Heat and Thermodynamics

Given, length of rod at $30^{\circ} \mathrm{C}$ is 30 cm

Rise in temperature

$ \begin{aligned} \Delta T & =105^{\circ}-30{ }^{\circ} \mathrm{C} \\ & =75^{\circ} \mathrm{C} \end{aligned} $

Increase in length,

$ \begin{aligned} \Delta l & =0.027 \mathrm{~cm} \\ & =0.027 \times 10^{-2} \mathrm{~m} \end{aligned} $

We know that, $\Delta l=l \alpha \Delta T$

Thus, $0.027 \times 10^{-2}=0.30 \times \alpha \times 75$

$ \begin{aligned} \Rightarrow \quad \alpha & =\frac{0.027 \times 10^{-2}}{0.30 \times 75} \\ & =0.0012 \times 10^{-2} \\ & =12 \times 10^{-6} \rho \mathrm{C} \end{aligned} $

Heat energy required for 10 kg to bring to $0^{\circ} \mathrm{C}$ will be, $Q_1=m s \Delta T$

$ \begin{aligned} & =10 \times 0.5 \times 10 \times 10^3 \\ & =50 \times 10^3 \mathrm{cal} \end{aligned} $

Heat energy required for 10 kg ice at $0^{\circ} \mathrm{C}$ to convert into water at $0^{\circ} \mathrm{C}$ will be.

$ \begin{aligned} Q_2 & =m L \\ & =10 \times 10^3 \times 80 \\ & =8 \times 10^5 \mathrm{cal} \end{aligned} $

Total heat $Q=Q_1+Q_2$

$ \begin{aligned} & =50 \times 10^3+800 \times 10^3 \\ & =850 \times 10^3 \mathrm{cal} \\ & =357 \times 10^4 \mathrm{~J} \end{aligned} $

We know that, ${ }^{\circ} \mathrm{C}=\left({ }^{\circ} \mathrm{F}-30\right) \times \frac{5}{9}$

Given ${ }^{\circ} \mathrm{F}=2 \times\left({ }^{\circ} \mathrm{C}\right)$

Thus, ${ }^{\circ} \mathrm{C}=\left(2^{\circ} \mathrm{C}-32\right) \times \frac{5}{9}$

$ \begin{aligned} & { }^{\circ} \mathrm{C} \times \frac{9}{5}=2{ }^{\circ} \mathrm{C}-32 \\ & 2^{\circ} \mathrm{C}-{ }^{\circ} \mathrm{C} \times \frac{9}{5}=32 \\ \Rightarrow & \frac{{ }^{\circ} \mathrm{C}}{5}=32 \Rightarrow{ }^{\circ} \mathrm{C}=160^{\circ} \end{aligned} $

Hence, Fahrenheit $\left({ }^{\circ} \mathrm{F}\right)=2^{\circ} \mathrm{C}$

$ =2 \times 160=320^{\circ} \mathrm{F} $

The heat $Q$ is converted into the internal energy and work. According to the first low of thermodynamics,

$ \begin{aligned} & \Delta Q=\Delta U+\Delta W \\ & n c_p d T=n c_V d T+W \end{aligned} $

The fraction of heat converted into

$ \begin{aligned} & \text { internal energy } \frac{\Delta U}{\Delta Q}=\frac{\Delta Q-\Delta W}{\Delta Q} \\ & =\frac{n c_p d T-\left(n c_p d T-n c_V d T\right)}{n c_p d T} \\ & =\frac{n c_V d T}{n c_p d T}=\frac{c_V}{c_p}=\frac{3}{5} \quad\left(\because r=\frac{5}{3}\right) \end{aligned} $

Therefore percentage of heat energy used

$ =\frac{3}{5} \times 100 \%=60 \% $

$ \begin{aligned} &\text { Average energy of an oscillator is }\\ &\begin{aligned} \langle E\rangle & =k_B T \\ & =1.38 \times 10^{-23} \times 300 \\ & =4.14 \times 10^{-21} \mathrm{~J} \end{aligned} \end{aligned} $

