Heat and Thermodynamics

Let $n_m$ be the number of moles of the monoatomic ideal gas.

For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$, and the adiabatic constant is

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

For process $a \rightarrow b$ (Isothermal Process):

The temperature remains constant: $T_a = T_b = T_0 \rightarrow \Delta T_{ab} = 0$

So, the change in internal energy is zero: $\Delta U_{ab} = 0$

The volume changes from $V_1$ to $V_2 (V_2 > V_1)$. So, it's an isothermal expansion.

The work done by the gas during isothermal process is given as $W = nRT \ln\left(\frac{V_f}{V_i}\right)$, where $n$ is the amount of gas, $T$ is the temperature, $V_f$ is the final volume and $V_i$ is the initial volume.

According to the first law of thermodynamics,

$Q_{ab} = \Delta U_{ab} + W_{ab} = 0 + n_m R T_0 \ln\left(\frac{V_2}{V_1}\right)$

Since $V_2 > V_1$, $Q_{ab} > 0$. Which means heat is absorbed by the gas ($Q_{absorbed} = Q_{ab}$).

For the process $b \rightarrow c$ (Isochoric Process):

The volume is kept constant, $V_c = V_b = V_2 \rightarrow \Delta V = 0$

So, the work done is zero, $W_{bc} = 0$.

Using first law of thermodynamics the heat transferred directly equals the change in internal energy:

$Q_{bc} = \Delta U_{bc} = n_m C_v (T_c - T_b) = \frac{3}{2} n_m R (T_c - T_0)$

For the process $c \rightarrow a$ (Adiabatic Process):

As in adiabatic process there is no heat is exchanged with the surroundings $Q_{ca} = 0$.

The state changes from $(V_2, T_c)$ back to $(V_1, T_0)$ and completes the cycle.

For a reversible adiabatic process ($c \rightarrow a$), the relationship between temperature $T$ and volume $V$ is

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

Applying this equation at states $c$ and $a$:

$T_c V_c^{\gamma - 1} = T_a V_a^{\gamma - 1}$

Substituting $V_c = V_2$, $V_a = V_1$, and $T_a = T_0$:

$T_c V_2^{\gamma - 1} = T_0 V_1^{\gamma - 1} \rightarrow \frac{T_0}{T_c} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1}$

For a monoatomic gas, $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.

Therefore :

$\frac{T_0}{T_c} = \left(\frac{V_2}{V_1}\right)^{2/3}$

When $\frac{v_2}{v_1} = 8$

$\frac{T_a}{T_c} = \frac{T_a}{T_c} = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$

Which means $T_a = 4\,T_c$, i.e., the temperature of the gas at state a is exactly 4 times the temperature at state c.

Thus, option (C) is correct.

Process $a \rightarrow b$ is isothermal ($T_a = T_b$). Using the ideal gas law $(PV = n_mRT)$:

$P_a\,V_a = P_bV_b \rightarrow P_aV_1 = P_bV_2$

$\frac{P_a}{P_b} = \frac{V_2}{V_1}$

If $\frac{v_2}{v_1} = 8$, then:

$\frac{P_a}{P_b} = 8 \rightarrow P_a = 8\,P_b$

So, the pressure of the gas at a is 8 times the pressure of the gas at b. Thus, option (D) is incorrect.

Since $T_a = T_0$ and $T_c = T_0/4$ when $\frac{v_2}{v_1} = 8$, the temperature decreases during the constant-volume process $b \rightarrow c$. The heat exchanged in process $b \rightarrow c$ is:

$Q_{bc} = \frac{3}{2}\,n_m\,R\,(T_c - T_b) = \frac{3}{2}\,n_m\,R\left(\frac{T_0}{4} - T_0\right) = \frac{3}{2}\,n_m\,R\left(-\frac{3}{4}T_0\right) = -\frac{9}{8}\,n_m\,R\,T_0$

The negative sign indicates that heat is released by the system

$Q_{\text{released}} = |Q_{bc}| = 1.125\,n_m\,R\,T_0$

The heat absorbed during the isothermal process $a \rightarrow b$:

