Heat and Thermodynamics

$\begin{aligned} & \mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right) \\ & K\left(P_0, 3 V_0, 3 T_0\right) \\ & \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \\ & \mathrm{L}\left(2 \mathrm{P}_{\circ}, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \\ & \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \\ & \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \\ & \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \\ & \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ell \mathrm{n}\left(\frac{1}{2}\right)=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\ & \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\ & \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT}_0 \\ & \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT}_0 \ln 2 ; \Delta \mathrm{U}=0 \\ & \mathrm{WD}_{\text {nes }}=-2 \mathrm{nRT}_0 \ln 2 \\ & \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2 \\ & \end{aligned}$

1. Formula for Sound Velocity:

- The velocity of sound, denoted as $ v $, is determined by the equation $ v = \sqrt{\frac{\gamma R T}{M}} $, where:

- $ \gamma $ is a factor given by $ \gamma = 1 + \frac{2}{n} $.

- $ R $ is the gas constant.

- $ T $ is the temperature.

- $ M $ is the molar mass of the gas.

2. Relation of $ \gamma $ with $ n $:

- The value of $ \gamma $ is inversely related to $ n $. This means as $ n $ increases, $ \gamma $ decreases.

3. Internal Energy Relation:

- The internal energy, denoted as $ U $, for a mole of gas is given by $ U = \frac{n R T}{2} $.

4. Effect of Increasing $ n $ on Sound Velocity and Internal Energy:

When the value of $ n $ increases:

- The factor $ \gamma $ decreases, according to its formula.

- As a result, the velocity of sound $ v $ also decreases, as it is dependent on $ \sqrt{\gamma} $.

- Conversely, the internal energy $ U_n $ increases with the increase in $ n $, as evident from the internal energy formula.

Therefore, in summary, a higher value of $ n $ leads to a lower velocity of sound and higher internal energy in the context of these formulas.

The expansion of an ideal gas can be analyzed using the thermodynamic relations for adiabatic and isothermal processes.

1. Adiabatic Expansion: For an adiabatic process, we use the relation involving temperatures and volumes at the initial and final states, which is given by :
   
   $ TV^{\gamma-1} = \text{constant} $
   
   Therefore, for the adiabatic expansion from $(T_A, V_0)$ to $(T_f, 5V_0)$, we have :
   
   $ T_A V_0^{\gamma-1} = T_f (5V_0)^{\gamma-1} $
   
   Simplifying, we get : $ T_A = T_f \times 5^{\gamma-1} $

2. Isothermal Expansion : In an isothermal process, the temperature remains constant. So, $ T_B = T_f $ for the isothermal expansion from $(T_B, V_0)$ to $(T_f, 5V_0)$.

To find the ratio $ T_A / T_B $, we substitute the expressions for $ T_A $ and $ T_B $ :

$ \frac{T_A}{T_B} = \frac{T_f \times 5^{\gamma-1}}{T_f} $

$ \frac{T_A}{T_B} = 5^{\gamma-1} $

Therefore, the ratio $ T_A / T_B $ is $ 5^{\gamma-1} $, which corresponds to Option A.

$\lambda_m T=b$

For option $(\mathrm{P})$ temperature is minimum hence $\lambda _\mathrm{m}$ will be maximum.

$\lambda_m=\frac{b}{T}=\frac{2.9 \times 10^{-3}}{2000}=1.45 \times 10^{-6} \mathrm{~m}=1450 \mathrm{~nm}$

$\begin{aligned} & \sin \theta=\frac{\lambda}{d} \\\\ & 2 \theta=\text { width of central maximum } \\\\ \Rightarrow & \text { width } \propto \lambda . \\\\ \Rightarrow & \text {maximum width of }(\mathrm{P}) \\\\ & \mathrm{P} \rightarrow 3\end{aligned}$

$\begin{aligned} & \Rightarrow \text { For option }(\mathrm{Q}),\mathrm{T}=3000 \\\\ & \\ & \lambda_\mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000} \\\\ & \lambda_\mathrm{m}=\frac{2.9}{3} \times 10^{-6} \\\\ & =0.96 \times 10^{-6} \\\\ & =966.6 \mathrm{~nm} \\\\ & \mathrm{P}_{3000}=6 \mathrm{~A}(3000)^4 \\\\ & \mathrm{P}_{6000}=6 \mathrm{~A}(6000)^4 \\\\ & \frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16} \\\\ & \mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{6000} \\\\ & \mathrm{Q}-4\end{aligned}$

$\begin{aligned} \Rightarrow & \text { For }(\mathrm{R}), \mathrm{T}=5000 \mathrm{~K} \\\\ & \lambda_\mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6} \\\\ & =580 \mathrm{~nm} \\\\ & \text { Visible to human eyes } \\\\ & \mathrm{R}-2\end{aligned}$