We know that, ideal gas equation is given gy

$ \begin{gathered} p V=n R T \\or\,\,\,\frac{1}{T}=\frac{n R}{p V} \end{gathered} $

Since, $p$ is constant

$ p d V=n R d T \Rightarrow \frac{d V}{d T}=\frac{n R}{p} $

coefficient of volume expansion will be

$ \begin{aligned} \alpha_V & =\frac{d V}{V d T}=\frac{n R}{p V}=\frac{1}{T} \\ \Rightarrow \quad \alpha_V & =\frac{1}{300 \mathrm{~K}} \quad\left(\because 23^{\circ} \mathrm{C}=300 \mathrm{~K}\right) \\ & =33 \times 10^{-4} \mathrm{~K}^{-1} \end{aligned} $

Given, temperature,

$ T_1=60^{\circ} \mathrm{C}=273+60=333 \mathrm{~K} $

Let initial mass be $M_1=m$ and air escaped be $M_2=M-\frac{M}{4}=\frac{3 M}{4}$

If pressure and volume of gas remain constant.

Then, $M T=$ Constant

$ \begin{aligned} \frac{T_1}{T_2} & =\frac{M_1}{M_2}=\left(\frac{M}{\frac{3 M}{4}}\right)=\frac{4}{3} \\ \Rightarrow \quad T_2 & =\frac{4}{3} T=\frac{4}{3} \times 333 \\ & =444 \mathrm{~K}=171^{\circ} \mathrm{C} \end{aligned} $

Given mass of water, $m_w=100 \mathrm{~g}$

Change in temperature, $(\Delta T)=90^{\circ} \mathrm{C}- 24^{\circ} \mathrm{C}=66^{\circ} \mathrm{C}$

From the principle of calorimetry

Heat gain by water $=$ Heat lost by steam

$ \begin{aligned} & m_w S \Delta T \\ = & m_{\text {steam }}(540)+\left(m_{\text {steam }}\right)+\left(m_{\text {steam }}(100-90)\right. \\ \Rightarrow & 100 \times 1(66)=m(540)+m(10) \\ \Rightarrow & 550 m=100 \times 66 \\ \Rightarrow \quad & m=12 \mathrm{~g} \end{aligned} $

For an isothermal process, the temperature, will constant Volume is increasing therefore, pressure will decrease i.e., $p \propto \frac{1}{V}$

In chamber $A, \Delta p_A=\left(p_A\right)_i-\left(p_A\right)_f$

$ \begin{aligned} & =\frac{n_A R T}{V}-\frac{n_A R T}{2 V} \\ & =\frac{n_A R T}{2 V} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

In chamber $B, \Delta p_B=\left(p_B\right)_i-\left(p_B\right)_t$

$ =\frac{n_B R T}{V}-\frac{n_B R T}{2 V}=\frac{n_B R T}{2 V} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

According to question, $\frac{\Delta p_A}{\Delta p_B}=\frac{2}{3}$

$ \Rightarrow \quad \frac{1}{1.5}=\frac{n_A}{n_B}=\frac{2}{3} \frac{m_A / M}{m_B / M}=\frac{2}{3} $

or   $ 3 m_A=2 m_B $

Initial Setup :

First rod: length $L$, coefficient of expansion $2\alpha$

Second rod: length $2L$, coefficient of expansion $\alpha$

Composite Rod:

Total initial length = $L + 2L = 3L$

Expansion Calculation:

Expansion of the First Rod:

$ \Delta L_1 = L(2\alpha)\Delta T $

Expansion of the Second Rod:

$ \Delta L_2 = 2L(\alpha)\Delta T $

Total Expansion:

Combine the expansions:

$ \Delta L_{\text{total}} = 2L\alpha\Delta T + 2L\alpha\Delta T = 4L\alpha\Delta T $

Final Step:

Given that the total increase in length of the composite rod is the sum of the individual expansions, equate the change to find the overall coefficient of linear expansion, $\alpha_{\text{total}}$:

$ 4L\alpha\Delta T = 3L\alpha_{\text{total}}\Delta T $

Solving for $\alpha_{\text{total}}$:

$ \alpha_{\text{total}} = \frac{4}{3}\alpha $

Thus, the coefficient of linear expansion of the composite rod is $\frac{4}{3}\alpha$.