$Q_{ab} = n_m\,R\,T_0\,\ln\left(\frac{V_2}{V_1}\right) = n_m\,R\,T_0\,\ln(8) = n_m\,R\,T_0\,\ln(2^3) = 3\,n_m\,R\,T_0\,\ln(2)$

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

$Q_{ab} \approx 3\,n_m\,R\,T_0(0.7) = 2.1\,n_m\,R\,T_0$

So, heat released in $b \rightarrow c$ is $Q_{\text{released}} = 1.125\,n_m\,R\,T_0$

Heat absorbed in $a \rightarrow b$ is $Q_{\text{absorbed}} = 2.1\,n_m\,R\,T_0$

Hence, the heat released in process $b \rightarrow c$ is smaller than the heat absorbed in process $a \rightarrow b$. Thus, option (A) is correct.

The net efficiency $\eta$ of the cycle is,

$\eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}$

Here, $Q_{\text{in}} = Q_{ab} = n_m\,R\,T_0\,\ln\left(\frac{v_2}{v_1}\right)$ and $Q_{\text{out}} = |Q_{bc}| = \frac{3}{2}\,n_m\,R\,(T_0 - T_c)$.

Also, $T_c = T_0\left(\frac{V_1}{V_2}\right)^{\gamma - 1}$.

Substituting into $Q_{\text{out}}$,

$Q_{\text{out}} = \frac{3}{2} n_m R T_0 \left[ 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right]$

Substituting both heat expressions into the efficiency formula:

$\eta = 1 - \frac{\frac{3}{2} n_m R T_0 \left[ 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right]}{n_m R T_0 \ln\left( \frac{V_2}{V_1} \right)}$

$\eta = 1 - \frac{\frac{3}{2} \left[ 1 - \left( \frac{V_1}{V_2} \right)^{2/3} \right]}{\ln\left( \frac{V_2}{V_1} \right)}$

For a fixed volume ratio $V_2/V_1$, the efficiency $\eta$ does not depend on the temperature of the isothermal process.

Thus, option (B) is correct.

Therefore, the correct options are (A), (B), and (C)

Carnot Engine:

Hot reservoir temperature ($T_H$) = 1000 K

Efficiency ($\eta$) = 0.4

Heat extracted from the hot reservoir ($Q_H$) = 150 J per cycle

Heat Pump:

Coefficient of Performance (COP) = 10

Work input ($W_{pump}$) = Work output from the engine ($W_{engine}$)

Hot reservoir temperature ($T_{H,hp}$) = 300 K

Let's analyze each statement:

Statement A: Work extracted from the Carnot engine in one cycle is 60 J.

The efficiency of an engine is defined as the ratio of the work output to the heat input from the hot reservoir.

$ \eta = \frac{W_{engine}}{Q_H} $

We are given $\eta = 0.4$ and $Q_H = 150$ J. We can find the work done by the engine:

$ W_{engine} = \eta \times Q_H = 0.4 \times 150 \text{ J} = 60 \text{ J} $

This statement is correct.

Statement B: Temperature of the cold reservoir of the Carnot engine is 600 K.

For a Carnot engine, the efficiency can also be expressed in terms of the temperatures of the hot and cold reservoirs:

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

where $T_C$ is the temperature of the cold reservoir and $T_H$ is the temperature of the hot reservoir (both in Kelvin).

We know $\eta = 0.4$ and $T_H = 1000$ K. We can solve for $T_C$:

$ 0.4 = 1 - \frac{T_C}{1000} $

$ \frac{T_C}{1000} = 1 - 0.4 = 0.6 $

$ T_C = 0.6 \times 1000 \text{ K} = 600 \text{ K} $

This statement is also correct.

Statement C: Temperature of the cold reservoir of the heat pump is 270 K.

