Heat and Thermodynamics

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.

The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:

$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

In this case, we are given that the initial temperature is $200 \, K$ and the final temperature is $800 \, K$. Let the r.m.s speed at $200 \, K$ be $v_0$, then:

$v_0 = \sqrt{\frac{3R \cdot 200}{M}}$

Now, we want to find the r.m.s speed at $800 \, K$, let's call this $v_1$:

$v_1 = \sqrt{\frac{3R \cdot 800}{M}}$

Now, divide $v_1$ by $v_0$:

$\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}$

Simplify the expression:

$\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$

So, $v_1 = 2v_0$.

Hence, the r.m.s speed of the gas at $800 \, K$ will be 2 times the r.m.s speed of the gas at $200 \, K$.

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The average rate of cooling can be represented as:

$\frac{T_1-T_2}{t} = K \left(\frac{T_1+T_2}{2} - T_s\right)$

where:

• $T_1$ and $T_2$ are the initial and final temperatures of the body,
• $t$ is the time it takes for the body to cool from $T_1$ to $T_2$,
• $T_s$ is the temperature of the surroundings, and
• $K$ is a constant of proportionality.

In the first 7 minutes, the body cools from $60^\circ C$ to $40^\circ C$, and the surrounding temperature is $10^\circ C$. So, the first equation is:

$\frac{60-40}{7} = K \left(\frac{60+40}{2} - 10\right) \tag{1}$

In the next 7 minutes, the body cools from $40^\circ C$ to $T^\circ C$, with the same surrounding temperature. So, the second equation is:

$\frac{40-T}{7} = K \left(\frac{40+T}{2} - 10\right) \tag{2}$

Solving equation (1) for $K$ and substituting into equation (2) will give the temperature $T$ after the next 7 minutes:

$T = 28^\circ C$

The speed of sound in a gas is given by the formula:

$v = \sqrt{\frac{\gamma RT}{M}}$

where $v$ is the speed of sound, $\gamma$ is the adiabatic index, $R$ is the universal gas constant, $T$ is the temperature, and $M$ is the molar mass of the gas.

For diatomic gases, such as hydrogen (H₂) and oxygen (O₂), the adiabatic index $\gamma$ is the same, approximately equal to $\frac{7}{5}$, and the temperature is given as the same for both gases.

Let's denote the speed of sound in hydrogen as $v_\text{H}$ and in oxygen as $v_\text{O}$. The ratio of the speeds can be calculated as:

$\frac{v_\text{H}}{v_\text{O}} = \sqrt{\frac{M_\text{O}}{M_\text{H}}}$

The molar mass of hydrogen (H₂) is $2\, \text{g/mol}$, and the molar mass of oxygen (O₂) is $32\, \text{g/mol}$.

Now, we can calculate the ratio of the speeds:

$\frac{v_\text{H}}{v_\text{O}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

This means the ratio of the speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is $4:1$.

Therefore, the correct answer is $4:1$.

The rate of increase of internal energy of a system can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

In this case, the heat being supplied to the system is 1000 W and the work being done by the system is 200 W.

Therefore, the rate at which the internal energy of the system increases is:

1000 W (heat supplied) - 200 W (work done) = 800 W

So, the correct answer is 800 W.

In thermodynamics, the pressure-temperature graph for an ideal gas kept at constant volume (isochoric process) is a straight line. This is because for an ideal gas, the pressure is proportional to the temperature (as described by the ideal gas law, 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).

It is clear from graph that for all the gases lines of graphs meet at same value.

At $\mathrm{x}$-axis (temperature axis) $\mathrm{P}$ is zero but temperature is negative and it will be equal to 0 K or $-273^{\circ} \mathrm{C}$.

$\eta = 1 - {{{T_{\mathrm{cold}}}} \over {{T_{\mathrm{hot}}}}}$

${T_{\mathrm{cold}}} = \mathrm{OK} \Rightarrow \eta = 1$ (max)

A is correct.

R is also correct and explains A.