For a gas with a constant temperature, we examine the relationship between the volume ($V$) and pressure ($p$). Specifically, we are interested in the slope of a graph where $\log_e V$ is plotted on the X-axis, and $\log_e p$ is plotted on the Y-axis.

Given the ideal gas equation:

$ pV = nRT $

Taking the natural logarithm of both sides, we have:

$ \log_e(pV) = \log_e(nRT) $

This expands to:

$ \log_e p + \log_e V = \log_e(nRT) $

Since $nRT$ is a constant (as both the amount of gas and the temperature are constant), we can rearrange the equation as follows:

$ \log_e p = -\log_e V + \log_e(nRT) $

This equation is in the form of a straight line, where:

$ Y = mx + c $

By comparing:

$ \log_e p = (-1)\log_e V + \log_e(nRT) $

We identify that the slope ($m$) is $-1$.

Therefore, the slope of the graph where $\log_e V$ is on the X-axis and $\log_e p$ is on the Y-axis is $-1$.

Given:

Initial temperature of the ideal gas, $ T_0 = 127^{\circ} \mathrm{C} = 400 \, \mathrm{K} $

Initial volume, $ V_0 $

Final volume, $ V_1 = \frac{8}{27} V_0 $

Heat capacity ratio, $ \gamma = \frac{5}{3} $

The process described is adiabatic, meaning no heat is exchanged with the surroundings. For an adiabatic process involving an ideal gas, the relation $ T V^{\gamma-1} = \text{constant} $ holds.

Initial State:

$ T_0 V_0^{\gamma-1} = T_0 V_0^{5/3-1} \quad \text{(equation i)} $

Final State:

$ T_1 V_1^{\gamma-1} = T_1 \left(\frac{8}{27} V_0\right)^{5/3-1} \quad \text{(equation ii)} $

By dividing equation (i) by equation (ii), we have:

$ \frac{T_0}{T_1} = \frac{V_1^{\gamma-1}}{V_0^{\gamma-1}} = \left(\frac{8}{27}\right)^{5/3-1} = \left(\frac{8}{27}\right)^{2/3} $

This simplifies to:

$ \frac{T_0}{T_1} = \frac{4}{9} $

Solving for $ T_1 $:

$ T_1 = \frac{400 \times 9}{4} = 900 \, \mathrm{K} $

The rise in temperature is then calculated as:

$ T_1 - T_0 = 900 \, \mathrm{K} - 400 \, \mathrm{K} = 500 \, \mathrm{K} $

Given an insulating cylinder containing 4 moles of an ideal diatomic gas, when heat $ Q $ is supplied, 2 moles of the gas molecules dissociate. The temperature of the gas remains constant. We need to find the value of $ Q $.

Initially, the gas has 4 moles of an ideal diatomic gas. When supplied with heat, 2 moles dissociate into monoatomic gas, effectively converting 2 moles of diatomic gas into 4 moles of monoatomic gas.

Degrees of Freedom

Diatomic gas: $ f = 5 $

Monoatomic gas: $ f = 3 $

Internal Energy Calculation

The internal energy $ E $ for gas is given by:

$ E = n \times \frac{f}{2} \times RT $

Where $ n $ is the number of moles, $ f $ is the degree of freedom, $ R $ is the universal gas constant, and $ T $ is the temperature.