For $(S), T=10,000 K$

$\lambda_m T=b \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{10,000}=290 \mathrm{~nm}$

$ \begin{aligned} \phi & =4 e \mathrm{~V} \\\\ \Rightarrow \lambda_{\mathrm{T}} & =\frac{1240}{4}=310 \mathrm{n} \mathrm{m} \end{aligned} $

To emit photoelectron

$ \begin{aligned} & \lambda_0 < \lambda_{\mathrm{T}} \\\\ & \Rightarrow \mathrm{S} \rightarrow 1 \end{aligned} $

$U = ML - P\Delta V$

$ = {10^{ - 3}} \times 2250 - {10^2}kP \times ({10^{ - 3}} - {10^{ - 6}})$ m³

= 2.25 kJ $-$ 0.1 kJ

= 2.15 kJ

I $\to$ P

(II) ${C_V} = {{5R} \over 2}$ (rigid diatomic)

For isobaric expansion

$V \propto T$

${{{V_1}} \over {{V_2}}} = {{{T_1}} \over {{T_2}}} \Rightarrow {V \over {3V}} = {{500} \over {{T_2}}} \Rightarrow {T_2} = 1500\,K$

$\Delta U = n{C_V}\Delta T = 0.2 \times {{5 \times 8} \over 2} \times (1500 - 500)$ J = 4 kJ

II $\to$ R

(III) Adiabatic expansion $\left( {\gamma = {5 \over 3}} \right)$

${P_1}V_1^\gamma = {P_2}V_2^\gamma \Rightarrow $ (2 kPa) $ \times V_0^{5/3} = {P_2} \times {\left( {{{{V_0}} \over 8}} \right)^{5/3}}$

${P_2} = 64$ kPa

$\Delta U = n{C_V}\Delta T$

$ = {{3nR\Delta T} \over 2} = {3 \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2} \times \left( {64 \times {1 \over {3 \times 8}} - 2 \times {1 \over 3}} \right)$

$ = {3 \over 2} \times \left( {{8 \over 3} - {2 \over 3}} \right) = 3$ kJ

III $\to$ T

(IV) For isobaric expaision,

$\Delta U = n{C_V}\Delta T = {7 \over 2}\,nR\Delta T$

$\Delta Q = n{C_P}\Delta T = {9 \over 2}\,nR\Delta T$

${{\Delta U} \over {\Delta Q}} = {7 \over 9}$

$\Delta U = {7 \over 9}\Delta Q = 7$ kJ

IV $\to$ Q

$\gamma = {3 \over 2}$

${T_0}{V_0}^{{3 \over 2} - 1} = {T_R} \times {\left( {{{{V_0}} \over 2}} \right)^{{3 \over 2} - 1}}$

${{{T_R}} \over {{T_0}}} = \sqrt 2 $

$Q = \Delta U = \Delta {U_1} + \Delta {U_2}$

$\Delta {U_2} = 1 \times 2R \times (\sqrt 2 {T_0} - {T_0}) + 2R{T_0}(\sqrt 2 - 1)$

For $\Delta$U₂ : Using mole conservation of left side.

${{{p_0}{V_0}} \over {R{T_0}}} = {{2\sqrt 2 {p_0}{{3{V_0}} \over 2}} \over {R{T_L}}} \Rightarrow {T_L} = 3\sqrt 2 {T_0}$

$\Delta {U_1} = 1 \times 2R \times (3\sqrt 2 {T_0} - {T_0}) = 2R{T_0}(3\sqrt 2 - 1)$

$Q = 2R{T_0}(\sqrt 2 - 1) + 2R{T_0}(3\sqrt 2 - 1) = 2R{T_0}(4\sqrt 2 - 2)$

${{\Delta Q} \over {R{T_0}}} = 2(4\sqrt 2 - 1)$

Process 1

p = constant, Volume increases and temperature also increases

Q = W + $\Delta$U

$\therefore$ W = positive, $\Delta$U = positive

$\Rightarrow$ Heat is positive and supplied to the gas.

Process 2

V = constant, Pressure decreases

T $\propto$ pV [as V = constant]

$\Rightarrow$ Temperature decreases

W = 0

$\Delta$T is negative and $\Delta$U = ${f \over 2}$nR$\Delta$T

$\therefore$ $\Delta$U is also negative

Q = $\Delta$U + W

$\therefore$ Heat is negative and rejected by gas.

Process 3

p = constant, Volume decreases

$\Rightarrow$ Temperature also decreases

W = p$\Delta$V = negative

$\Delta$U = ${f \over 2}$nR$\Delta$T = negative

$\therefore$ Heat is negative and rejected by gas.

