Heat and Thermodynamics

To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.

In an adiabatic process, $PV^\gamma = \text{constant}$, where $\gamma = \frac{C_P}{C_V}$, $P$ is the pressure, $V$ is the volume, $C_P$ is the heat capacity at constant pressure, and $C_V$ is the heat capacity at constant volume.

Also, the ideal gas equation is $PV = nRT$, where $n$ is the amount of substance, $R$ is the ideal gas constant, and $T$ is the absolute temperature. This can be rearranged to express $P$ in terms of $T$ and $V$, resulting in $P = \frac{nRT}{V}$.

Given that $P = kT^3$, we can equate this to the expression we got from the ideal gas law: $\frac{nRT}{V} = kT^3$. Simplifying, we get $V = \frac{nR}{kT^2}$, which shows that $V$ is inversely proportional to $T^2$.

Now, let's recall the adiabatic condition $PV^\gamma = \text{constant}$. Substituting the proportionalities $P \propto T^3$ and $V \propto T^{-2}$ into this equation, it becomes $T^3(T^{-2})^\gamma = \text{constant}$, simplifying to $T^{3-2\gamma} = \text{constant}$.

For the expression $T^{3-2\gamma} = \text{constant}$ to hold true in an adiabatic process, where the only variable is temperature $T$, the exponent must equal to zero (as the quantity of $T$ to some power equals to a constant suggests that the change in $T$ does not alter the value of the expression). This implies $3 - 2\gamma = 0$, solving for $\gamma$:

$3 - 2\gamma = 0 \implies 2\gamma = 3 \implies \gamma = \frac{3}{2}$

Therefore, the ratio of $\frac{C_P}{C_V}$ for the gas is $\frac{3}{2}$, which matches with Option B $\frac{3}{2}$.

The mean free path $\lambda$ is the average distance covered by a molecule between two successive collisions. It is given by the formula:

$\lambda = \frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

Where:

• $n$ is the number density of the molecules, i.e., the number of molecules per unit volume,
• $d$ is the diameter of the molecules, and
• $\sqrt{2}$ arises from considering the relative motion of a pair of particles in a gas, taking into account that either particle can be moving towards the other, which effectively doubles the cross-sectional area through which they can collide.

This formula shows that the mean free path is inversely proportional to the number density $n$ of the molecules and the square of the diameter $d$ of the molecules. It also takes into account that collisions are more likely as the cross-sectional area of the molecules increases, or as the density of the molecules increases.

Therefore, the correct answer is:

Option A: $\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

$\begin{aligned} \Delta Q & =\Delta U+w \\ & =\pi(140)^2 \times 10^3 \times 10^{-6} \\ & =61.6 \mathrm{~J} \end{aligned}$

The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \propto \sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:

$Z_1 = Z_0 \sqrt{\frac{T_1}{T_0}}$

where:

• $Z_1$ is the collision frequency at temperature $T_1$,
• $Z_0$ is the initial collision frequency at temperature $T_0$,
• $T_1$ is the final absolute temperature, and
• $T_0$ is the initial absolute temperature.

To solve the given problem, we first need to convert the provided temperatures from Celsius to Kelvin (since absolute temperature in Kelvin should be used):

• $T_0 = 27^{\circ}C = 27 + 273 = 300$ K
• $T_1 = 127^{\circ}C = 127 + 273 = 400$ K

Using the formula for collision frequency change and substituting the given values:

$Z_1 = Z_0 \sqrt{\frac{400}{300}} = Z_0 \sqrt{\frac{4}{3}}$

This can be simplified to:

$Z_1 = Z_0 \times \frac{2}{\sqrt{3}}$

Thus, the collision frequency of the system at $127^{\circ}C$ is $\frac{2}{\sqrt{3}}$ times the collision frequency at $27^{\circ}C$, making Option B the correct answer:

$\frac{2}{\sqrt{3}} \mathrm{Z}$

$\begin{aligned} & w=\frac{-n R}{\gamma-1}(\Delta T) \\ &=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\ &=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\ &=R T(2-\sqrt{2}) \\ & \therefore \quad T V_\gamma^{-1}=\text { cons. } \\ & T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\ & T_f=\frac{T}{\sqrt{2}} \end{aligned}$

For non-linear polyatomic molecules, both translational and rotational degree of freedom have same value and is equal to 3.