The coefficient of performance (COP) for a heat pump is defined as the ratio of the heat delivered to the hot reservoir ($Q_{H,hp}$) to the work input ($W_{pump}$). For an ideal (Carnot) heat pump operating between temperatures $T_{C,hp}$ and $T_{H,hp}$, the COP is also given by:

$ \text{COP} = \frac{T_{H,hp}}{T_{H,hp} - T_{C,hp}} $

We are given the COP of the heat pump is 10 and its hot reservoir temperature $T_{H,hp} = 300$ K. Let's assume this heat pump is operating ideally with this COP. We can find $T_{C,hp}$:

$ 10 = \frac{300 \text{ K}}{300 \text{ K} - T_{C,hp}} $

Let's solve for $T_{C,hp}$:

$ 10 \times (300 - T_{C,hp}) = 300 $

$ 3000 - 10 T_{C,hp} = 300 $

$ 10 T_{C,hp} = 3000 - 300 = 2700 $

$ T_{C,hp} = \frac{2700}{10} \text{ K} = 270 \text{ K} $

This statement is correct, assuming the heat pump is ideal.

Statement D: Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

The COP of a heat pump is also defined as the ratio of the heat delivered to the hot reservoir ($Q_{H,hp}$) to the work input ($W_{pump}$).

$ \text{COP} = \frac{Q_{H,hp}}{W_{pump}} $

We know the COP is 10 and the work input to the heat pump is the work output from the engine, which we found to be $W_{pump} = W_{engine} = 60$ J.

So, the heat supplied to the hot reservoir of the heat pump is:

$ Q_{H,hp} = \text{COP} \times W_{pump} = 10 \times 60 \text{ J} = 600 \text{ J} $

This statement says the heat supplied is 540 J, which does not match our calculation of 600 J.

This statement is incorrect.

Based on our analysis, statements A, B, and C are correct.

Steeper curve (B) is adiabatic

Adiabatic $\Rightarrow \mathrm{PV}^v=$ const.

Or $\mathrm{P}\left(\frac{\mathrm{T}}{\mathrm{P}}\right)^v=$ const.

$\frac{\mathrm{T}^v}{\mathrm{P}^{v-1}}=\text { const. }$

Curve (A) is isothermal

$\mathrm{T}=$ const.

$\mathrm{PV}=$ const.

For adiabatic process $(\mathrm{A} \rightarrow \mathrm{B})$

$ \begin{aligned} & \mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^\gamma=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^\gamma \\\\ & 10^5 \times(0.8)^{\frac{5}{3}}=3 \times 10^5\left(\mathrm{~V}_{\mathrm{B}}\right)^{\frac{5}{3}} \\\\ & \Rightarrow \mathrm{V}_{\mathrm{B}}=0.8 \times\left(\frac{1}{3}\right)^{0.6}=0.4 \end{aligned} $

Work done in process $\mathrm{A} \rightarrow \mathrm{B}$

$ \begin{aligned} & \mathrm{W}_{\mathrm{AB}}=\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\gamma-1} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=\frac{10^5 \times 0.8-3 \times 10^5 \times 0.4}{\frac{5}{3}-1} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=-60 \mathrm{~kJ}=\Rightarrow\left|\mathrm{W}_{\mathrm{AB}}\right|=60 \mathrm{~kJ} \end{aligned} $

Work done in process $\mathrm{B} \rightarrow \mathrm{C}$ (Isothermal process)

$ \begin{aligned} & \mathrm{W}_{\mathrm{BC}}=\mathrm{nRT} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=3 \times 10^5 \times 0.4 \ell \mathrm{n} \frac{0.8}{0.4} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=84 \mathrm{~kJ} \end{aligned} $

Work done in process $\mathrm{C} \rightarrow \mathrm{A}$

$ \mathrm{W}_{\mathrm{CA}}=\mathrm{P} \Delta \mathrm{V}=0 \quad(\because \Delta \mathrm{V}=0) $

So total work done in the process $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$

$ \begin{aligned} & \mathrm{W}_{\mathrm{ABC}}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}}=-60+84+0 \\\\ & \mathrm{~W}_{\mathrm{ABC}}=24 \mathrm{~kJ} \end{aligned} $

So correct options are (B,C,D)

Here, A = 64 mm², T = 2500K (where, A = surface area of filament, T = temperature of filament, d is distance of bulb from observer and R_e = radius of pupil of eye)

Point source, d = 100 m, R_e = 3 mm

(a) P = $\sigma $AeT^$-$4

= 5.67 $ \times $ 10^$-$8 $ \times $ 64 $ \times $ 10^$-$6 $ \times $ 1 $ \times $ (2500)⁴