Process 4

V = constant, Pressure increases

W = 0 (as V = constant)

pV = nRT $\Rightarrow$ Temperature increase

$\Rightarrow$ $\Delta$U = ${f \over 2}$nR$\Delta$T is positive

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

1-2 process is isothermal and 2-3 process is isochoric.

(I) ${W_{1 \to 2}} = nRI\,\ln {{{V_f}} \over {{V_i}}} = 1 \times R{{{T_0}} \over 3}\,\ln {{{V_2}} \over {{V_1}}}$

W_{2$ \to $3} = 0 (Isochoric process)

${W_{1 \to 2 \to 3}}$ = ${W_{1 \to 2}}$ + ${W_{2 \to 3}}$

= ${{R{T_0}} \over 3}$ ln 2(I$ \to $P)

(II) $\Delta U = {f \over 2}nR({T_f} - {T_i})$

$\Delta {U_{1 \to 2 \to 3}} = {3 \over 2}\left[ {{T_0} - {{{T_0}} \over 3}} \right]$

$ = R{T_0}$ (II$ \to $R)

(III) ${Q_{1 \to 2 \to 3}} = \Delta {U_{1 \to 2 \to 3}} + {W_{1 \to 2 \to 3}}$

(First law of thermodynamics)

$ = R{T_0} + {{R{T_0}} \over 3}\ln 2$

${{R{T_0}} \over 3}[3 + \ln 2]$ (III$ \to $T)

(IV) ${Q_{1 \to 2}} = \Delta {U_{1 \to 2}} + {W_{1 \to 2}} = 0 + {{R{T_0}} \over 3}ln2 = {{R{T_0}} \over 3}ln2$ (IV$ \to $P)

(I) ${W_{1 \to 2 \to 3}} = {W_{1 \to 2}} + {W_{2 \to 3}} = {P_0}[2{V_0} - {V_0}] + 0 = {P_0}{V_0}$

${W_{1 \to 2 \to 3}} = {P_0}{V_0} = {{R{T_0}} \over 3}$ (I$ \to $Q)

${U_{1 \to 2 \to 3}} = {3 \over 2}\left[ {{{3{P_0}} \over 2} \times 2{V_0} - {P_0}{V_0}} \right] = {3 \over 2} \times 2{P_0}{V_0} = 3{P_0}{V_0} = R{T_0}$(II$ \to $R)

(III) ${Q_{1 \to 2 \to 3}} = {U_{1 \to 2 \to 3}} + {W_{1 \to 2 \to 3}} = R{T_0} + {{R{T_0}} \over 3} = {{4R{T_0}} \over 3}$(III$ \to $S)

(IV) ${Q_{1 \to 2}} = n{C_p}\Delta T = n{5 \over 2}R({T_2} - {T_1}) = {5 \over 2}[{P_0}2{V_0} - {P_0}{V_0}] = {5 \over 2}{P_0}{V_0} = {5 \over 2}{{R{T_0}} \over 3} = {5 \over 6}{R_0}{T_0}$ (IV$ \to $U)

Process I is adiabatic, therefore, $\Delta$Q = 0

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

Thus, W < 0 since volume of gas is decreasing $\Rightarrow$ $\Delta$U > 0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.

Thus, the correct mapping is P $\to$ 3.

Process II is isobaric, therefore, pressure remains constant

$\Rightarrow$ W = P$\Delta$V = 3P₀[3V₀ $-$ V₀] = 6P₀V₀

Therefore, work done by gas in process II is 6P₀V₀.

Thus, the correct mapping is Q $\to$ 4.

Process III is isochoric, therefore, volume remains constant

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

and work done by gas is zero.

Thus, the correct mapping is R $\to$ 1.

Process IV is isothermal process, therefore temperature remains constant

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = W and $\Delta$ = 0

Thus, the correct mapping is S $\to$ 2.

Therefore, option (C) is correct.

$\because$ dW = PdV

and there is no change in volume in isochoric process.

$\therefore$ dW = 0

Adiabatic process is used as a correction in the determination of the speed of sound in an ideal gas. The p-V diagram of this process is shown in (Q). The work done on the gas in an adiabatic process is given by ${W_{1 \to 2}} = {{{p_2}{V_2} - {p_1}{V_1}} \over {\gamma - 1}}$.

Compare the given relation, $\Delta$U = $\Delta$Q $-$ p$\Delta$V, with the first law of thermodynamics, $\Delta$Q = $\Delta$U + $\Delta$W, to get work done by the gas as $\Delta$W = p$\Delta$V. This is the work done in an isobaric process. The p-V diagram of this process is shown in (P). The work done in this process is ${W_{1 \to 2}} = \int {pdV = p{V_1} - p{V_2}} $.

