Kinetic Theory Test

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Kinetic Theory Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

K.E. of molecular motion appears as:

A
Pressure

B
P.E.

C
Temperature

D
All of the above

Solution

In the kinetic theory of gasses, increasing the temperature of a gas increases the average kinetic energy of the molecules, causing increased motion. This increased motion increases the outward pressure of the gas, an expected result from the ideal gas equation $$PV=NkT$$

Hence, the kinetic energy of the molecular motion appears as temperature since it varies with change in temperature.

Hence, the correct option is $$C$$
Question 2
1 / -0

A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by $$\Delta$$ T, the weight becomes 150 gm. $$\Delta$$ T should be :

A
$$27^{\circ}$$

B
$$\dfrac{27^{\circ}} {4}$$ C

C
$$300^{\circ}$$

D
$$327^{\circ}$$

Solution

Initially mass of air is 200 - 100 = 100gm. finally mass of air is 150 - 100 = 50 gm. As there is a hole in the wall, pressure inside the container will remain constant = $$P_{0}$$
$$ PV = nRT \Rightarrow T \propto \dfrac{1} {n}$$
as number of moles of gas is halved, the temperature should be doubled (in K)
$$T_{i} = 300K$$ So, $$T_{f} = 600 K \Rightarrow \Delta T = 300 K = 300^{\circ}C$$
Question 3
1 / -0

The relation between the internal energy $$U$$ and adiabatic constant $$\gamma$$ is

A
$$\displaystyle U=\frac{PV}{\gamma-1}$$

B
$$\displaystyle U=\frac{PV^{\gamma}}{\gamma-1}$$

C
$$\displaystyle U=\frac{PV}{\gamma}$$

D
$$\displaystyle U=\frac{\gamma}{PV}$$

Solution

$$PV =nRT$$
$$ U =( N_0n)f(\dfrac{1}{2}kT)$$ .....(1) where
f is the no of degrees of freedom,
$$N_0$$ is Avogadro's number
n is the no of moles
k is the Boltzman's constant
T is the absolute temp
f is related to $$\gamma$$ by $$\gamma = \dfrac{2+f}{f} $$ where $$\gamma$$ is the adiabatic exponent.
$$\therefore \gamma -1 = \dfrac{2+f}{f} -1=\dfrac{2}{f}$$
$$ \Rightarrow f =\dfrac{2}{\gamma -1}$$ .....(2)
Substituting f from(2) in (1)
$$U =( N_0n)f(\dfrac{1}{2}kT)= ( N_0n)\dfrac{2}{\gamma -1}(\dfrac{1}{2}kT)=( N_0n)\dfrac{2}{\gamma -1}(\dfrac{1}{2}\dfrac{R}{N_0}T)=\dfrac{nRT}{\gamma -1}=\dfrac{PV}{\gamma -1}$$
Question 4
1 / -0

Energy of all molecules of a monatomic gas having a volume V and pressure P is $$\frac{3}{2} PV$$. The total translational kinetic energy of all molecules of a diatomic gas at the same volume and pressure is

A
$$\dfrac{1}{2} PV$$

B
$$\dfrac{3}{2} PV$$

C
$$\dfrac{5}{2} PV$$

D
$$3 PV$$

Solution

The no of translational degrees of freedom for mono atomic or diatomic gas is same and is 3.
Hence, translational KE will be same at same temp for both gases.
Mono atmic gas has only translational degree of freedom.
Hence, translational kinetic energy for diatomic gas is $$\dfrac{3}{2}PV$$
Question 5
1 / -0

A gas mixture consists of $$2\:mol$$ of oxygen and $$4\:mol$$ of argon at temperature $$T$$. Neglecting all vibrational modes, the total internal energy of the system is

A
$$4RT$$

B
$$15RT$$

C
$$9RT$$

D
$$11RT$$

Solution

$$O_2$$ molecule has 3 translational and 2 rotational degrees i.e total of 5 degrees of freedom.
Since Argon is a noble gas, it consists only of atoms and the energy is purely translational kinetic energy.
Total internal energy: $$U = 2 \times \dfrac{5}{2}RT + 4 \times \dfrac{3}{2}RT = 11RT$$
Question 6
1 / -0

A cylinder of capacity $$20\:L$$ is filled with $$H_2$$ gas. The total average kinetic energy of translatory motion of its molecules is $$1.5\times 10^5\:J$$. The pressure of hydrogen in the cylinder is

A
$$2\times 10^6\:N/m^2$$

B
$$3\times 10^6\:N/m^2$$

C
$$4\times 10^6\:N/m^2$$

D
$$5\times 10^6\:N/m^2$$

Solution

The degree of freedom of a diatomic gas ($${H}_{2}$$) is =5
out of which 3 are related to translatory motion and 2 are rotatory.
according to Maxwell's Law of Equipartition Energy total energy is equally shared among all degree of freedom.
if total energy is $${E}_{t}$$, then translatory motion energy $$E = \dfrac{3}{5}{E}_{t}$$
$$\dfrac{3}{5}{E}_{t} = 1.5 \times {10}^{5}$$
$${E}_{t} = 2.5 \times {10}^{5} J$$
total energy of diatomic gas $${E}_{t} = \dfrac{5}{2}PV$$
$$P = \dfrac{2{E}_{t}}{5V}$$
$$V = 20 \times {10}^{-3}$$ because $$1 {m}^{3} = 1000 L$$
$$P = \dfrac{2 \times 2.5 \times {10}^{5}}{5 \times 20 \times {10}^{-3}}$$
solve for $$P$$
$$P = 5 \times {10}^{6} N/{m}^{2}$$
answer is D.
Question 7
1 / -0

According to Kinetic theory of gases, molecules are:

A
perfectly inelastic particles in random motion

B
perfectly elastic particles in random motion

C
perfectly inelastic particles at rest

D
perfectly elastic particles at rest
Solution
Assumptions of Kinetic Theory of Gases
All gases are made up of molecules which are constantly and persistently moving in random directions.
The separation between the molecules is much greater than the size of molecules.
When a gas sample is kept in a container, the molecules of the sample do not exert any force on the walls of the container during the collision.
The time interval of collision between two molecules, and between a molecule and the wall is considered to be very small.
All the collisions between molecules and even between molecules and wall are considered to be elastic.
All the molecules in a certain gas sample obey Newton’s laws of motion.
If a gas sample is left for a sufficient time, it eventually comes to a steady state. The density of molecules and the distribution of molecules are independent of position, distance and time.
Hence, the correct option is $$(B)$$
Question 8
1 / -0

A box contains $$N$$ molecules of a perfect gas at temperature $$T_1$$ and pressure $$P_1$$. The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is $$P_2$$ and temperature is $$T_2$$, then

A
$$P_2=P_1$$, $$T_2=T_1$$

B
$$P_2=P_1$$, $$\displaystyle T_2=\frac{T_1}{2}$$

C
$$P_2=2P_1$$, $$T_2=T_1$$

D
$$P_2=2P_1$$, $$\displaystyle T_2=\frac{T_1}{2}$$

Solution

$$P_1V=NkT_1$$ ....(1)
$$P_2V=2NkT_2$$ .....(2)
Dividing (2) by (1)
$$\dfrac{P_2}{P_1}= \dfrac{2T_2}{T_1}$$ ....(3)
Since KE remains same, $$N \times \dfrac{3}{2}kT_1 = 2N \times \dfrac{3}{2}kT_2$$
$$\Rightarrow T_2 = \frac{T_1}{2}$$ ....(4)
Substituting (4) in (3)
$$\frac{P_2}{P_1}=1$$
$$\Rightarrow P_2= P_1$$
Question 9
1 / -0

Gaseous hydrogen contained initially under standard conditions in a sealed vessel of volume $$V= 5L$$ was cooled by $$\Delta T = 55 K$$. The internal energy of the gas will change by

A
$$-255 J$$

B
$$200 J$$

C
$$250 J$$

D
$$-200 J$$

Solution

Volume of vessel $$V = 5L = 0.005$$ $$m^3$$
Standard conditions : $$P = 1$$ atm $$ = 1.01325 \times 10^5$$ Pa $$T = 273.15 K$$
Using Ideal gas equation : $$PV = nRT$$
$$\therefore$$ $$(1.01325 \times 10^5)\times 0.005 = nR \times 273.15 $$ $$\implies nR = 1.8547$$

Degree of freedom for diatomic gas $$f = 5$$
Change in internal energy $$\Delta U = \dfrac{f}{2}nR\Delta T = \dfrac{5}{2} \times 1.8547 \times (-55) = -255$$ $$J$$
Question 10
1 / -0

Two containers of equal volume contain the same gas at pressures $$P_1$$ and $$P_2$$ and absolute temperatures $$T_1$$ and $$T_2$$, respectively. On joining the vessels, the gas reaches a common pressure $$P$$ and common temperature $$T$$. The ratio $$\displaystyle\frac{P}{T}$$ is equal to

A
$$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$$

B
$$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$$

C
$$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$$

D
$$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$$

Solution

$$P_1V=n_1RT_1$$
$$P_2V=n_2RT_2$$
where V is the volume of each vessel.
When the vessels are joined, $$P(2V)=(n_1 + n_2)RT$$
$$ \therefore \dfrac{P}{T}= \dfrac{1}{2}\dfrac{(n_1 + n_2)R}{V}=\dfrac{1}{2}(\dfrac{P_1}{T_1} + \dfrac{P_2}{T_2})=\dfrac{P_1}{2T_1} + \dfrac{P_2}{2T_2}$$

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Kinetic Theory Test - 33

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Kinetic Theory Test - 33

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Question 1

Pressure

P.E.

Temperature

All of the above

Solution

Question 2

$$27^{\circ}$$

$$\dfrac{27^{\circ}} {4}$$ C

$$300^{\circ}$$

$$327^{\circ}$$

Solution

Question 3

$$\displaystyle U=\frac{PV}{\gamma-1}$$

$$\displaystyle U=\frac{PV^{\gamma}}{\gamma-1}$$

$$\displaystyle U=\frac{PV}{\gamma}$$

$$\displaystyle U=\frac{\gamma}{PV}$$

Solution

Question 4

$$\dfrac{1}{2} PV$$

$$\dfrac{3}{2} PV$$

$$\dfrac{5}{2} PV$$

$$3 PV$$

Solution

Question 5

$$4RT$$

$$15RT$$

$$9RT$$

$$11RT$$

Solution

Question 6

$$2\times 10^6\:N/m^2$$

$$3\times 10^6\:N/m^2$$

$$4\times 10^6\:N/m^2$$

$$5\times 10^6\:N/m^2$$

Solution

Question 7

perfectly inelastic particles in random motion

perfectly elastic particles in random motion

perfectly inelastic particles at rest

perfectly elastic particles at rest

Solution

Question 8

$$P_2=P_1$$, $$T_2=T_1$$

$$P_2=P_1$$, $$\displaystyle T_2=\frac{T_1}{2}$$

$$P_2=2P_1$$, $$T_2=T_1$$

$$P_2=2P_1$$, $$\displaystyle T_2=\frac{T_1}{2}$$

Solution

Question 9

$$-255 J$$

$$200 J$$

$$250 J$$

$$-200 J$$

Solution

Question 10

$$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$$

$$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$$

$$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$$

$$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$$

Solution

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