Initial Internal Energy $ E_i $

For 4 moles of diatomic gas:

$ E_i = 4 \times \frac{5}{2} \times RT = 10RT $

Final Internal Energy $ E_f $

After dissociation, we have:

2 moles of diatomic gas:

$ 2 \times \frac{5}{2} \times RT = 5RT $

4 moles of monoatomic gas:

$ 4 \times \frac{3}{2} \times RT = 6RT $

Total final internal energy:

$ E_f = 5RT + 6RT = 11RT $

Change in Internal Energy

The change in internal energy is:

$ \Delta E = E_f - E_i = 11RT - 10RT = RT $

Thus, the value of $ Q $ required to maintain constant temperature during the dissociation process is $ RT $.

Let's analyze the problem step by step.

For the object with mass $2m$:

Initial temperature: $T$

Final temperature: $2T$

Temperature change: $\Delta T = 2T - T = T$

The amount of heat required is given by:

$Q = \text{mass} \times c \times \Delta T = 2m \cdot c \cdot T$

So, we have:

$Q = 2m c T$

For the object with mass $m$:

Initial temperature: $2T$

Let the final temperature be $T_f$

Temperature change: $\Delta T = T_f - 2T$

The same heat $Q$ is supplied, so:

$Q = m \cdot c \cdot (T_f - 2T)$

Since both expressions for $Q$ are equal, set them equal to each other:

$2m c T = m c (T_f - 2T)$

Cancel the common factors $m$ and $c$ (assuming they are non-zero):

$2T = T_f - 2T$

Now, solve for $T_f$:

$ \begin{aligned} T_f &= 2T + 2T \\ &= 4T \end{aligned} $

Thus, when the same heat is supplied to the object of mass $m$, its temperature raises to $4T$.

The correct answer is Option C: $4T$.

To convert temperatures between the Celsius scale and the new X scale, we can assume a linear relationship. Here's how to determine the conversion:

Recognize the given points:

Melting point of ice: $0^\circ \text{C} = 20^\circ \text{X}$

Boiling point of water: $100^\circ \text{C} = 110^\circ \text{X}$

Express the relationship in the form:

$X = a + b \cdot C,$

where $a$ is the offset and $b$ is the scale factor.

Substitute the known temperatures:

For $0^\circ \text{C}:$

$20 = a + b \cdot 0 \quad \Rightarrow \quad a = 20.$

For $100^\circ \text{C}:$

$110 = 20 + b \cdot 100.$

Solve for $b$:

$110 - 20 = 100b \quad \Rightarrow \quad 90 = 100b \quad \Rightarrow \quad b = \frac{90}{100} = 0.9.$

Write the complete conversion formula:

$X = 20 + 0.9 \cdot C.$

Convert $40^\circ \text{C}$ to the new X scale:

$X = 20 + 0.9 \times 40 = 20 + 36 = 56.$

Therefore, on the new scale, $40^\circ \text{C}$ is indicated as $56^\circ \text{X}.$ This corresponds to Option B.

In an isobaric process, the pressure of the system remains constant. For such a process, the heat supplied to a diatomic ideal gas can be expressed as:

$ \Delta Q = n C_p \Delta T $

Where:

$ n $ represents the number of moles of the ideal gas.

$ C_p $ is the specific heat at constant pressure.

$ \Delta T $ is the change in temperature.

The work done during an isobaric process is calculated using:

$ \Delta W = p \Delta V $

Substituting the ideal gas law $ pV = nRT $ into the work done expression, we get:

$ \Delta W = p(V_2 - V_1) = nR(T_2 - T_1) = nR \Delta T $

The ratio of the work done $\Delta W$ to the heat supplied $\Delta Q$ is:

$ \frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} $

For a diatomic gas, the specific heat at constant pressure is:

$ C_p = \frac{7}{2} R $

Substituting this into the ratio, we find:

$ \frac{W}{Q} = \frac{R}{\frac{7}{2} R} = \frac{2}{7} $

Therefore, the percentage of the heat supplied that is converted into work is:

$ = \frac{2}{7} \times 100 \approx 28.6\% $

For any gas molecule, the translational degrees of freedom are determined by its ability to move in three-dimensional space. This means:

Translational degrees of freedom = 3 (movement along the x, y, and z axes).