Rate of heat flow from P to Q,

${{dQ} \over {dt}} = {{2KA(T - 10)} \over 1}$

Rate of heat flow from Q to S,

${{dQ} \over {dt}} = {{KA(4000 - T)} \over 1}$

At steady state, state rate of heat flow is same

$ \therefore $ ${{2KA(T - 10)} \over 1} = KA(400 - T)$

or $2T - 20 = 400 - T$ or $3T = 420$

$ \therefore $ $T = 140^\circ $



Temperature of junction is $140^\circ $C.

Temperature at a distance x from end P is ${T_x} = (130x + 10^\circ )$

Change in length dx is suppose dy.

Then, $dy = \alpha dx({T_x} - 10)$

$\int_0^{\Delta y} {dy} = \int_0^1 {\alpha dx(130x + 10 - 10)} $

$\Delta y = \left[ {{{\alpha {x^2}} \over 2} \times 130} \right]_0^1$

$\Delta y = 1.2 \times {10^{ - 5}} \times 65$

$\Delta y = 78.0 \times {10^{ - 5}}$ m = 0.78 mm

In the first process : ${p_i}p_i^\gamma = {p_f}V_f^\gamma $



$ \Rightarrow {{{p_i}} \over {{p_f}}} = {\left( {{{{V_f}} \over {{V_i}}}} \right)^\gamma } \Rightarrow 32 = {8^\gamma }$

$\gamma = {5 \over 3}$ ... (i)

For the two step process,

$W = {p_i}({V_f} - {V_i}) = {10^5}(7 \times {10^{ - 3}})$

$W = 7 \times {10^2}J$

$\Delta U = {f \over 2}({p_f}{V_f} - {p_i}{V_i})$

$ = {1 \over {\gamma - 1}}\left( {{1 \over 4} \times {{10}^2} - {{10}^2}} \right)$

$\Delta U = - {3 \over 2}.{3 \over 4} \times {10^2} = - {9 \over 8} \times {10^2}J$

$Q - W = \Delta U$

$ \Rightarrow Q = 7 \times {10^2} - {9 \over 8} \times {10^2}$

$ = {{47} \over 8} \times {10^2}J = 588J$

Heat generated in device in 3 h

= Time $\times$ power

= 3 $\times$ 3600 $\times$ 3 $\times$ 10³ = 324 $\times$ 10⁵ J

Heat used to heat water = ms $\Delta$ $\theta$

= 120 $\times$ 1 $\times$ 4.2 $\times$ 10³ $\times$ 20J

Heat absorbed by coolant

Pt = 324 $\times$ 10⁵ $-$ 120 $\times$ 1 $\times$ 4.2 $\times$ 10³ $\times$ 20J

Pt = (325 $-$ 100.8) $\times$ 10⁵ J = 223.2 $\times$ 10⁵ J

P = ${{223.2 \times {{10}^5}} \over {3600}}$ = 2067 W

Effective area of incident light = $\pi$r²

Total energy per second = 912 $\times$ $\pi$r²

From Stefan's law, we have

E = Ae $\sigma$ (T⁴ $-$ T$_0^4$)

For black body emissivity, e = 1.

A = 4$\pi$r²

912 $\times$ $\pi$r² = 4$\pi$r² $\times$ 5.7 $\times$ 10^$-$8 [T⁴ $-$ 300⁴]

$({T^4} - {300^4}) = {{912} \over {4 \times 5.7 \times {{10}^{ - 8}}}} = 40 \times {10^8}$

${T^4} = 40 \times {10^8} + {3^4} \times {10^8}$

${T^4} = (90 + 81) \times {10^8} \Rightarrow 121 \times {10^8}$

$T = {(121)^{1/4}} \times 100 = \sqrt {11} \times 100 \Rightarrow 3.3 \times 100$

T = 330 K

T is final temperature, when equilibrium is reached

Heat gained = Heat lost

$n{C_p}\Delta T = n{C_v}\Delta T$

$2 \times {{7R} \over 2} \times (T - 400) = 2 \times {{3R} \over 2} \times (700 - T)$

$7T - 2800 = 2100 - 3T$

$10T = 4900$

$T = 490K$

Heat lost by monoatomic gas = Heat gained by diatomic gas

$2 \times {5 \over 2}R \times (700 - T) = 2 \times {5 \over 2}R(T - 400)$

$3500 - 5T = 7T - 2800$

$12T = 6300$ or, $T = 525K$

Work done by monoatomic gas:

$ = {n_1}R\Delta T$ ($\because$ PV = nRT)

$ = 2 \times R \times (700 - 525)$ ($\Delta$T is negative)

$ = - 350R$

Work done by diatomic gas $ = {n_2}R\Delta T$

$ = 2R(525 - 400) = 250R$. (There is gain in temperature)

The net work done = $-$350 + 250 = $-$100R and $\Delta$T is positive.

