ \Rightarrow & \Delta Q & =\Delta W & {[\because \Delta T=0] } \end{aligned} $

Hence, heat supplied to an ideal gas under isothermal condition will result in external work done by the system.

Change in volume due to increase of temperature $(\Delta T)$ is given by

$ \begin{equation*} \Delta V=\alpha_V V \Delta T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

where, $\alpha_V=$ coefficient of volume expansion

and $\quad V=$ initial volume of gas.

For an ideal gas, we know that

$ \begin{aligned} & & p V & =n R T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & & p \Delta V & =n R \Delta T \\ \Rightarrow & & \Delta V & =\frac{n R \Delta T}{p} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

$ \begin{aligned} &\text { From Eqs. (i) and (iii), we get }\\ &\begin{aligned} & \alpha_V \cdot V \Delta T=\frac{n R \Delta T}{p} \\ \Rightarrow & \alpha_V=\frac{n R}{p V}=\frac{n R}{n R T} \quad \text { [from Eq. (ii)] } \\ \Rightarrow & \alpha_V=\frac{1}{T} \end{aligned} \end{aligned} $

According to Wien's displacement law,

$ \begin{equation*} \lambda_m \cdot T=b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

where, $\lambda_m=$ wavelength corresponding to maximum radiation of energy from black body,

$ b=\text { Wien's constant } $

and $T=$ temperature.

∴ From Eq. (i), we get

$ \lambda_m \propto \frac{1}{T} \Rightarrow \lambda_m \propto T^{-1} $

Efficiency of Carnot's engine $C_1$ is given as $\eta_1=1-\frac{T_2}{T_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $T_2=$ temperature of sink and $\quad T_1=$ temperature of source. Similarly, efficiency of second Carnot's engine, $\quad \eta_2=1-\frac{T_3}{T_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

The efficiency of combined engine ( $C_1$ and $C_2$ ) is given as

$ \begin{aligned} \eta & =\eta_1+\eta_2=1-\frac{T_2}{T_1}+1-\frac{T_3}{T_2} \\ & =2-\left(\frac{T_2}{T_1}+\frac{T_3}{T_2}\right) \end{aligned} $

At high pressure, the volume occupied by gas molecules does not approach to zero and at low temperature, the contribution of intermolecular forces becomes significant. Hence, at high pressure and low temperature, all gases deviate from gas laws.

Heat absorbed by a solid is given by

$ \Delta Q=m s \Delta T $

where, $s=$ specific heat of solid.

$ \begin{aligned} \therefore \quad s=\frac{\Delta Q}{m \cdot \Delta T} & =\frac{50 \times 1000}{2 \times\left(70^{\circ}-20^{\circ}\right) \mathrm{C}} \\ {[\because \Delta Q} & =50 \mathrm{~kJ}=50 \times 1000 \mathrm{~J}] \\ & =500 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body,

⇒ Rate of heat transfer $\propto$ Surface area of body

$ \Rightarrow \quad \frac{Q_2}{Q_1}=\frac{S_2}{S_1} \Rightarrow Q_2=Q_1 \times \frac{S_2}{S_1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, volume remains same in stretching, so

$ \begin{aligned} \pi r_1^2 l_1 & =\pi r_2^2 l_2 \\ \Rightarrow \quad l_2 & =\frac{r_1^2}{r_2^2} \cdot l_1=\frac{(2.5)^2}{(0.5)^2} \times 5=125 \mathrm{~cm} \end{aligned} $

So, area before stretching,

$ S_1=2 \pi r_1\left(l_1+r_1\right) $

and area after stretching,

$ S_2=2 \pi r_2\left(l_2+r_2\right) $

Hence by Eq. (i), we get

$ \begin{aligned} Q_2 & =Q_1 \times \frac{S_2}{S_1}=Q_1 \times \frac{2 \pi r_2\left(l_2+r_2\right)}{2 \pi r_1\left(l_1+r_1\right)} \\ & =Q_1 \times \frac{r_2\left(l_2+r_2\right)}{r_1\left(l_1+r_1\right)} \\ & =\frac{1 \times 0.5 \times 10^{-2}(125+0.5) \times 10^{-2}}{2.5 \times 10^{-2}(5+2.5) \times 10^{-2}} \\ & =3.346=3.35 \mathrm{~W} \end{aligned} $

The given situation is shown in the following diagram

Work done in expansion (process I)

$ \begin{aligned} & =\int_{V_i}^{V_f} p d V=\int_{V_0}^{3 V_0} \frac{p_0 V}{V_0} \cdot d V \\ & =\frac{p_0}{V_0}\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0}=\frac{p_0}{2 V_0}\left(9 V_0^2-V_0^2\right)=4 p_0 V_0 \end{aligned} $

In process II, there is no work done as

$V=$ constant.