$\begin{aligned} & P M=\rho R T \\ & \Rightarrow \frac{P}{T}=\left(\frac{R}{m}\right) \rho=\text { slope } \end{aligned}$

So from given curve, $\rho_1>\rho_2>\rho_3$

The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

$R_t = R_0(1 + \alpha t)$

where:

• $R_t$ is the resistance at temperature $t$,


• $R_0$ is the resistance at 0°C (ice point),


• $\alpha$ is the temperature coefficient of resistance, and


• $t$ is the temperature in degrees Celsius.

In this question, we are given:

• Resistance at the ice point ($R_{0} = 8 \Omega$),


• Resistance at the steam point ($R_{100} = 10 \Omega$).

To find the temperature coefficient of resistance ($\alpha$), we use the resistance values at the ice and steam points:

$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$

Now, to find the resistance $R_t$ at $400 ^{\circ}C$, we substitute $\alpha$, $R_0$, and $t = 400$ into the formula:

$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$

Hence, the resistance of the platinum wire at $400^{\circ}C$ is $16 \Omega$. The correct option is:

Option B: 16 $\Omega$

To find the increase in temperature on the Fahrenheit scale, we use the relationship between the Celsius and Fahrenheit temperature scales. The formula to convert Celsius to Fahrenheit is:

$F = \frac{9}{5}C + 32$

However, since we are interested in the increase in temperature, we can ignore the "+ 32" part of the formula, because this constant does not affect the change in temperature, only the absolute temperatures. Thus, to find the increase in temperature on the Fahrenheit scale, we can use:

$\Delta F = \frac{9}{5}\Delta C$

Given that the increase in temperature is $40^{\circ}C$, we can substitute this value into the equation:

$\Delta F = \frac{9}{5} \times 40^{\circ}C$

$\Delta F = 9 \times 8$

$\Delta F = 72^{\circ}F$

Therefore, the increase in temperature on the Fahrenheit scale is $72^{\circ}F$. So, the correct answer is:

Option C $72^{\circ} \mathrm{F}$

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To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:

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

where:

• $ v_{\text{rms}} $ is the root mean square velocity of the gas molecules,
• $ k $ is the Boltzmann constant,
• $ T $ is the absolute temperature in Kelvin, and
• $ m $ is the mass of one molecule of the gas.

Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.

Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen. The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.

We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules. Using the rms velocity formula and considering the ratio of masses:

$ \frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{m_{\text{O₂}}}{m_{\text{H₂}}}} $

We know the mass m is proportional to the molecular weight for each gas, so we can substitute:

$ \frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{M_{\text{O₂}}}{M_{\text{H₂}}}} $

Now we plug in the values:

$ \frac{2 \text{ km/s}}{v_{\text{rms, O₂}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 $

Solving for $ v_{\text{rms, O₂}} $:

$ v_{\text{rms, O₂}} = \frac{2 \text{ km/s}}{4} = 0.5 \text{ km/s} $

Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.

The correct answer is Option D : 0.5.

To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).

Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.

For a monoatomic ideal gas, the molar specific heat at constant volume is:

$ C_{V,m1} = \frac{3}{2}R $

For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:

$ C_{V,m2} = \frac{5}{2}R $

The weighted average of the molar specific heat for the mixture is:

$ C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2} $

Putting the values into the equation, we get:

$ C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6} $

$ C_{V,m} = \frac{(3R + 15R)}{8} $

$ C_{V,m} = \frac{18R}{8} $

$ C_{V,m} = \frac{9}{4}R $

Therefore, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$. The correct answer is:

Option D

$ \frac{9}{4}R $

To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $ W $ is given by the integral of the pressure $ P $ with respect to the volume $ V $:

$ W = \int_{V_1}^{V_2} P dV $

Given the relationship $ PV^{\frac{3}{2}} = K $, we can solve for $ P $:

$ P = \frac{K}{V^{\frac{3}{2}}} $

Now, substitute $ P $ into the work integral and evaluate it:

$ W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV $

To integrate this, we'll use the power rule for integration. The integral of $ V^{-\frac{3}{2}} $ is $ -2V^{-\frac{1}{2}} $, so the work done is:

$ W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right) $

We can rewrite $ V^{-\frac{1}{2}} $ as $ \frac{1}{\sqrt{V}} $:

$ W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $

Since $ P_2V_2^{\frac{3}{2}} = K $ and $ P_1V_1^{\frac{3}{2}} = K $, we can express $ K $ in terms of $ P_1 $ and $ V_1 $ (or $ P_2 $ and $ V_2 $, but we’ll use $ P_1 $ and $ V_1 $ for now):

$ W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $

$ W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}} $

$ W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1 $

Since $ \frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}} $, and recalling $ P_2V_2^{\frac{3}{2}} = K $ once more:

$ W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1 $

$ W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1 $

Putting $ P_1V_1^{\frac{3}{2}} $ back as $ K $:

$ W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1 $

$ W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1 $

$ W = -2P_2V_2 + 2P_1V_1 $

So the work done is:

$ W = 2P_1V_1 - 2P_2V_2 $

Therefore, the correct answer matching the given options is:

Option D: $ 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) $

Shortcut :

The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:

$ v = \sqrt{\gamma \frac{R T}{M}} $

Where:

• $ v $ = speed of sound in the gas
• $ \gamma $ = adiabatic index (ratio of specific heats, $ C_p/C_v $)
• $ R $ = universal gas constant
• $ T $ = temperature in Kelvin
• $ M $ = molar mass of the gas in kilograms per mole (kg/mol)

Given:

• $ R = 8.3 \, J \cdot mol^{-1} \cdot K^{-1} $


• $ \gamma = 1.4 $ (For diatomic gases such as oxygen)


• $ T = 273.15 \, K $ (Standard temperature, 0°C in Kelvin)


• Molar mass of Oxygen ($ O_2 $) $ M = 32 \times 10^{-3} \, kg/mol $ (32 g/mol converted to kg/mol)

Plugging these values into the formula:

$ v = \sqrt{1.4 \times \frac{8.3 \times 273.15}{32 \times 10^{-3}}} $

Calculating the values inside the square root:

$ v = \sqrt{1.4 \times \frac{2268.745}{0.032}} $

$ v = \sqrt{1.4 \times 70896.40625} $

$ v = \sqrt{99304.96875} $

$ v \approx 315 \, m/s $

To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $\frac{1}{2} RT$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.

An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom. Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has $3 \times \frac{1}{2} RT = \frac{3}{2} RT$ of energy.

Oxygen, on the other hand, is a diatomic molecule. This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom. So, each mole of oxygen has $5 \times \frac{1}{2} RT = \frac{5}{2} RT$ of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.

Hence, the total internal energy ($U$) of the gas mixture is:

$ U = (\text{Energy per mole of Ar} \times \text{Number of moles of Ar}) + (\text{Energy per mole of O}_{2} \times \text{Number of moles of O}_{2}) $

$ U = \left(\frac{3}{2} RT \times 8\right) + \left(\frac{5}{2} RT \times 6\right) $

$ U = \left(12 RT + 15 RT\right) $

$ U = 27 RT $

Thus, the total internal energy of the system is $27 RT$.

The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions.

Isobaric processes follow the equation $PV = nRT$, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature.

When we rearrange the equation in terms of V (Volume), we get $V = \left(\frac{nR}{P}\right)T$. The slope of the volume-temperature graph in such a scenario is given by $\frac{nR}{P}$. This implies that the slope is inversely proportional to the pressure (i.e., slope $\propto \frac{1}{P}$).

From this relationship, if one process has a higher slope compared to another, its corresponding pressure must be lower. Observing the given figure and applying this understanding, if Process 2 has a greater slope than Process 1, it indicates that $P_2 < P_1$.

