Kinetic Theory - Class 11 Medical Physics - Extra Questions

Let density and rate of diffusion of gas $$'A'$$ is $$d_{A}$$ and $$r_{A}$$ and let density and rate of diffusion of gas $$'A'$$ is $$d_{B}$$ and $$r_{B}$$
$$\therefore d_{A} = 4d_{B} \therefore r_{A}\propto \dfrac {1}{\sqrt {d_{A}}}, r_{B}\propto \dfrac {1}{\sqrt {d_{B}}}$$
$$\therefore \dfrac {r_{A}}{r_{B}} = \dfrac {\dfrac {1}{\sqrt {d_{A}}}}{\dfrac {1}{\sqrt {d_{B}}}}$$ i.e., $$\dfrac {r_{A}}{r_{B}} = \dfrac {\sqrt {d_{A}}}{\sqrt {d_{B}}} = \sqrt {\dfrac {d_{B}}{d_{A}}}$$
i.e., $$\dfrac {r_{A}}{r_{B}} = \sqrt {\dfrac {1}{4}} = \dfrac {1}{2}$$
$$r_{A} : r_{B} = 1 : 2$$

1. Charles' Law is a special case of the ideal gas law. It states that the volume of a fixed mass of a gas is directly proportional to the temperature. This law applies to ideal gases held at a constant pressure
2. It states that, for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature, assuming in a closed system.
   

$$PV=RT$$

As implied by equation, the positive sign is equivalent to heat being added to the gas. 

We note that $$\Delta =n\left(\dfrac 32 R \right)\Delta T$$ according to the distance in. Also $$\Delta E_{int}=nC_V \Delta V$$ can be used for each of these processes. Finally , we note equation  leads to $$C_P= C_V+R\cong 8.0\ cal/mol-K$$ after we convert joules to calories in the ideal gas constant value $$R\cong 2.0\ cal/mol-K$$. The first law of thermodynamic $$Q=\Delta E_{int}+W$$ applies to each process.
Consider volume process with $$\Delta T=50\ K$$ and $$n=3.0\ mol$$.
The change in the translation kinetic energy is $$\Delta K=(3.0)\left(\dfrac 32(2.0)\right)(50)=450\ cal$$
Adiabatic process with $$\Delta T=50\ K$$ and $$n=3.0\ mol$$

The work done in a constant pressure process is $$W=p\Delta V$$. Therefore, $$E=(25\ N/m^2) (1.8\ m^3 -3.0\ m^3)=-30\ J$$.
The sign conventions discussed in the textbook for $$Q$$ indicate that we should write $$-75\ J$$ for the energy that leaves the system in the form of heat. Therefore, the first law of thermodynamics leads to
$$\Delta E_{int}=Q-W=(-75\ J)-(-30\ J)=-45\ J$$. 

(a)

Triple point of water, $$T=273.16 K$$.

At this temperature, pressure in thermometer $$A, P_A=1.25 \times 10^5\ Pa$$ 

Let $$T_1$$ be the normal melting point of sulphur. 

At this temperature, pressure in thermometer $$A, P_1 = 1.797 \times 10^5\ Pa$$
According to Charles law, we have the relation:

$$P_A/T=P_1/T_1$$

$$T_1=\dfrac{1.797 \times 10^5 \times 273.16}{1.25 \times 10^5}=392.69\ K$$

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K. 

(b)

At triple point 273.16 K, the pressure in thermometer B, $$P_B=0.2 \times 10^5\ Pa$$ 
At temperature $$T_1$$, the pressure in thermometer B, $$P_2=0.287 \times 10^5\ Pa$$ 
According to Charles law, we can write the relation: 
$$P_B/T=P_1/T_1$$

$$0.2 \times 10^5/273.16=0.287 \times 10^5/T_1$$
 $$T_1=(\dfrac{0.287 \times 10^5}{0.2 \times 10^5})\times 237.16=391.98\ K$$

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

Take a small burning candle. Cover the burning candle with a glass jar. After few minutes the candle is extinguished. As the supply of air is stopped due to glass jar the burning of candle is also stopped. This experiment proves that air supports burning.