For rotational degrees of freedom:

In a linear molecule (like a polyatomic linear gas molecule), rotation about the axis of the molecule doesn't contribute significantly because the moment of inertia about that axis is negligible.

Thus, only the rotations about the two axes perpendicular to the molecular axis matter.

So, rotational degrees of freedom = 2.

Now, the ratio of translational to rotational degrees of freedom is:

$\frac{3 \text{ (translational)}}{2 \text{ (rotational)}} = 3:2.$

Hence, the correct answer is:

Option D: $3:2$

n = 0.5

T = 300

For v_rms to be doubled T' = 4 $\times$ 300 = 1200

$\Rightarrow$ Heat transferred

$ = (0.5)\left( {{5 \over 2}} \right)(8.32)(900)$

$ = 9360$ J

$1 - {{{T_L}} \over {{T_H}}} = 0.5$ ...... (1)

$1 - {{{T_L} - 40} \over {{T_H}}} = 0.65$ .... (2)

$ \Rightarrow {T_H} = {{800} \over 3}\,K \simeq 266.7\,K$

Average momentum $ = \left\langle {\overrightarrow P } \right\rangle = 0$

${v_{rms}} = \sqrt {{{3RT} \over M}} $

If temperature is doubled and oxygen atoms are used then

$v{'_{rms}} = \sqrt {{{3R(2T)} \over {M/2}}} = 4\,{v_{rms}}$

${\eta _{net}} = {{{W_1} + {W_2}} \over {{Q_1}}}$

${\eta _{net}} = {{{W_1}} \over {{Q_1}}} + {{{W_2}} \over {{Q_1}}}$

${\eta _{net}} = {\eta _1} + {{{W_2}} \over {{Q_2}}} \times {{{Q_2}} \over {{Q_1}}}$

${\eta _{net}} = {\eta _1} + ({\eta _2})(1 - {\eta _1})$

$1 - {\eta _1} < 1$

$ \Rightarrow {\eta _{net}} < {\eta _1} + {\eta _2}$

Statement A is incorrect, statement B is correct by equipartition of energy. Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH₄ is a polyatomic gas so it has 6 degree of freedom. So only B and D are correct.

$450 - T = {{dQ} \over {dt}} \times {{{l_1}} \over {{K_1}{A_1}}}$

$T - 0 = {{dQ} \over {dt}} \times {{{l_2}} \over {{K_2}{A_2}}}$

So, ${{450 - T} \over T} = {{{K_2}{A_2}{l_1}} \over {{K_1}{A_1}{l_2}}} = 9 \times {1 \over 2} \times 2 = 9$

$450 - T = 9T$

$ \Rightarrow T = 45^\circ C$

From Newton's cooling law ${{dQ} \over {dt}} = - k(T - {T_s})$ the statement A is correct.

For B

$U = \sigma eA{T^4}$

So, ${{{U_1}} \over {{U_2}}} = {\left( {{{283} \over {293}}} \right)^4} \simeq {1 \over {1.15}}$

Statement B is correct

For C

$\eta = 1 - {{{T_1}} \over {{T_2}}} = 1 - {{100} \over {400}} = {3 \over 4}$

So, efficiency is 75% C is correct

For D

From Newton's law of cooling ${{dQ} \over {dt}} = - k(T - {T_s})$

The statement is wrong.

${v_{rms}} = \sqrt {{{3RT} \over {{M_0}}}} $ because T is same

v_rms will be same so, A is correct D is incorrect

${{{P_1}} \over {{P_2}}} = {{{n_1}R{T_1}/{V_1}} \over {{n_2}R{T_2}/{V_2}}} = {{{n_1}} \over {{n_2}}} = {1 \over 4}$

B is correct

C is incorrect

${{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$

$ \Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$

$ \Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}$

$ \simeq 61.1 \times {10^{ - 5}}$ kg/s

${C_V} = {{nR} \over 2}$

And ${C_P} = {{nR} \over 2} + R$

$ \Rightarrow {{{C_V}} \over {{C_P}}} = {{{{nR} \over 2}} \over {{{nR} \over 2} + R}} = {n \over {n + 2}}$

