Kinetic Theory Test

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Kinetic Theory Test - 45

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Question 1
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The temperature of a body is an indicator of :

A
The total energy of the moelcules of an object

B
The average energy of the molecules of an object

C
The total velocity of the molecules of the object

D
The average kinetic energy of the molecules of an object

Solution

the average kinetic energy of a body is directly proportional to the temperature of the body, hence temperature of the body is an indicator of the total energy of the molecules of an object.
Question 2
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If the pressure of the gas contained in a vessel is increased by $$0.4$$%, when heated through $$\displaystyle { 1 }^{ \circ }C$$. What is the initial temperature of the gas?

A
$$\displaystyle 250K$$

B
$$\displaystyle { 250 }^{ \circ }C$$

C
$$\displaystyle 2500K$$

D
$$\displaystyle { 25 }^{ \circ }C$$

Solution

For an Ideal Gas,
$$PV=nRT$$
$$\implies V\Delta P =nR\Delta T$$
$$\implies \dfrac{\Delta P}{P}=\dfrac{\Delta T}{T}=0.004$$
$$\implies T=\dfrac{\Delta T}{0.004}=\dfrac{1K}{0.004}=250K$$
Question 3
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A closed container of volume V contains an ideal gas at pressure P and Kelvin Temperature T. The temperature of the gas is changed to 4T due to passing heat to the container. Choose correct pairs of pressure and volume after change of the temperature.

A
$$P\;and \;V$$

B
$$\displaystyle P\;and \frac{1}{2}V$$

C
$$4\;P\;and \;4\;V$$

D
$$4\;P\;and \;V$$

E
$$\displaystyle \frac{1}{4}\;P\;and\;V$$

Solution

As the container is closed, thus the volume of the gas remains constant i.e $$V$$
Using ideal gas equation, $$PV = nRT$$ $$\implies$$ $$\dfrac{P}{T} = \dfrac{nR}{V} =constant$$
$$\therefore$$ $$\dfrac{P}{T} =\dfrac{P'}{T'}$$ where $$T' = 4T$$

OR $$\dfrac{P}{T} =\dfrac{P'}{4T}$$ $$\implies P' = 4P$$
Hence the final pressure and volume of the gas is $$4P$$ and $$V$$ respectively.
Question 4
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The number of vibrational degrees of freedom for a $$CO_2$$ molecule is

A
4

B
5

C
6

D
9

Solution

$$Answer:-$$ A
There are always $$3N$$ total independent degrees of freedom for a molecule, where $$N$$ is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the $$x, y, z$$ directions.
Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.
The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have $$3N$$ total degrees of freedom, but you can set aside $$3$$ of them as translation of the whole molecule in each of the $$x, y, z$$ directions, leaving $$(3N-3)$$ degrees of freedom.
Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the $$x, y, z$$ directions. But for linear molecules like $$CO_2$$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the $$3N$$ total. So while for non-linear molecules there are $$(3N-3-3) = (3N-6)$$ degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are $$(3N-3-2) = (3N-5)$$ degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for $$any$$ of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that $$none$$ of the atoms move due to the "rotation".
So since $$CO_2$$ has three atoms and is linear, it has ($$3\times3 - 5 = 4 $$)degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes.
Question 5
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A gas in a flexible container initially has volume of 320 L. If we decrease the pressure of the gas from 4.0 atm to 1.0 atm and decrease the temperature form 273 degrees Celsius to 0 degrees Celsius, what is the new volume of the gas?

A
0 L

B
320 L

C
160 L

D
640 L

E
80 L

Solution

Given : $$V_1 = 320$$ L , $$P_1 = 4.0$$ atm, $$P_2 = 1.0$$ atm
Initial temperature, $$T_1 = 273 ^oC =546$$ K
Final temperature, $$T_2 = 0 ^oC =273$$ K
Using $$\dfrac{P_2V_2}{T_2} = \dfrac{P_1 V_1}{T_1}$$

$$\therefore$$ $$\dfrac{1.0 \times V_2}{273} = \dfrac{4.0 \times 320}{546}$$ $$\implies V_2 = 640$$ L
Question 6
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An ideal gas is heated in a container that has a fixed volume. Identify which of the following will increase as a result of this heating?
I. The pressure against the walls of the container
II. The average kinetic energy of the gas molecules
III. The number of moles of gas in the container

A
I only

B
I and II only

C
II and III only

D
II only

E
III only

Solution

Ideal gas equation : $$PV = nRT$$ $$\implies P = \dfrac{nRT}{V}$$
Number of moles of gas (n) remains constant because the volume of gas is constant.
As the container has fixed volume i.e $$V =constant$$ $$\implies P \propto T$$ .........(1)
Due to heating, the temperature of the gas increases.
Equation (1) implies that the pressure-exerting against the wall by gas molecules also increases.
Average kinetic energy of molecules $$K.E = \dfrac{3}{2}k_{B}T$$ $$\implies$$ $$K.E \propto T$$
Thus average kinetic of the gas molecules increases with the increase in temperature.
Hence option B is correct.
Question 7
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A closed container of a gas is heated to a high temperature. What do you predict will happen?

A
The gas particles will repel each other

B
The gas particles inside the container will move slower and slower

C
The pressure inside the container will decrease

D
The pressure inside the container will increase

Solution

When temperature increases, gas particles absorb energy and move far apart from each other hence pressure inside the closed container increases.

Hence, option $$D$$ is correct.
Question 8
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At $$\displaystyle { 10 }^{ \circ }C$$, the value of the density of a fixed mass of an ideal gas divided by its pressure is $$'x'$$, at $$\displaystyle { 110 }^{ \circ }C$$ this ratio is:

A
$$\displaystyle \frac { 10 }{ 110 } x$$

B
$$\displaystyle \frac { 383 }{ 283 } x$$

C
$$\displaystyle \frac { 110 }{ 10 } x$$

D
$$\displaystyle \frac { 283 }{ 383 } x$$

Solution

Let the ratio of density and pressure be $$r$$.

From Ideal Gas Equation, $$\rho =\dfrac { PM }{ RT } $$

$$r=\dfrac { \rho }{ P } =\dfrac { M }{ RT } $$

$$\therefore \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } =\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { 273+10 }{ 273+110 } =\dfrac { 283 }{ 383 } $$

$${ r }_{ 2 }=\dfrac { 283 }{ 383 } x$$
Question 9
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Identify the best graph which represents the relationship between the average kinetic energy of the molecules of a gas and its temperature?

A

B

C

D

E

Solution

Average kinetic energy of gas molecules $$K = \dfrac{3}{2}k_B T$$ $$\implies K \propto T$$
Thus K-T graph is a straight line having a finite slope and passing through origin.
Hence option B is correct.
Question 10
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For polyatomic molecules having 'f' vibrational modes, the ratio of two specific heats, $$\dfrac{C_p}{C_v}$$ is ............

A
$$\dfrac{1+f}{2+f}$$

B
$$\dfrac{2+f}{3+f}$$

C
$$\dfrac{4+f}{3+f}$$

D
$$\dfrac{5+f}{4+f}$$

Solution

By the law of equipartition of energy, for one mole of polyatomic gas
$$C_p=(4+f)R \ and \ C_v=(3+f)R$$
$$\therefore \dfrac{C_p}{C_v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$$