($ \because $ e = 1 for black body)

= 141.75 W

(b) Power reaching to the eye

= ${P \over {4\pi {d^2}}} \times (\pi R_e^2)$

$ = {{141.75} \over {4\pi \times {{(100)}^2}}} \times \pi \times {(3 \times {10^{ - 3}})^2}$

$ = 3.189375 \times {10^{ - 8}}$ W

(c) ${\lambda _m}T = b$

${\lambda _m} \times 2500 = 2.9 \times {10^{ - 3}}$

$ \Rightarrow {\lambda _m} = 1.16 \times {10^{ - 6}}$ = 1160 nm

(d) Power received by one eye of observer = $\left( {{{hc} \over \lambda }} \right) \times {{\Delta N} \over {\Delta t}}$

where, ${{\Delta N} \over {\Delta t}}$ = number of photons.

entering into eye per second.

$ = 3.189375 \times {10^{ - 8}}$

$ = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1740 \times {{10}^{ - 9}}}} \times {{\Delta N} \over {\Delta t}}$

$ \Rightarrow {{\Delta N} \over {\Delta t}} = 2.79 \times {10^{11}}$

${\gamma _{mix}} = {{{n_1}{C_{p1}} + {n_2}{C_{p2}}} \over {{n_1}{C_{v1}} + {n_2}{C_{v2}}}} = {8 \over 5}$

$W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$

${P_0}{V_0}^{8/5} = {P_2}{\left( {{{{V_0}} \over 4}} \right)^{8/5}}$

${P_2} = 9.2{P_0}$

$W = {{{P_0}{V_0} - 9.2{P_0}{{{V_0}} \over 4}} \over {3/5}} = - 13R{T_0}$

$ \therefore $ $|W| = 13R{T_0}$

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

${T_2} = {T_1}{(2)^{6/5}} = 23{T_0}$

Average kinetic energy of gas mixture

= $nC{v_{mix}}{T_2} = 23R{T_0}$

P-V graph of the given V-T graph is shown below.



(d) $\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{3 \to 4}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{T_0}} \over {{T_0}/2}} = 2$

(b) $\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{2 \to 3}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{C_p}} \over {{C_V}}} = {5 \over 3}$

(a) ${W_{cycle}} = {p_0}{V_0} = nR\left[ {{{{T_0}} \over 2}} \right]$

Note For ideal gas equation,

pV = nRT

$ \therefore $ (pV)₄ = (nRT)₄

or p₀V₀ = $nR{{{T_0}} \over 2}$ = $R{{{T_0}} \over 2}$

as n = 1

(c) Wrong as no adiabatic process is involved.

In process I, volume is changing. Therefore, it is not isochoric. Therefore 'A' is incorrect.

In process II, q = $\Delta$U + W . $\Delta$U = 0 as temperature is constant. Therefore, q = W. Here W = P(V_f $-$ V_i) is positive therefore q is positive i.e., gas absorbs heat. Therefore 'B' is correct.

For process IV, q = $\Delta$U + W. Here $\Delta$ = 0 and W is negative (volume decreases). Therefore q is negative i.e., gas releases heat. 'C' is correct.

For an isobaric process, V $\propto$ T i.e., we will get a straight inclined line in T-V graph. Therefore I and II are NOT isobaric. 'D' is correct.

Initially both the compartments has same pressure as they are in equilibrium.

Suppose spring is compressed by x on heating the gas.

Let A be the area of cross-section of piston. As gas is ideal monoatomic, so

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ...... (i)

Force on spring by gas = kx

$\therefore$ ${P_2} = {P_1} + {{kx} \over A}$ ...... (ii)

Case I : When ${V_2} = 2{V_1}$, ${T_2} = 3{T_1}$

From Eqn. (i)

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(2{V_1})} \over {3{T_1}}} \Rightarrow {P_2} = {3 \over 2}{P_1}$

Putting this value in eqn. (ii) we get

${3 \over 2}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 2}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1} - {V_1}} \over A} = {{{V_1}} \over A}$

Energy stored in the spring

$ = {1 \over 2}k{x^2} = {1 \over 2}(kx)(x) = {{{P_1}{V_1}} \over 4}$

So, option (a) is correct.