From the given figure, it can be concluded that process FG is isothermal and process FH is adiabatic.

${\gamma _{mono}} = 1 + {2 \over f} = {5 \over 3}$

${P_0}V_G^{5/3} = 32{P_0}V_0^{5/3}$

${V_G} = {(32)^{3/5}}{V_0} = 8{V_0}$

$\Delta {W_{GE}} = {P_0}({V_0} - 32{V_0}) = - 31{P_0}{V_0}$

Thus, the correct mapping is (P) $\to$ (4).

$\Delta {W_{GH}} = {P_0}(8{V_0} - 32{V_0}) = - 24{P_0}{V_0}$

Thus, the correct mapping is (Q) $\to$ (3).

$\Delta {W_{FH}} = {{{P_0}(8{V_0}) - 32{P_0}{V_0}} \over {1 - (5/3)}} = {{ - 24{P_0}{V_0}} \over { - (2/3)}} = 36{P_0}{V_0}$

Thus the correct mapping is (R) $\to$ (2).

$\Delta {W_{FG}} = 32R{T_0}\ln 32 = 32R{T_0}\ln {2^5} = 160R{T_0}\ln 2 = 160{P_0}{V_0}\ln 2$

Thus, the correct mapping is (S) $\to$ (1).

To determine the ratio of the densities of the two non-reactive monoatomic ideal gases, we can use the ideal gas law and the given information. The ideal gas law is given by:

$ PV = nRT $

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the gas constant
• T is the temperature

Given that the gases are monoatomic and non-reactive, their molecular masses are directly proportional to their atomic masses. Let the atomic masses of the two gases be $M_1$ and $M_2$, with $M_1 : M_2 = 2 : 3$.

The density ($\rho$) of a gas is given by the formula:

$ \rho = \frac{m}{V} $

Where $m$ is the mass and $V$ is the volume. Using the ideal gas law, we can relate density to pressure and temperature.

From the ideal gas law:

$ PV = nRT $

Since $ n = \frac{m}{M} $, where $M$ is the molar mass, we have:

$ P = \frac{mRT}{MV} $

Rearranging to solve for density ($\rho\ = \frac{m}{V}$):

$ \rho = \frac{PM}{RT} $

Thus, the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to the temperature. Given that the ratio of their partial pressures $P_1 : P_2 = 4 : 3$, we can write:

$ \rho_1 = \frac{P_1 M_1}{RT} $

$ \rho_2 = \frac{P_2 M_2}{RT} $

Taking the ratio of the densities:

$ \frac{\rho_1}{\rho_2} = \frac{\left(\frac{P_1 M_1}{RT}\right)}{\left(\frac{P_2 M_2}{RT}\right)} = \frac{P_1 M_1}{P_2 M_2} $

Substituting the given ratios, $ P_1 : P_2 = 4 : 3 $ and $ M_1 : M_2 = 2 : 3 $, into the equation, we get:

$ \frac{\rho_1}{\rho_2} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9} $

Therefore, the ratio of their densities is $ \frac{8}{9} $, which corresponds to option D. Consequently, the answer is:

Option D: 8 : 9

Let L and A be length and area of cross-section of each block.

$\therefore$ Thermal resistance of block 1 is

$\therefore$ ${R_1} = {L \over {\kappa A}}$

Thermal resistance of block 2 is

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

In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is

${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$ ...... (i)

$\therefore$ Rate of heat flow in configuration I is

$\therefore$ ${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$ ..... (ii)

In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is

${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$ ...... (iii)

${R_P} = {1 \over 3}{L \over {\kappa A}}$

$\therefore$ Rate of same amount of heat flow in configuration II is

$\therefore$ ${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$ ...... (iv)

Divide (ii) by (iv), we get

${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} \times {{2\kappa A} \over {3L}} = {2 \over 9}$ (Using (i) and (iii))

$t' = {2 \over 9}t = {2 \over 9} \times 9s = 2s$

Here's a breakdown of how to solve this problem:

Understanding the Concepts

• Ideal Gas: An ideal gas is a theoretical gas that follows the ideal gas law perfectly. The ideal gas law is a relationship between pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$) of a gas:

$PV = nRT$, where $R$ is the ideal gas constant.

• Heat Capacity: Heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius (or 1 Kelvin). For an ideal gas, we can use the specific heat capacity at constant volume ($C_v$) or at constant pressure ($C_p$).