In process III, work done is given as

$ \begin{aligned} W & =p_0 \int_{V_i}^{V_f} d V=p_0 \int_{3 V_0}^{V_0} d V \\ & =p_0\left(V_0-3 V_0\right)=-2 p_0 V_0 \end{aligned} $

So, total work done in entire process is

$ W_{\text {Total }}=4 p_0 V_0+\left(-2 p_0 V_0\right)=2 p_0 V_0 $

A rigid diatomic molecule can store energy only by two possible ways of rotation, so it has only 2 rotational degrees of freedom.

Given, $C_V=12.6 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

$ R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} $

By Mayer's relation,

$ \begin{aligned} C_p & =C_V+R=12.6+8.314 \\ & =20.914 \approx 20.9 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned} $

The given situation is shown below

In steady state, heat flow rate through $A$ and $B$ must be equal.

If $T=$ temperature of common surface, then

$ \begin{aligned} Q_A & =Q_B \\ \frac{K_A\left(75^{\circ} \mathrm{C}-T\right)}{X_A} & =\frac{K_B\left(T-50^{\circ} \mathrm{C}\right)}{X_B} \end{aligned} $

here, $K_A=\frac{K_B}{2}$ and $X_A=2 X_B$

$ \begin{aligned} & \Rightarrow \quad \frac{K_B\left(75^{\circ} \mathrm{C}-T\right)}{2 \times 2 X_B}=\frac{K_B\left(T-50^{\circ} \mathrm{C}\right)}{X_B} \\ & \Rightarrow \quad 75^{\circ} \mathrm{C}-T=4 T-200^{\circ} \mathrm{C} \\ & \Rightarrow \quad 5 T=275^{\circ} \mathrm{C} \Rightarrow T=55^{\circ} \mathrm{C} \end{aligned} $

Work done in an adiabatic expansion is

$ \Delta W=\frac{\mu R \Delta T}{\gamma-1} $

For 1 mole of monoatomic gas,

$ \begin{aligned} & \Delta T=20^{\circ} \mathrm{C}, \gamma=\frac{5}{3} \\ \therefore & W=\frac{1 \times R \times 20}{\left(\frac{5}{3}-1\right)}=30 R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For 6 moles of rigid diatomic gas,

$ \begin{aligned} \Delta T & =20^{\circ} \mathrm{C}, \gamma=\frac{7}{5} \\ \Delta W^{\prime} & =\frac{6 \times R \times 20}{\left(\frac{7}{5}-1\right)}=\frac{5 \times 6 \times R \times 20}{2} \\ & =10 \times 30 \mathrm{R}=10 \mathrm{~W} \end{aligned} $

[from Eq. (i)]

For a gas with $n$ degree of freedom,

$ C_p=\left(\frac{n}{2}+1\right) \cdot R \quad \text { and } C_V=\frac{n}{2} R $

$ \Rightarrow \quad \frac{C_p}{C_V}=\frac{\left(\frac{n}{2}+1\right) R}{\frac{n}{2} \cdot R}=\left(\frac{n+2}{n}\right) $

Let E' be the new radiant energy.

According to Stefan's law,

E $\propto$ T⁴

$\Rightarrow$ ${{{E_2}} \over {{E_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$

$\Rightarrow$ ${{E'} \over E} = {\left\{ {{{(T/4)} \over T}} \right\}^4} = {\left( {{1 \over 4}} \right)^4}$

or, $E' = {E \over {256}}$

In adiabatic process,

pV^$\gamma$ = K

where, K is a constant.

Given that, $\gamma = {4 \over 3} \Rightarrow p{V^{4/3}} = K$

Taking logarithm on both sides, we get

$\log p + {4 \over 3}\log V = \log K$

$ \Rightarrow {{\Delta p} \over p} + {4 \over 3}{{\Delta V} \over V} = 0$

$ \Rightarrow {{\Delta V} \over V} = - {3 \over 4}{{\Delta p} \over p}$

Now, % change in volume,

${{\Delta V} \over V} \times 100 = - {3 \over 4}\left( {{{\Delta p} \over p} \times 100} \right)$

It is given that, pressure decreases by ${3 \over 4}$%,

i.e. ${{\Delta p} \over p} \times 100 = - {3 \over 4}$

$\therefore$ ${{\Delta V} \over V} \times 100 = - {3 \over 4} \times \left( { - {3 \over 4}} \right) = {9 \over {16}}$%

Hence, the volume increases by ${9 \over {16}}$%.

Efficiency of Carnot engine is given by

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

Given, T₁ = 600 K

T₂ = 300 K

T'₁ = 1600 K and T'₂ = x K

Carnot engine has same efficiency between 600 K to 300 K and 1600 K to x K, so

$\left( {1 - {{{T_2}} \over {{T_1}}}} \right) = \left( {1 - {{T{'_2}} \over {T{'_1}}}} \right)$

Substituting all the given values in above equation, we get

$1 - {{300} \over {600}} = 1 - {x \over {1600}}$

$1 - {1 \over 2} = 1 - {x \over {1600}}$

${x \over {1600}} = 1 - {1 \over 2} = {1 \over 2}$

$ \Rightarrow x = {{1600} \over 2} = 800$ K