$\mathrm{KE}=\frac{\mathrm{f}}{2} \mathrm{kT}$

$\begin{array}{ll} \mathrm{f}_1=3, & \mathrm{f}_2=5 \\ \mathrm{n}_1=3, & \mathrm{n}_2=2 \end{array}$

$\begin{aligned} & \mathrm{f}_{\text {mixture }}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{9+10}{\mathrm{f}}=\frac{19}{5} \\ & \gamma_{\text {mixture }}=1+\frac{2 \times 5}{19}=\frac{29}{19}=1.52 \end{aligned}$

For process $\mathrm{A}$

$\begin{aligned} & \log P=\gamma \log \mathrm{V} \Rightarrow \mathrm{P}=\mathrm{V}^\gamma,(\gamma>1) \\ & P V^{-\gamma}=\text { Constant } \end{aligned}$

$C_A=C_V+\frac{R}{1+\gamma}$ ..... (i)

Likewise for process $\mathrm{B} \rightarrow P V^{-1}=$ Constant

$\begin{aligned} & C_B=C_v+\frac{R}{1+1} \\ & C_B=C_v+\frac{R}{2} \quad \text{..... (ii)} \\ & C_P=C_v+R \quad \text{..... (iii)} \end{aligned}$

By (i), (ii) & (iii)

$C_P>C_B>C_A>C_v$ [No answer matching]

$\begin{aligned} & \sqrt{\frac{3 R T}{2}}=\sqrt{\frac{3 R(320)}{32}} \\ & T=\frac{320}{16}=20 \mathrm{~K} \end{aligned}$

$\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT} \\ & \mathrm{N}=\text { Total no. of molecules } \\ & \mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT} \\ & 1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T} \\ & 1.01 \times 10^5=2 \times 10^2 \times \mathrm{T} \\ & \mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K} \end{aligned}$

$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}$

For diatomic gas $\mathrm{f}_{\mathrm{eq}}=5$

$\begin{aligned} & 5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2} \\ & 5 \mathrm{~N}+10=6 \mathrm{~N}+6 \\ & \mathrm{~N}=4 \end{aligned}$

Work done $\mathrm{AB}=\frac{1}{2}(8000+4000)$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3=\left(6000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Work done $\mathrm{BC}=-\left(4000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Total work done $=2000$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3$

$\begin{aligned} =2 \times 10^3 & \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}} R T_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}} \mathrm{RT}_{\mathrm{B}}} \\ & \text { Given } \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \\ & \text { And } \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{1 / 2}{1 / 32}=16 \end{aligned}$

For an adiabatic process, the following relation holds:

$P V^{\gamma} = \text{constant}$

where P is the pressure, V is the volume, and $\gamma = \frac{C_p}{C_v}$.

We are given that the pressure is proportional to the cube of the absolute temperature:

$P \propto T^3$.

Using the ideal gas law, $PV = nRT$, we can rewrite this as:

$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$.

Substituting this into the adiabatic relation, we get:

$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$.

Simplifying, we have:

$P^{1-\gamma} T^{2\gamma} = \text{constant}$.

Since P is proportional to $T^3$, we can write:

$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$.

This simplifies to:

$T^{3-3\gamma + 2\gamma} = \text{constant}$.

For this equation to hold, the exponent of T must be zero. Therefore:

$3 - 3\gamma + 2\gamma = 0$.

Solving for $\gamma$, we get:

$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$.

Therefore, the correct answer is Option B.

$\begin{aligned} & {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \\ & \text { And }[\mathrm{V}]=[\mathrm{b}] \\ & \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}] \end{aligned}$

The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom. For a diatomic molecule such as oxygen ($O_2$), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of $\frac{1}{2} kT$ per molecule, where $k$ is the Boltzmann constant and $T$ is the temperature in kelvins.

However, since we are dealing with moles, we'll use the universal gas constant $R$ instead of the Boltzmann constant $k$, because $R = k \cdot N_A$ where $N_A$ is the Avogadro constant (the number of molecules in a mole). Therefore, the average energy per mole for each degree of freedom is $\frac{1}{2}RT$.

To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:

For diatomic oxygen:

Given that temperature $ T $ is $ 27^{\circ} \mathrm{C} $, we first convert it to kelvins:

Now we plug in the values for $ R $ and $ T_{\text{K}} $:

When we calculate this, we find:

So the total kinetic energy of 1 mole of oxygen at $27^{\circ} \mathrm{C}$ is approximately 6232.5 J.