• Constant Volume Process: In a constant volume process, the volume of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_v\Delta T$

• Constant Pressure Process: In a constant pressure process, the pressure of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_p\Delta T$

• Molar Heat Capacities of Helium: Helium is a monatomic gas. For monatomic gases, we have:


• $C_v = \frac{3}{2}R$


• $C_p = \frac{5}{2}R$

Solving the Problem

1. Identify the Process: Since the balloon is fully expandable, the pressure remains constant. This is a constant pressure process.


2. Calculate the Heat: Use the formula for heat at constant pressure:

$Q = nC_p\Delta T$

1. Substitute the Values:

• $n = 2 \text{ moles}$


• $C_p = \frac{5}{2}R = \frac{5}{2}(8.31 \text{ J/mol.K})$


• $\Delta T = 35^\circ C - 30^\circ C = 5 K$

1. Solve for Q:

$Q = (2 \text{ moles}) \times \left(\frac{5}{2} \times 8.31 \text{ J/mol.K}\right) \times (5 K) $

$Q = 207.75 \text{ J}$

Answer:

The amount of heat required to raise the temperature of the helium gas is approximately 208 J. Therefore, the correct answer is Option D.

The root mean square (rms) speed of a gas is given by the formula :

$v_{rms} = \sqrt{\frac{3kT}{m}}$

where :

• $k$ is the Boltzmann constant
• $T$ is the temperature in Kelvin
• $m$ is the mass of one molecule of the gas

Given that the atomic mass of helium is 4 amu and that of argon is 40 amu, we can find the rms speed ratio by comparing the rms speeds of helium and argon, as temperature $T$ and Boltzmann constant $k$ are the same for both gases :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{m_{\text{argon}}}{m_{\text{helium}}}}$

Substituting the masses :

$m_{\text{helium}} = 4 \text{ amu}$

$m_{\text{argon}} = 40 \text{ amu}$

we get :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$

Thus, the ratio of the rms speeds of helium to argon is approximately 3.16.

The correct answer is :

Option D : 3.16

Let the steady state temperature of the middle plate to T₀. In the steady state, the heat radiated per unit time by the middle plate is equal to the heat received per unit time by it. The middle plate radiates heat from both the surfaces and receives the heat radiated by the first and the third plates.

Stefan-Boltzmann law gives the rate of heat loss and heat gain by the middle plate as

$d{Q_{out}}/dt = \sigma AT_0^4 + \sigma AT_0^4$,

$d{Q_{in}}/dt = \sigma A{(2T)^4} + \sigma A{(3T)^4}$.

The steady state condition, $d{Q_{out}}/dt = d{Q_{in}}/dt$

$\sigma A{(2T)^4} + \sigma A{(3T)^4} = \sigma 2A{({T_0})^4}$

$16{T^4} + 81{T^4} = 2{({T_0})^4}$

$97{T^4} = 2{({T_0})^4}$

${({T_0})^4} = {{97} \over 2}{T^4} = {\left( {{{97} \over 2}} \right)^{1/4}}T$

Initially

V₁ = 5.6 l, T₁ = 273 K, P₁ = 1 atm,

$\gamma = {5 \over 3}$ (For monoatomic gas)

The number of moles of gas is $n = {{5.6l} \over {22.4l}} = {1 \over 4}$

Finally (after adiabatic compression)

V₂ = 0.7 l

For adiabatic compression ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$

$\therefore$ ${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} \right)^{\gamma - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} \right)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$

Work done during an adiabatic process is

$W = {{nR[{T_1} - {T_2}]} \over {(\gamma - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} \right]}} = - {9 \over 8}R{T_1}$

Negative sign shows that work is done on the gas.

Process A $\to$ B,

It is a isobaric process

P = constant, V $\propto$ T

$\because$ V_B < V_A $\Rightarrow$ T_B < T_A

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence internal energy decreases.

$\Delta$Q = nC_P$\Delta$T = $-$ve

Hence heat is lost.

$\Delta$W = nR$\Delta$T = $-$ve

Hence, work is done on the gas.

Process, B $\to$ C,

It is a isochoric process.

V = constant, P $\propto$ T

$\because$ P_C < P_B $\Rightarrow$ T_C < T_B

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence, internal energy decreases.

$\Delta$W = 0, $\Delta$Q = $\Delta$U = $-$ve

Hence, heat is lost.

Process C $\to$ D,

It is a isobaric process.

P = constant, V $\propto$ T

$\because$ V_D > V_C $\Rightarrow$ T_D > T_C

$\Delta$U = nC_V$\Delta$T = +ve

Hence internal energy increases.

$\Delta$Q = nC_P$\Delta$T = +ve

Hence, heat is gained.