Change in internal energy,

$\Delta U = {f \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2}\left( {{3 \over 2}{P_1} \times 2{V_1} - {P_1}{V_1}} \right) = 3{P_1}{V_1}$

So, option (b) is correct.

Case II : When ${V_2} = 3{V_1}$ and ${T_2} = 4{T_1}$

From Eqn. (i),

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(3{V_1})} \over {4{T_1}}} \Rightarrow {P_2} = {4 \over 3}{P_1}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1}} \over A}$

From eqn. (ii),

${4 \over 3}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 3}$

Gas is heated very slowly so pressure on the other compartment remains same.

Work done by gas = Work done by gas on atmosphere + Energy stored in spring.

${W_g} = {P_1}Ax + {1 \over 2}k{x^2} = {P_1}(2{V_1}) + {1 \over 2}\left( {{{{P_1}A} \over 3}} \right)\left( {{{2{V_1}} \over A}} \right)$

$ = 2{P_1}{V_1} + {1 \over 3}{P_1}{V_1} = {7 \over 3}{P_1}{V_1}$

So, option (c) is correct.

Heat supplied to the gas,

$\Delta Q = {W_g} + \Delta U$

$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}({P_2}{V_2} - {P_1}{V_1})$

$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}\left( {{4 \over 3}{P_1} \times 3{V_1} - {P_1}{V_1}} \right)$

$ = {7 \over 3}{P_1}{V_1} + {9 \over 2}{P_1}{V_1} = {{41} \over 6}{P_1}{V_1}$

So, option (d) is incorrect.

The internal energy of one mole of an ideal gas at temperature T is given by $U = {f \over 2}RT$, where f is the degrees of freedom of the gas molecule. The degress of freedom of the gas molecule. The degrees of freedom for hydrogen (diatomic) and helium (monatomic) gases are f_H2 = 5 and f_He = 3, respectively. Thus, ${U_{{H_2}}} = {5 \over 2}RT$ and ${U_{He}} = {3 \over 2}RT$. The total internal energy of the gas mixture is

${U_{total}} = {U_{{H_2}}} + {U_{He}} = {5 \over 2}RT + {3 \over 2}RT = 4RT$.

The mixture contains two moles of the gases. The internal energy per mole of the mixture is ${U_{mix}} = {U_{total}}/2 = 2RT$.

The specific heat at constant volume is given by ${C_v} = dU/dT$. Thus, the specific heats at constant volume for helium and the mixture are

${C_{v,He}} = d{U_{He}}/dT = {3 \over 2}R$, and

${C_{v,mix}} = d{U_{mix}}/dT = 2R$

The specific heats at constant pressure, ${C_p} = {C_v} + R$, for these gases are

${C_{p,He}} = {C_{v,He}} + R = {5 \over 2}R$, and

${C_{p,mix}} = {C_{v,mix}} + R = 3R$.

The ratio of specific heats, $\gamma = {C_p}/{C_v}$, are ${\gamma _{He}} = 5/3$ and ${\gamma _{mix}} = 3/2$. The speed of sound, in a gas of molecular mass M, is given by ${v_s} = \sqrt {\gamma RT/M} $. The molecular mass of the gas mixture is

${M_{mix}} = {{{n_{{H_2}}}{M_{{H_2}}} + {n_{He}}{M_{He}}} \over {{n_{{H_2}}} + {n_{He}}}}$

$ = {{(1)(2) + (1)(4)} \over {1 + 1}} = 3$ g/mol,

where n_H2 = 1 and n_He = 1 are the number of moles of hydrogen and helium in the gas mixture. The ratio of the speeds of sound in the gas mixture and helium is

${{{v_{s,mix}}} \over {{v_{s,He}}}} = {{\sqrt {{\gamma _{mix}}RT/{M_{mix}}} } \over {\sqrt {{\gamma _{He}}RT/{M_{He}}} }} = \sqrt {{{{\gamma _{mix}}} \over {{\gamma _{He}}}}{{{M_{He}}} \over {{M_{mix}}}}} $

$ = \sqrt {{{(3/2)(4)} \over {(5/3)(3)}}} = \sqrt {{6 \over 5}} $

The rms speed of the atoms/molecules is given by ${v_{rms}} = \sqrt {3RT/M} $. The ratio of the rms speed of helium atoms to that of hydrogen molecules is

${{{v_{rms,He}}} \over {{v_{rms,{H_2}}}}} = {{\sqrt {3RT/{M_{He}}} } \over {\sqrt {3RT/{M_{{H_2}}}} }} = \sqrt {{{{M_{{H_2}}}} \over {{M_{He}}}}} = \sqrt {{2 \over 4}} = \sqrt {{1 \over 2}} $.