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

$ = 7 \times {{3R} \over 2} \times 40$

$ = 3486\,J$

$P{V^\gamma }=$ constant

$ \Rightarrow P{V^\gamma } = (P'){\left( {{v \over 8}} \right)^\gamma }$ where $\gamma = 5/3$

$ \Rightarrow P' = 32P$

Molar mass $M - {{2 \times 4 + n \times 1} \over {2 + n}}$ ..... (i)

Also, $\gamma = {{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}} \over {{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}} = {{2 \times 5R + n \times 7R} \over {2 \times 3R + n \times 5R}}$

$ \Rightarrow \gamma = {{10 + 7n} \over {6 + 5n}}$ ...... (ii)

Given that ${V_{rms}} = \sqrt 2 \,{V_{sound}}$

$ \Rightarrow \sqrt {{{3RT} \over M}} = \sqrt 2 \sqrt {{{\gamma RT} \over M}} $

$ \Rightarrow \gamma = {3 \over 2}$

$ \Rightarrow n = 2$

${\eta _1} = 1 - {{420} \over {720}} = {{300} \over {720}}$

And ${\eta _2} = 1 - {{320} \over {1220}} = {{900} \over {1220}}$

$ \Rightarrow {{{\eta _1}} \over {{\eta _2}}} = {{300} \over {720}} \times {{1220} \over {900}}$

$ \simeq 0.56$

Let AB is isothermal process and BC is adiabatic process then for AB process

P_AV_A = P_BV_B

$\Rightarrow$ P_B = 10⁷ Nm^$-$2

For process BC

P_BV$_B^r$ = P_C V$_C^r$

P_C = 3.536 x $\times$ 10⁶ Pa

Because KE $\propto$ T so A is correct, B is incorrect, statement C cannot be said, statement D is contradicting itself, statement E is incorrect (Isothermal process)

So no answer correct (Bonus)

If the statement of D would have been.

"Pressure of gas increases with increase in temperature at constant volume, "then statement D would have been correct, so in that case answer would have been 'A'.

Here volume is constant.

$\therefore$ ${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}$

$ \Rightarrow {{100} \over {273}} = {{180} \over {{T_2}}}$

$ \Rightarrow {T_2} = {{180} \over {100}} \times 273$

$ = 1.8 \times 273$

$ = 491\,K$

For monoatomic gas degree of freedom f = 3 and $\gamma$ = ${5 \over 3}$

Here for gas,

Initial pressure (P₁) = 75 kPa

Initial volume (V₁) = 1200 cm³

Final volume (V₂) = 150 cm³

Final pressure (P₂) = ?

For adiabatic process,

${P_1}V_1^\gamma = {P_2}V_2^\gamma $

$ \Rightarrow 75 \times {(1200)^\gamma } = {P_2}{(150)^\gamma }$

$ \Rightarrow {P_2} = 75 \times {\left( {{{1200} \over {150}}} \right)^{{5 \over 3}}}$

$ = 75 \times {(8)^{{5 \over 3}}}$

$ = 75 \times 32$ kPa

= 2400 kPa

Work done in adiabatic process,

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

$ = {{2400 \times {{10}^3} \times 150 \times {{10}^{ - 6}} - 75 \times {{10}^3} \times 1200 \times {{10}^{ - 6}}} \over {1 - {5 \over 3}}}$

$ = {{(2400 \times 150 - 75 \times 1200) \times {{10}^{ - 3}}} \over { - {2 \over 3}}}$