$\Delta$W = nR$\Delta$T = +ve

Hence, work is done by the gas.

Process, D $\to$ A

According to ideal gas equation

${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$

Hence, it is a isothermal process.

$\therefore$ $\Delta$U = 0

$\because$ V_A < V_D

$\Delta$W = $-$ve, hence work done is done on the gas.

$\Delta$Q = $\Delta$W = $-$ve, hence heat is lost.

In an ideal gas, the average force of attraction between the molecules and volume of the molecules (in comparison to volume of the gas) are negligibly small. These conditions are satisfied for a real gas when pressure is low and temperature is high.

(A)$\to$(P), (Q), (T)

$\bullet$ Case (P) : $U=\frac{1}{2}CV^2$

$\bullet$ Case (Q) : Since the work is done on the system, the internal energy increases.

$\bullet$ Case (T) : time varying $\overrightarrow B $ creates an induced $\overrightarrow E $ causing current to flow. This dissipates heat in the loop: $H=L^2(Rt)$.

(B)$\to$(Q) : As explained above.

(C)$\to$(S)

$\bullet$ Case (S) : Mass defect is converted into energy which is released.

(D)$\to$(S) : As explained above.

As the ideal gas expands in vacuum, no work is done i.e., w = 0

Container is insulated, so no heat lost or gained i.e., Q = 0.

According to the first law of thermodynamics

$\Delta$U = Q + w

$\therefore$ $\Delta$U = 0

Therefore, no change in the temperature of the gas.

(b) $\to$ ($p, r$) :

Given, PV$^2$ = constant ... (i)

For ideal gas, ${{PV} \over T}$ = constant .... (ii)

From (i) and (ii) V $\mu$ T = constant

As gas expands, its volume increase and temperature decreases.

$\therefore$ (p) is the correct.

Also, Q = nC$\Delta$T .... (i)

Where C = molar specific heat

For a polytropic process,

$C = {C_V} + {R \over {1 - n}}$ and

PV$^n$ = constant

Here, PV$^2$ = constant, where n = 2

$\therefore$ $C = {C_V} + {R \over {1.2}} = {C_V} - R$

For monoatomic gas,

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

$\therefore$ $C = {3 \over 2}R - R = {R \over 2}$

Substituting this value in (1), we get

$Q = n \times {R \over 2} \times \Delta T$

Temperature decreases i.e., $\Delta$T is negative. Therefore, Q is negative. This is turn mean that heat is lost by the gas during the process.

So (r) is the correct option.

(C) $\to$ (P, S)

V$^{1/3}$ $\times$ T = constant

$\Rightarrow$ As the gas expands and volume increases, the temperature decreases. Therefore (P) is the correct option.

In the process, $x = {4 \over 3}$

$\therefore$ $C = {C_V} + {R \over {1 - {4 \over 3}}} = {3 \over 2}R + {{3R} \over { - 1}}$

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

$\therefore$ $Q = n\left( { - {3 \over 2}R} \right)\Delta t$

As $\Delta t$ = $-$negative ($-$ve), Q = positive (+ve). That means heat is gained by the gas during the process.

(s) is correct.

(d) $\to$ (q, s)

$\Delta T = {{\Delta (PV)} \over {nR}}$

$\Delta (PV)$ = Positive $\therefore$ $\Delta$T = positive

$\therefore$ Temperature increase (Q) is the correct option.

From graph it is clear that, during the process the pressure of the gas increases which shows that the internal energy of the gas has increased. Also, the volume increases which means work is done by the system which needs energy. We conclude from the above fact that gas gains heat during the process.

(S) is correct.

Given, PT$^2$ = constant ..... (i)

For an ideal gas,

$\frac{\mathrm{PV}}{\mathrm{T}}$ = constant ...... (ii)

From above two equation, after eliminating P.

$\frac{\mathrm{V}}{\mathrm{T^3}}$ = Constant

$\Rightarrow$ V = KT$^3$, where $k$ = constant.

$\frac{d\mathrm{V}}{\mathrm{V}}=3\frac{d\mathrm{T}}{\mathrm{T}}$

$ \Rightarrow dV = \left( {{3 \over T}} \right)VdT$ ..... (iii)

Change in volume due to thermal expansion is given by $dV=Vydt$ ..... (iv)

Where, $y$ = coefficient of volume expansion

From equation (iii) and (iv), we have,

$VydT = \left( {{3 \over T}} \right)VdT$

$ \Rightarrow y = {3 \over T}$

Statement 1 :

Total translational kinetic energy $=\frac{3}{2} n \mathrm{RT}$

where,

n = no. of moles

R = gas constant

T = Temperature

but, PV = nRT

$\therefore$ K.E. = 1.5 PV

Statement 2 : Molecules of a gas collide with each other and the velocities of the molecules change due to collision.

