Heat and Thermodynamics

We know that ${v_{rms}} = \sqrt {{{3{k_B}T} \over m}} $. Given, ${k_B} = 1.4 \times {10^{ - 23}}$ J/K; T = 300 K; m = 7 $\times$ 10^$-$27 kg. Therefore,

${v_{rms}} = \sqrt {{{3 \times 1.4 \times {{10}^{ - 23}} \times 300} \over {7 \times {{10}^{ - 27}}}}} $

$ = \sqrt {{{3 \times 300 \times 14 \times {{10}^{ - 23}}} \over {7 \times 10 \times {{10}^{ - 27}}}}} $

$ = \sqrt {{{{3^2} \times {{10}^2} \times 2 \times {{10}^4}} \over {10}}} = 3 \times 10 \times {10^2} \times \sqrt {{2 \over {10}}} $

$ = 3 \times \sqrt {10} \times {10^2} \times \sqrt 2 = 3 \times \sqrt 2 \times \sqrt 5 \times {10^2} \times \sqrt 2 $

$ = 3 \times 2 \times \sqrt 5 \times {10^2}$

${v_{rms}} = 6 \times {10^2} \times \sqrt 5 = 13.41 \times {10^2}$ or ${v_{rms}} = 1.34 \times {10^3}$

Given that two Carnot engines A and B are operated in series. Efficiency of a Carnot engine $\eta = 1 - {{{T_2}} \over {{T_1}}}$; where T₁ is temperature of source and T₂ is temperature of sink. Therefore,

${\eta _A} = 1 - {{{T_{2A}}} \over {{T_{1A}}}} = 1 - {T \over {600}}$; T is heat rejected by A

and ${\eta _B} = 1 - {{{T_{2B}}} \over {{T_{1B}}}} = 1 - {{100} \over T}$; T is heat rejected by A and received by B

Now, ${{{\eta _B}} \over {{\eta _A}}}{{1 - {{100} \over T}} \over {1 - {T \over {600}}}} = \left( {{{T - 100} \over T}} \right) \times {{600} \over {600 - T}}$

Let T = 350 K, therefore,

${{{\eta _B}} \over {{\eta _A}}} = {{350 - 100} \over {350}} \times {{600} \over {600 - 350}} = {{250} \over {350}} \times {{600} \over {25}} = {{600} \over {350}}$

$ \Rightarrow {{{\eta _B}} \over {{\eta _A}}} = {{12} \over 7}$

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.

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}} $.

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}} $.

If the junction temperature is $T,$ then

${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$

${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$

$ \Rightarrow 200 - 2T = 2T + T$

$ \Rightarrow T = {40^ \circ }C$

$\therefore$ ${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$

where, ${C_v} = $ molar specific heat at constant volume.
For $BC,$ $\Delta T = - 200\,K$
$\therefore$ $\Delta {U_{BC}} = - 500R$

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

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

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

$7T - 2800 = 2100 - 3T$

$10T = 4900$

$T = 490K$