We can assume that the process is isothermal because the temperature of the surrounding remains constant

So, by applying Boyle’s law

$${{P}_{s}}{{V}_{s}}={{P}_{d}}{{V}_{d}}.....(I)$$

Where, $$s$$ = surface of water

$$d$$ = depth

 Given that,

Depth of lake $$=h$$

Volume $${{V}_{s}}=3{{V}_{d}}$$

Now,  

  $$ {{P}_{s}}=75\,cm\,Hg $$

 $$ {{P}_{d}}=75+\dfrac{h}{10} $$

Now, put the value of equation(I)

  $$ 75\times 3=75+\dfrac{h}{10} $$

 $$ 225-75=\dfrac{h}{10} $$

 $$ h=1500\,cm $$

 $$ h=15\,m $$

Hence, the depth of lake is $$15\ m$$

We note that $$\Delta =n\left(\dfrac 32 R \right)\Delta T$$ . Also $$\Delta E_{int}=nC_V \Delta V$$ can be used for each of these processes. Finally , we note equation  leads to $$C_P= C_V+R\cong 8.0\ cal/mol-K$$ after we convert joules to calories in the ideal gas constant value $$R\cong 2.0\ cal/mol-K$$. The first law of thermodynamic $$Q=\Delta E_{int}+W$$ applies to each process.
Consider volume process with $$\Delta T=50\ K$$ and $$n=3.0\ mol$$.
we have $$W=300\ cal$$

We note that change in energy can be written as$$\Delta =n\left(\dfrac 32 R \right)\Delta T$$. Also $$\Delta E_{int}=nC_V \Delta V$$ can be used for each of these processes. Finally , we note equation  leads to $$C_P= C_V+R\cong 8.0\ cal/mol-K$$ after we convert joules to calories in the ideal gas constant value $$R\cong 2.0\ cal/mol-K$$. The first law of thermodynamic $$Q=\Delta E_{int}+W$$ applies to each process.
Consider volume process with $$\Delta T=50\ K$$ and $$n=3.0\ mol$$.
$$W=p\Delta V$$ for constant pressure processes, so (using the ideal law)
$$W=n R\Delta T=(3.0)(2.0)(50)=300\ cal$$
The first law gives $$Q=(900+300)\ cal=1200\ cal$$. 

Since the pressure (and also the number of moles) don't change in this process, then the volume is simply proportional to be (absolute) temperature. Thus, the final temperature is $$1/4$$ of the initial temperature. The answer is $$90\ K$$.

Since the pressure is constant (and the number of moles is presumed constant), the ideal gas law in ratio from leads to
$$T_2=T_1 \left(\dfrac {V_2}{V_1}\right)=(300\ K)\left(\dfrac {1.8\ m^3}{3.0\ m^3}\right)=1.8\times 10^2 K$$.
It should be noted that this is consistent with the gas being monatomic (that is, if one assume $$C_V =\dfrac 32 R$$ and , one arrives at this same value for the final temperature,)

According to Boyle's law; at a constant temperature the volume of gas in inversely proportional to pressure ie it is valid only for constant mass of gas. But in this case when air is jumped continuously in tyre, so number of moles of air increases. So Boyle's or gas laws does not valid in this case.