But statement 2 is not a correct explanation of statement 1 .

When the piston is pulled out slowly, the pressure drop produced inside the cylinder is almost instantaneously neutralized by the air entering from outside into the cylinder. So, the pressure inside the cylinder is P$_0$ throughout the slow pulling process.

At equilibrium, P = P

$\Rightarrow$ P = P$_0$ + (L$_0$ $-$ H)$\rho$g ..... (i)

$ \Rightarrow {P_0}\pi (\pi {R^2}{L_0}) = P[\pi {R^2}({L_0} - H)]$

$ \Rightarrow P = {{{L_0}{P_0}} \over {{L_0} - H}}$ ..... (ii)

From (i) as (ii)

$ \Rightarrow {{{L_0}{P_0}} \over {{L_0} - H}} = {P_0} + ({L_0} - H)\rho g$

$ \Rightarrow {L_0}{P_0} = {P_0}({L_0} - H) + {({L_0} - H)^2}\rho g$

$ \Rightarrow \rho g{({L_0} - H)^2} + {P_0}({L_0} - H) - {L_0}{P_0} = 0$

$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}, \mathrm{~W}=\int \mathrm{P} d \mathrm{~V} $

Along JK : V is constant, P is decreasing and

$\mathrm{P} \propto \mathrm{T}, \mathrm{T}$ should also decrease.

As $\mathrm{PV}=n \mathrm{RT}$.

$ \begin{aligned} \therefore & & \mathrm{W} & =P \Delta V=0 \\ & & \Delta \mathrm{U} & =\Delta \mathrm{Q}<0 \quad \text { (negative) } \end{aligned} $

$\Rightarrow$ Along KL : $\mathrm{P}=$ constant, i.e., isobaric process.

$\therefore n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T}$ is increasing. So temperature should also increase.

$ \begin{aligned} & \therefore \Delta \mathrm{W}=\mathrm{PdV}>0, \Delta \mathrm{U}=n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{~T}>0 \text { and } \\ & \mathrm{Q}=m c \Delta T>0 . \end{aligned} $

$\therefore$ Volume increases, $\Delta \mathrm{W}>0$

$\Rightarrow$ Along LM : MV = constant, $\Delta \mathrm{V}=0$ [isochoric process]

both P increases and T increases.

$ \therefore W=0, \Delta U>0 \text { and } Q>0 \quad[\Delta Q=\Delta U] $

$\Rightarrow$ Along $\mathrm{MJ}: \mathrm{V}$ is decreasing, $\therefore \Delta \mathrm{W}<0$

T is also decreasing, $\Delta \mathrm{V}<0$ and $\Delta \mathrm{Q}<0$

P increases.

(A) Heat supplied to the cylinder is used up in if raising temperature of cylinder $=m s \Delta \mathrm{T}$

(i) $\mathrm{Q}_{1}=m s \Delta \mathrm{T}=1 \times 400(\mathrm{~T}-20) \text {...(i) }$

S = 400 J kg$^{-10}$C$^{-1}$

(ii) doing work $=\mathrm{P}_{\text {atom }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

Work done $=\mathrm{P}_{\mathrm{atm}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

$\mathrm{Q}_{2}=10^{5} \times\left(\mathrm{V}_{1} \gamma \Delta \mathrm{T}\right)$

$\mathrm{Q}_{2}=10^{5} \times \frac{\text { mass }}{\text { density }} \times \gamma(\mathrm{T}-20)$

$\mathrm{Q}_{2}=10^{5} \times \frac{1}{900} \times 9 \times 10^{-5} \times(\mathrm{T}-20)$

$\mathrm{Q}_{2}=0.001(\mathrm{~T}-20)$ ...... (ii)

$\therefore \quad \mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$ Heat supplied.

(Given $\mathrm{Q}=20000 \mathrm{~J}$ )

400.$(T-20)+0.001(T-20)=20000$

$400.001(\mathrm{~T}-20)=2000$

$\mathrm{T}-20=\frac{20000}{400.001}=50$ (approx)

$\mathrm{T}=50+20=70^{\circ} \mathrm{C}$ ...... (iii)

(B) Workdone by cylinder

$\begin{aligned} \mathrm{W} & =\mathrm{Q}_{2}=0.001(70-2 a) \\ & =0.001(50) \\ \mathrm{W} & =0.05 \mathrm{~J} .\end{aligned}$

(C) Change in internal energy

$\begin{aligned} \Delta \mathrm{U} & =20000-0.05 \\ & =19999.95 \mathrm{~J}=1.9 \times 10^{4} \mathrm{~J} \end{aligned}$