According to definition of specific heat capacity,

$C = {{\Delta Q} \over {m\Delta T}}$

$ \Rightarrow \Delta Q = mC\Delta T$ $\therefore$ ${{\Delta Q} \over {\Delta t}} = mC{{\Delta T} \over {\Delta t}}$

Rate of heat absorbed $R = {{\Delta Q} \over {\Delta t}}$

$ \Rightarrow {{\Delta Q} \over {\Delta t}} \propto C$

(a) In 0-100 K,

C increases with T but not linearly. So R increases but not linearly.

(b) As $\Delta Q = mC\Delta T$

$Q = m\int {C\Delta T} $

= m area under C-T curve

From the graph it is clear that area under C-T is more in 400-500 K than in 0-100 K.

Therefore, heat absorbed in 0-100 K is less than in 400-500 K.

(c) In 400-500 K,

C remains constant so there is no change in R.

(d) In 200-300 K,

C increases so R increases.

The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., T_A = T_B = T₀, which makes U_A = U_B.

The work done in the isothermal process AB is

W_AB = nRT ln(V_B/V_A)

= nRT₀ ln(4V₀/V₀) = p₀V₀ ln4.

Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.

Hence, the correct options are (a) and (b).

Note:

If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,

p_C = p_B = RT_B / V_B

= RT₀/(4V₀) = (RT₀/V₀)4 = p₀/4.

Apply the ideal gas equation for the states A and C to get

${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$.

We observe the following:

$\bullet$ For the path B $\to$ C $\to$ D, the volume (V) decreases; $\Delta$W < 0.

$\Delta U < 0 \Rightarrow \Delta Q < 0$

$\bullet$ For the process A $\to$ B $\to$ C, $\Delta$W is the area of the semicircle and $\Delta W \ne 0$.

$\bullet$ For the process A $\to$ B (semicircle), it cannot be an isothermal process (whose PV-diagram is a rectangular hyperbola).

$\bullet$ For (clockwise) process A $\to$ B $\to$ C $\to$ D $\to$ A, $\Delta$W is the area enclosed, which is positive.

We know that ${C_P} - {C_V} = R$ is same for all gases. For a diatomic gas, we have

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

Therefore, ${C_P} = {C_V} + R = {{7R} \over 2}$

For a monoatomic gas, we have

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

${C_P} = {{5R} \over 2}$

Therefore,

${C_P} + {C_V} = 6R$ (diatomic)

${C_P} + {C_V} = 4R$ (monoatomic)

${{{C_P}} \over {{C_V}}} = {7 \over 5} = 1.4$ (diatomic)

${{{C_P}} \over {{C_V}}} = {5 \over 3} = 1.67$ (monoatomic)

${C_P}\,.\,{C_V} = {{35{R^2}} \over 4}$ (diatomic)

${C_P}\,.\,{C_V} = {{15{R^2}} \over 4}$ (monoatomic)

Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

The reason given in question is self explanatory. With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase. Reflected energy per unit time will be zero since black body has zero reflectivity and hence it will remain constant.

The bodies which radiate energy in the form of photons can be determined using Kirchoff's

law of radiation. When these photons reach another surface, they may be absorbed, reflected or transmitted. Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

Therefore, the quantity of radiation absorbed by the black body in unit time will increase, since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increases, Black body radiates more energy in unit time in the visible spectrum, the reflected energy in unit time by the black body remains same.

Hence, the option (A), (B), (C) and (D) are the correct answer.

$(b,c)$ are wrong. As first law is applicable to a cyclic process and concept of entropy is introduced by the second law.