Volume of $$1$$ molecule
$$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }=\cfrac { 4 }{ 3 } \times 3.14\times { \left( { 10 }^{ -10 } \right)  }^{ 3 }$$
$$r=1\mathring { A } ={ 10 }^{ -10 }m(Given)$$
volume of $$1$$ molecule $$=4\times 1.05\times { 10 }^{ -30 }{ m }^{ 3 }=4.20\times { 10 }^{ -30 }{ m }^{ 3 }\quad $$
Number of mole in $$0.5g$$ $${H}_{2}$$ gas $$=\cfrac { 0.5 }{ 2 } =0.25$$ ($$\therefore$$ $${H}_{2}$$ has $$2$$ mole)
Volume of $${H}_{2}$$ molecules in $$0.25$$ mole
$$=0.25\times 6.023\times { 10 }^{  }\times 4.20\times { 10 }^{ -30 }{ m }^{ 3 }$$
$$=1.05\times 6.023\times { 10 }^{ -23-30 }=6.324\times { 10 }^{ -23-30 }=6.3\times { 10 }^{ -7 }{ m }^{ 3 }\quad $$
volume of $${H}_{2}$$ molecules $$=6.3\times { 10 }^{ -7 }{ m }^{ 3 }\quad $$
Now for ideal gas at constant temperature
$${ P }_{ i }{ V }_{ i }={ P }_{ f }{ V }_{ f }\quad { V }_{ f }=\cfrac { { P }_{ i }{ V }_{ i } }{ { P }_{ f } } =\cfrac { 1 }{ 100 } \times { \left( 3\times { 10 }^{ -2 } \right)  }^{ 3 }$$ ($$\because$$ vol. of cube $${V}_{i}={(side)}^{3};{P}_{i}=1atm$$ at NTP)
$${ V }_{ f }=\cfrac { 27\times { 10 }^{ -5 } }{ 100 } =2.7\times { 10 }^{ -5-2 }=2.7\times { 10 }^{ -7 }{ m }^{ 3 }$$
Hence on compression the volume of the gas of the order of nuclear force of interaction will play the role, as in kinetic theory of gas molecules do not interact each other so gas will not obey the ideal gas behaviour

Let the initial volume, $$V = V_{0}$$
Final volume, $$V = 2.V_{0}$$
Initial temperature, $$T_{i} = 40^\circ C$$
$$= 40 + 273$$
$$313 \,K$$
Let final temperature $$= T_{f}$$
For an adiabatic process,
$$TV^{\gamma - 1} = constant$$
$$\therefore T_{i}V_{i}^{\gamma - 1} = T_{f} \,V_{f}^{\gamma - 1}$$
$$\therefore (3 / 3) . (V_0)^{\dfrac{5}{3} - 1} = T_{f} (2V_0)^{\dfrac{5}{3} - 1}$$
$$\therefore T_{f} = \dfrac{3 / 3}{2^{\dfrac{2}{3}}}$$
$$\therefore T_{f} = 197.18 \,K$$
Therefore, on adiabatic expansion, the temperature of the gas drops to $$197.18 \,K.$$

We have, $$P . V^{2}$$ = constant
i.e. $$P_{i}V V_{i}^{2} = P_{f}V V_{f}^{2}$$ ........(1)
For ideal gas, $$\dfrac{PV}{T} = a$$ constant 
$$\dfrac{P_i}{f_i} = \dfrac{V_f}{T_f} \dfrac{T_i}{V_i}$$ ........(3)
Using equation (3) and (1),
$$\dfrac{V_f}{T_f} . \dfrac{T_i}{V_i} V_{i}^{2} = V_{f}^{2}$$
$$\Rightarrow T_{i}V_{i} = T_{f}V_{f}$$ or $$T_{f} = \dfrac{T_iV_i}{V_f}$$

Charles's law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature .

At depth, $$P =P_0 +p gh$$ and $$PV_i = nRT_i$$ 
At the surface, $$P_0V_f= nRT_f : \dfrac{P_0V_f}{(P_0+pgh)V_i}=\dfrac{T_f}{T_i}$$ 
Therefore,  $$V_f =V_i\dfrac{T_f}{T_i}(\dfrac{P_0+pgh}{P_0})$$ and 
$$V_f =1.00 cm^3(\dfrac{293 K}{278 K })\times (\dfrac{(1.013 \times 10^5 Pa)+(1030 kg/m^3)(9.80 m/s^2)(25.0 m)}{1.013 \times 10^5 Pa})$$
$$=3.68 cm^3$$

Along the process line : $$S = aT + C_V In \ T$$
or the specific heat is: $$C=  T \frac{dS}{dT} = aT + C_V$$
On the other hand  : $$dQ  = CdT  = C_V dT + pdV $$ for an ideal gas
Thus, $$pdV  = \frac{RT}{V}dV  = aTdT$$
or $$\frac{R}{a}\frac{dV}{V} = dT $$ or, $$\frac{R}{a} In \ V + constant = T$$
Using $$T = T_0$$ when $$V = V_0$$, we get, $$T = T_0+ \frac{R}{a}In \frac{V}{V_0}$$