$ = {{ - 810000 \times {{10}^{ - 3}}} \over 2}$

$ = - 405$ kJ

$\Delta D = D\alpha \Delta T$

$\Delta T = {{0.011} \over {6.230 \times 1.4 \times {{10}^{ - 5}}}}$

= 126.11$^\circ$C

$\Rightarrow$ T_f = T + $\Delta$T

= (27 + 126.11)$^\circ$C

= 153.11$^\circ$C

Comparing the area under the PV graph

A₃ > A₁ > A₂

$\Rightarrow$ W₃ > W₁ > W₂

Total number of moles are

$n = {n_{{H_2}}} + {n_{{O_2}}}$

$ = {{16} \over 2} + {{128} \over {32}}$

= 12 moles

Using $PV = nRT$

$V = {{nRT} \over P}$

$ = {{12 \times 8.31 \times 273.15} \over {{{10}^5}}}$ m³

= 0.27 m³ = 27 $\times$ 10⁴ cm³

$\Delta Q = n{C_v}\Delta T$ (Isochoric process)

$ = 2 \times {{3R} \over 2} \times 20$

$ = 498$ J

From the equation

$[a] \equiv [P{V^2}]$

$[b] \equiv [V]$

$ \Rightarrow \left[ {{a \over b}} \right] \equiv [PV]$

$\Delta$U_AB = 40 J as process is isochoric.

$\Delta$U_BC = + 50 (W_BC = $-$ 50 J)

U_C = U_A + $\Delta$U_AB + $\Delta$U_BC = 1650

For CA process,

Q_CA = $-$ 60 J

$\Delta$U_CA + W_CA = $-$60

$-$90 + W_CA = $-$ 60

$\Rightarrow$ W_CA = +30 J

The graph given is inconsistent with the statement BC may be adiabatic and CA cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.

As ${v_{rms}} = \sqrt {{{3RT} \over {{M_0}}}} $

T is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved.

So, $v{'_{rms}} = \sqrt {{{3RT \times 2{T_0}} \over {{M_0}/2}}} = 2{v_{rms}}$

So velocity of atomic oxygen is doubled.

$W = {{\mu R({T_2} - {T_1})} \over {1 - r}}$ for a polytropic process for adiabatic process r = $\gamma$

$\Rightarrow$ Statement I is true.

In an adiabatic process

$\Delta$U = $-$ $\Delta$W

$\Rightarrow$ If work is done on the gas

$\Rightarrow$ $\Delta$W is negative

$\Rightarrow$ $\Delta$U is positive or temperature increases

$\Rightarrow$ Statement II is true

As per ideal gas equation, $V = {{nR} \over P}T$

$\Rightarrow$ Slope of V-T graph is inversely proportional to P.

As m₂ > m₁ $\Rightarrow$ P₁ > P₂

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0 K.

B. The mean free path decreases on increasing the number density of the molecules as $\mu = {1 \over {\sqrt 2 \pi n{d^2}}} \Rightarrow \mu \propto {1 \over n}$.

C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.

D. K.E._avg per molecule per degree of freedom is ${1 \over 2}{k_B}T$.

${2 \over 5} \times {1 \over 2}m{v^2} = mL + ms\Delta T$

$ \Rightarrow {{{v^2}} \over 5} = 2.5 \times {10^4} + 125 + 200$

$ \Rightarrow {{{v^2}} \over 5} = 5 \times {10^4}$

$ \Rightarrow v = 500$ m/s

${P_1}V = {n_1}RT$

${P_2}V = {n_2}RT$

$\Rightarrow$ (100 kPa) V = (n₁ + n₂)RT

$ \Rightarrow {n_1} + {n_2} = {{(100\,kPa)(2000\,c{m^3})} \over {8.3 \times 300}}$ ..... (1)

Also, n₁ $\times$ 2 + n₂ $\times$ 32 = 0.76 ...... (2)

Solving (1) and (2),

n₁ = 0.06

n₂ = 0.02

$ \Rightarrow {{{n_1}} \over {{n_2}}} = 3$

$K{E_{avg}} = {3 \over 2}kT$ (At lower temperature)

As temperature is same for both the gases.

$\Rightarrow$ Both gases will have same average kinetic energy.

$ \Rightarrow {{{{(K{E_{avg}})}_{\arg on}}} \over {{{(K{E_{avg}})}_{oxygen}}}} = {1 \over 1}$