To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:

$\Delta U = m c_v \Delta T$

Where:

• $\Delta U$ is the change in internal energy,
• $m$ is the mass of the substance (in this case, air),
• $c_v$ is the specific heat at constant volume,
• $\Delta T$ is the change in temperature.

Given that:

• The mass of air, $m = 0.08 \, \text{kg},$
• The specific heat of air at constant volume, $c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C},$
• The change in temperature, $\Delta T = 5^{\circ} \text{C},$
• And the conversion factor from calories to Joules, $1 \, \text{cal} = 4.18 \, \text{J}.$

First, convert the specific heat from kcal to Joules:

$c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C} \times 1000 \, \text{cal/kcal} \times 4.18 \, \text{J/cal} = 710.6 \, \text{J/kg}^{\circ}\text{C}$

Now, substitute the values into the formula:

$\Delta U = 0.08 \times 710.6 \times 5$

$\Delta U = 284 \, \text{J}$

Thus, the change in internal energy is approximately 284 Joules.

To find the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$, we use the equation for the average kinetic energy of a molecule in terms of temperature:

$K_{\text{avg}} = \frac{3}{2} k_B T$

Where:

• $K_{\text{avg}}$ is the average kinetic energy
• $k_B$ is the Boltzmann constant, given as $1.38 \times 10^{-23} \, \text{J/K}$
• $T$ is the temperature in Kelvin

First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, so:

$K_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Substituting $K_{\text{avg}}$ and $k_B$ into the equation:

$\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Solving for $T$:

$T = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}$

After performing the calculations:

$T \approx 3200 \, \text{K}$

Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$ is approximately 3200 K.

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.

The relationship between the root-mean-square (rms) speed ($v_{rms}$) and the average speed ($v_{avg}$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:

Where:

• $R$ is the ideal gas constant
• $T$ is the temperature in Kelvin
• $M$ is the molar mass of the gas
• $\pi$ is the mathematical constant pi

In this problem, the rms speed of the oxygen molecule is given by:

Now, let's divide the expression for $v_{rms}$ by the expression for $v_{avg}$:

By simplifying the expression, we get:

Square both sides of the equation:

Now we will substitute the provided value of $\pi = \frac{22}{7}$:

By simplifying the expression, we get:

Now let's solve for $x$:

Multiplying both sides by $x$:

Finally, solving for $x$:

So, the value of $x$ is $\boxed{28}$.

To answer this question, we need to find the efficiency of a Carnot engine operating between the boiling and freezing points of water. The boiling point of water is 100°C (373 K) and the freezing point is 0°C (273 K). The efficiency of a Carnot engine is given by the formula:

Efficiency = $1 - \frac{T_{cold}}{T_{hot}}$

Plugging in the values for the boiling and freezing points of water:

Efficiency = $1 - \frac{273}{373} = 1 - 0.732 = 0.268$

So, the efficiency of a Carnot engine operating between these temperatures is approximately 26.8%.

Now, let's analyze the given options: A. efficiency more than 27% - This is incorrect, as the Carnot efficiency is 26.8% and no engine can be more efficient than a Carnot engine.

B. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures - This is correct, as real engines are less efficient than Carnot engines operating between the same temperatures.

C. efficiency equal to $27 \%$ - This is incorrect, as the Carnot efficiency is 26.8%, not 27%.

D. efficiency less than $27 \%$ - This is correct, as the efficiency is less than 27% (26.8%).

Thus, the correct answer is "B and D only."

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$Q = m \times L$

where $m = 1 \mathrm{~kg}$ and $L = 2257 \mathrm{~kJ/kg}$:

$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$

Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:

$W = -P \Delta V$

where $P$ is the atmospheric pressure and $\Delta V$ is the change in volume. We are given that the atmospheric pressure is $P = 1 \times 10^5 \mathrm{~Pa}$, and the change in volume is

$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$.

Now, we can calculate the work done:

$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$

Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:

$\Delta U = Q + W$

Substitute the values of $Q$ and $W$:

$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$

The change in internal energy of the system during the process is +2090 kJ.