To compute the mass of air Ieaving the room, we begin with the ideal gas law: 
$$P_0V =n_1RT_1 =\dfrac{m_1}{M}RT_1$$,
 As the temperature is increased at constant pressure, 
$$P_oV =n_2RT_2 =(\dfrac{m_2}{M})RT_2$$ 
Subtracting the two equations gives 
$$m_1-m_2=\dfrac{P_0VM}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})$$

Again, we have
$$pV = nkT$$
$$H_{2}$$ and $$He$$ have equal volumes. Since pressure are equal, we must have
$$n_{H}T_{H} = n_{He}T_{He}$$
or, $$\dfrac {T_{He}}{T_{H}} = \dfrac {n_{H}}{n_{He}} = 2$$
Here, $$T_{He} = 27 + 273 = 300\ K$$
$$\therefore T_{H} = \dfrac {1}{2} T_{He} = 150\ K = (150 - 273)^{\circ}C = -123^{\circ}C$$.

We assume that the pressure of the air in the bubble is essentially the same as the pressure in the surrounding water. If $$d$$ is the depth of the lake and $$\rho$$ is the density of water, then the pressure at the bottom of the lake is $$p_1=p_0+\rho gd$$, where $$p_0$$ is atmospheric pressure. Since $$p_1V_1=nRT_1$$, the number of moles of gas in the bubble is 
$$n=p_1V_1/ RT_1=( p_0+\rho gd)V_1/ RT_1$$
where $$V_1$$ is the volume of the bubble at the bottom of the lake and $$T_1$$ is the temperature there. At the surface of the lake pressure is $$p_0$$ and the volume of the bubble is $$V_2=nRT_2/p_0$$. We substitute for $$n$$ to obtain
$$V_2=\dfrac{T_2}{T_1}\dfrac{p_0+\rho gd }{p_0}V_1$$
$$=\left( \dfrac{293\ K}{277\ K}\right)\left( \dfrac{1.013 \times 10^5\ Pa+(0.998\times 10^3\ kg/m^3)(9.8\ m/s^2)(40\ m)}{1.013 \times 10^5\ Pa}\right)(20\ cm^3)$$

Since the pressure is constant the work is given by $$W=p(V_2-V_1)$$. The initial volume is $$V_1=(AT_1-BT_1^2)/p$$, where $$T_1=315\ K$$ is the initial temperature, $$A=24.9\ J/K$$ and $$B=0.00662\ J/K^2$$. The final volume is $$V_2=(AT_2-BT_2^2)/p$$, where $$T_2=315\ K$$. Thus
$$W=A(T_2-T_1)-B(T_2^2-T_1^2)$$
$$=(24.9\ J/K)(325\ K -315\ K)-(0.00662\ J/K^2)[ (325\ K)^2-(315\ K)^2]=207\ J$$.