K.E. per molecule $ = \left( {{f \over 2}KT} \right)$

${{\mathrm{average{{(K.E)}_{hydrogen}}}} \over {\mathrm{average{{(K.E)}_{oxygen}}}}} = {{{f_{\mathrm{hydrogen}}}} \over {{f_{\mathrm{oxygen}}}}} = 1$

$PT^2$ = constant

From $PV = nRT \Rightarrow {{{T^3}} \over V} = $ constant

${T^3} \propto V$ ..... (1)

$3{T^2}dT \propto dV$ ..... (2)

From (1) and (2)

${{3dT} \over T} = {{dV} \over V}$

$\therefore$ $\gamma = {1 \over V}{{dV} \over {dT}} = {3 \over T}$

Isothermal process $\Delta T=0$

$\Delta U=\frac{f}{2}nR\Delta T$

$\Delta U=0$

No change in internal energy

$\Delta Q=\Delta W$ (1$^{st}$ law)

$\Delta Q=+\mathrm{ve}$

$\Delta W=+\mathrm{ve}$

Heat required to raise the temperature of ice to 0$^\circ$C is

$ = {{60} \over {1000}}(2222.3)(12)$

$ = 16000.5$ J

$ \approx 16$ kJ

Heat required to melt ice completely

$ = \left( {{{600} \over {1000}}} \right)(336)$ kJ

$ = 201.6$ kJ

Energy left $ = (184 - 16) = 168$ kJ

$\therefore$ Partial ice will melt

$\therefore$ $168 = ({m_{ice\,melted}})336$

$0.5$ kg $ = ({m_{ice\,melted}})$

$\therefore$ ${m_{ice}}:{m_{water}} = 1:5$

${v_{rms}} = \sqrt {{{\alpha + 5} \over \alpha }} {v_{avg}}$

$\sqrt {{{3RT} \over m}} = \sqrt {{{\alpha + 5} \over 5}} \sqrt {{{8RT} \over {\pi m}}} $

${{3 \times \pi } \over 8} = {{\alpha + 5} \over \alpha }$

${{33} \over {28}} = {{\alpha + 5} \over \alpha }$

$\alpha = 28$

For isothermal process P-V graph is rectangular hyperbola

As the dotted line is an isobaric line which implies T₃ > T₂ > T₁ as volume is increasing.

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.

Given,

Steam temperature $=100^{\circ} \mathrm{C}$

Initial water mass $=150 \mathrm{~g}$

Initial temperature of water $=20^{\circ} \mathrm{C}$

Final temperature $=40^{\circ} \mathrm{C}$

Specific heat of water $=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$

Latent heat of steam $=540 \mathrm{cal} / \mathrm{g}$

Using principle of calorimetry.

Heat lost by steam = Heat gained by water

$ \begin{aligned} m L_{\text {vopourisation }}+m S_w \Delta T & =m_w S_w \Delta T \\ m \times 540+m \times 1 & \times(100-40) \\ & =150 \times 1 \times(40-20) \\ 540 m+60 m & =3000 \\ 600 m & =3000 \\ m & =5 \mathrm{~g} \end{aligned} $

Hence, total mass $=150+5=155 \mathrm{~g}$

Given,

Diameter of wooden wheel $=5.012 \mathrm{~m}$ Diameter of circular iron frame $=5 \mathrm{~m}$ Temperature, $T=27^{\circ} \mathrm{C}$

Length increases thermally is given by

$ \begin{aligned} \Delta L & =L_0 \alpha \Delta T \\ \text { where } \Delta L & =L_1-L_0 \\ \Delta T & =T_2-T_1 \end{aligned} $

$L_0=$ Initial diameter,$L_1=$ Final diameter [Diameter of circular frame, Diameter of wooden wheel]

$ \Delta T=\frac{\Delta L}{\alpha L_0} $

[ $\alpha$-Linear expansion coefficient $\left.=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\right]$

$ \begin{aligned} & T_2-T_1=\frac{L_1-L_0}{\alpha L_0} \\ & \left.T_2-27^{\circ} \mathrm{C}=\frac{5.012-5.000}{5 \times 1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}}\right] \\ & \Rightarrow \quad T_2=\frac{0.012}{5 \times 1.2 \times 10^{-5}}+27^{\circ} \mathrm{C} \\ & \quad=227^{\circ} \mathrm{C} \end{aligned} $

Hence, the required temperature is $227^{\circ} \mathrm{C}$.