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

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

where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molar mass of the gas molecules.

The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases. Their molar masses are:

1. Neon: $20.18\,\text{g/mol}$ (monoatomic)
2. Chlorine: $2 \times 35.45 = 70.90\,\text{g/mol}$ (diatomic)
3. Uranium hexafluoride: $238.03 + 6 \times 18.998 = 352.03\,\text{g/mol}$ (polyatomic)

Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:

$v_{rms} \propto \frac{1}{\sqrt{m}}$

The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:

$v{rms}(\text{mono}) > v{rms}(\text{dia}) > v_{rms}(\text{poly})$

The internal energy (U) of a gas depends on its degrees of freedom (f).

For a monatomic gas like neon, the degrees of freedom are f = 3 (translational). For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).

The internal energy for each component of the gas mixture can be calculated using the formula:

$U = \frac{f}{2}nRT$

where n is the number of moles, R is the universal gas constant, and T is the temperature.

For the 4 moles of neon (monatomic):

$U_{Ne} = \frac{3}{2} \cdot 4RT = 6RT$

For the 2 moles of oxygen (diatomic):

$U_{O_2} = \frac{5}{2} \cdot 2RT = 5RT$

Now, to find the total internal energy, we sum the internal energies of the individual components:

$U_{total} = U_{Ne} + U_{O_2} = 6RT + 5RT = 11RT$

Thus, the total internal energy of the system is 11RT.

An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings. This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.

Given the options:

Option A: There is no heat supplied to the system.

• This statement is TRUE. In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.

Option B: The temperature of the gas increases.

• This statement is TRUE. When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.

Option C: There is no change in the internal energy.

• This statement is NOT TRUE. In an adiabatic process, the change in internal energy of the system is equal to the work done on the system. When the gas is compressed, work is done on the gas, which increases its internal energy.

Option D: The change in the internal energy is equal to the work done on the gas.

• This statement is TRUE. As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

So, Option C is the statement that is not true for an adiabatic process.

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.

The final volume is $\frac{1}{8}$ of the initial volume, so $P_2 = P_1 \times \frac{V_1}{V_2} = P_1 \times 8 = 8P_1$.

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = \text{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.

Since $V_2 = \frac{V_1}{8}$, we have $P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) ^ \gamma = P_1 \times 8^\frac{5}{3} = P_1 \times 2^5 = 32P_1$.

The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1} = 4$.

Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).

Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).

Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (III) matches with (B).

Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom. So, option (IV) matches with (C).

The kinetic energy of an ideal gas is given by the equation:

$KE = \frac{3}{2} kT$

where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double. The original temperature is given as ($27^\circ C$), which is equal to (300 K) in absolute terms. Therefore, the final temperature ($T_f$) in kelvins is:

$T_f = 2 \cdot 300 K = 600 K$

Converting this back to degrees Celsius gives:

$T_f = 600K - 273 = 327 ^\circ C$

The efficiency of a Carnot engine is given by:

$\eta = 1 - \frac{T_{c}}{T_{h}}$

where:

• $\eta$ is the efficiency of the Carnot engine,
• $T_{c}$ is the temperature of the cold reservoir,
• $T_{h}$ is the temperature of the hot reservoir.

Note that the temperatures must be in Kelvin for this formula.

Here, the given temperatures are $127^{\circ}C$ and $27^{\circ}C$. Converting these to Kelvin gives:

$T_{h} = 127^{\circ}C + 273.15 = 400.15 K$ $T_{c} = 27^{\circ}C + 273.15 = 300.15 K$

So the efficiency of the Carnot engine is:

$\eta = 1 - \frac{300.15}{400.15} = 0.25$

The efficiency can also be defined as the work done divided by the heat supplied:

$\eta = \frac{W}{Q_{h}}$

where:

• $W$ is the work done by the engine,
• $Q_{h}$ is the heat transferred to the engine by the hot reservoir.

We can rearrange this equation to solve for $Q_{h}$:

$Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ$

Statement I: If heat is added to a system, its temperature must increase. This statement is not necessarily true. For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature. The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.