The main assumptions of kinetic theory of gases are:
1. A gas is made up of a large number of sub-microscopic particles called atoms. 
2. Gases consist of particles in constant random motion. They continue to move in a straight line until they collide with each other on the walls of their container. 
3. Gas pressure is due to the moleiMIeS colliding with the walls of the container. All of these collisions are perfectly elastic. This means that there is no change in energy of either the particles or the wall upon collision. So, no energy is lost or gained from collisions. 
4. No molecular forces are at work. The potential energy of molecules is zero. So, whole of the energy. in an ideal gas is kinetic energy only. There is no attraction or repulsion between the particles. 
5. The time taken for the collision is negligible as compared with the time between collisions. 6. The kinetic energy of a gas is a measure of its kelvin temperature. Each gas molecule has different speed but the temperature and kinetic energy of gas refer to the average of these speeds. 
7. The average kinetic energy of an atom of gas is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move. 
8. All gases at a given temperature have same average kinetic energy. 
9. fighter molecules of a gas move faster than the heavier molecules.
 According to the kinetic theory, the average kinetic energy of molecules of a gas is directly proportional to the absolute temperature of the gas. So, from equation(14.21),
$$\dfrac{1}{2} m\overset{-}{v^2} = \alpha T$$ ..........(14.24)
where $$\alpha$$ is a proportionality constant.
From equation (14.17), we get
$$PV = \dfrac{2}{3} N_A \alpha T$$
If temperature (T) is constant, then
$$PV = constant$$
i.e., the product of pressure and volume of gas at constant temperature is constant. This is Boyle's law.
From equation (14.25), if pressure (P) of a gas is constant, then
$$V \propto T$$ .........(14.26)
i.e., for a gas of a given mass, volume of gas is directly proportional to the temperature at constant pressure. This is Charles law.
From equation (14.17),
$$P = \dfrac{1}{3} \dfrac{mN}{V} \overset{-}{v^2}$$  .........(14.27)
Since, from equation (14.20),
$$\overset{-}{v^2} = \dfrac{3k_{B}T}{m}$$
By putting the value of $$\overset{-}{v^2}$$ in equation (14.27), we have
$$P = \dfrac{1}{3} \dfrac{mN}{V} \times \dfrac{3k_{B}T}{m}$$
$$P = \dfrac{Nk_{B}T}{V}$$
$$\Rightarrow P \propto T$$ .........(14.28)
So, at constant volume for a gas of given mass, the pressure of a gas is directly proportional to the temperature of gas. This is Gay Lussac’s law. 
According to Dalton’s law of partial pressure, in thermal equilibrium state, the total pressure of each gas is equal to the sum of the different pressures of each gas for a mixture of unreactive gases filled in a container, i.e., if $$P_1, P_2, P_3,...$$ respectively are the different pressure of gases. Then, the total pressure (P)
$$P = P_1 + P_2 + P_3+ .....$$   (14,41)
Let the gases be filled in a container with volume V in which the numb er of molecules are $$n_1, n_2, n_3,..$$ respectively, $$m_1, m_2, m_3,..$$ be the mass of each molecule, and $$\overset{-}{v_1^2}, \overset{-}{v_2^2}, \overset{-}{v_3^2},..$$ respectively be the mean square speed.Now, according to the kinetic theory, the pressure on the first gas
$$P_1 = \dfrac{1}{3} \dfrac{m_1n_1}{V} \overset{-}{v_1^2}$$
$$P_1V = \dfrac{1}{3} m_1n_1\overset{-}{v_1^2}$$ ........(14.42)
Similarly, for the second gas,
$$P_2V = \dfrac{1}{3}m_2n_2\overset{-}{v_2^2}$$ ........(14.43)
and for the third gas,
$$P_3V = \dfrac{1}{3} m_3n_3\overset{-}{v_3^2}$$ ........(14.44)
Similarly, we can write equation for other gases. Since the mixture of all-the gases are in thermal equilibrium state, i.e., the temperature is same. Then, average kinetic energy of each molecule of a gas will be equal, i.e.,
$$\dfrac{1}{2}m_1 \overset{-}{v_1^2} = \dfrac{1}{2} m_2 \overset{-}{v_2^2} = \dfrac{1}{2} m_3 \overset{-}{v_3^2} = ...... = \dfrac{1}{2} m\overset{-}{v^2}$$ ....(14.45)
Then, the sum of pressure of all gases
$$P_1 + P_2 + P_3 + ........... = \dfrac{1}{3} \dfrac{m_1n_1}{V} \overset{-}{v_1^2} + \dfrac{1}{3} \dfrac{m_2 n_2}{V} \overset{-}{v_2^2} + \dfrac{1}{3} \dfrac{m_3n_3}{V} \overset{-}{v_3^2} + ..$$   .....(14.46)
From equations (14.45) and (14.46),
$$P_1 + P_2 + P_3 + .... = \dfrac{1}{3} (n_1 + n_2 + n_3 + ..._ m\overset{-}{v^2}$$ ....(14.47)
If the total pressure of container is P and the total number of molecules are $$(n_1 + n_2 + n_3 + ...)$$, then the total applied pressure,
$$P = \dfrac{1}{3} (n_1 + n_2 + n_3 + ....) m\overset{-}{v^2}$$.......(14.48)
i.e., $$P = P_1 + P_2 + P_3 + ...$$
This is Dalton's law of partial pressure.