Kinetic Theory Test

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Kinetic Theory Test - 35

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Question 1
1 / -0

Directions For Questions

It is possible to make crystalline solids that are only one layer of atoms thick. Such two dimensional crystals can be created by depositing atoms on a very flat surface. The atoms in such a two dimensional crystal can move only within the plane of the crystal, At very low temperature the molar heat capacity of such a crystal is C$$_1$$ and at room temperature it is C$$_2$$

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What is C$$_2$$ in terms of R?

A
$$\displaystyle \frac{3}{2} R$$

B
2R

C
$$\displaystyle \frac{5}{2} R$$

D
3R

Solution

Number of degrees of freedom $$= $$ 2 translation + 2 rotation $$=$$ 4
$$\therefore \displaystyle C = \frac{R}{2} $$ (number of degrees of freedom) $$= 2R$$
Question 2
1 / -0

The African bombardier beetle stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen. The beetle's body has reservoirs of two different chemicals. When the beetle is disturbed, these chemicals are combined in a reaction chamber producing a compounds that is warmed from $$20^o$$C to $$100^o$$C by the heat of reaction. The high pressure produced allows the compound to the sprayed out at speeds 19 ms$$^{-1}$$ scaring away predators of all kinds. Assume specific heat of two chemicals and the spray to be same as that of water $$[4.19 \times 10^3 J (kgK)^{-1}]$$ and initial temperature of chemicals to be $$20^o$$C. How many times does the pressure increase?

A
1.5

B
2.3

C
1.9

D
1.28

Solution

$$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$$
or $$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{373}{293} = 1.28$$
Question 3
1 / -0

The correct graph between PV and P of one mol of gas at constant temperature will be

A

B

C

D

Solution

For an ideal gas $$PV=nRT$$
Thus for a constant temperature, $$PV=constant$$
Thus $$PV$$ does not change even when P changes.
Question 4
1 / -0

A gas is filled in a container at any temperature and at pressure 76 cm of Hg. If at the same temperature the mass of gas is increased by 50% then the resultant pressure will be

A
38 cm of Hg

B
76 cm of Hg

C
114 cm of Hg

D
152 cm of Hg

Solution

$$PV=nRT$$
$$PV=\dfrac{m}{M}RT$$
$$P=\dfrac{m}{M}\dfrac{RT}{V}$$ also given that P=76cm of Hg
Given that mass of the gas is increased by 50%
Now $$P_{1}=\left(\dfrac{m+\dfrac{m}{2}}{M}\right)\dfrac{RT}{V}$$
$$=\dfrac{3}{2}\dfrac{m}{M}\dfrac{RT}{V}$$
$$=\dfrac{3}{2}P$$
=$$\dfrac{3}{2}\times76$$
=$$114cm\ of\ Hg$$
Question 5
1 / -0

In the gas equation $$PV= RT$$, V is the volume of

A
1 mol of gas

B
1 g of gas

C
gas

D
1 litre of gas

Solution

For an Ideal Gas, $$PV=nRT$$
Here $$V$$ is the volume of n moles of gas.
Thus for $$PV=(1)RT$$, $$V$$ is the volume of 1 mol of gas.
Question 6
1 / -0

The mean kinetic energy of gas molecules is zero at

A
0$$^o$$C

B
-273$$^o$$C

C
100 K

D
100$$^o$$C

Solution

$$KE_{avg} =\dfrac{3}{2}kT$$ where T is in Kelvin.

$$KE_{avg} =0$$ at $$T=0^{\circ}K=-273^{\circ}C$$
Question 7
1 / -0

If number of gas molecules in a cubical vessel is increased from N to 3N, then its pressure and total energy will become

A
four times

B
three times

C
double

D
half

Solution

For an ideal gas $$ P \times V = N \times K\times T$$

where P is the pressure, V is volume of the cubical vessel,
N is no. of molecules, K Boltzmann's constant and T is temperature

So as N increases to 3N, pressure also increases by three times.

And the total energy for an ideal gas $$ E = \dfrac {3}{2} \times N \times K \times T $$
Hence the total energy also increases by three times.
Question 8
1 / -0

At what temperature will the mean molecular energy of a perfect gas be one-third of its value of 27$$^o$$C?

A
10$$^o$$C

B
10$$^1$$K

C
10$$^2$$K

D
10$$^3$$J

Solution

Mean $$KE \propto T$$
therefore at T=27 which in kelvin becomes 300 K
So KE will be 1/3 when temperature becomes 1/3 which is 300/3=100K
option (C)
Question 9
1 / -0

The pressure of a gas in a container is 10$$^{-11}$$ pascal at 27$$^o$$C. The number of molecules per unit volume of vessel will be

A
$$6 \times 10^{23} cm^{-3}$$

B
$$2.68 \times 10^{19} cm^{-3}$$

C
$$2.5 \times 10^{6} cm^{-3}$$

D
$$2400 cm^{-3}$$

Solution

$$10^5 Pa \equiv 6.023 \times 10^{23}$$ (equivalence equation according to information given in question)
$$10^{-11} Pa \equiv 6.023 \times 10^7$$
$$22400 cc \Rightarrow 6.023 \times 10^7$$
$$1 cc \Rightarrow \displaystyle \frac{6.023 \times 10^7}{2400} =2400 cm^{-3} $$
Question 10
1 / -0

The mean kinetic energy of a gas molecule at 27$$^o$$C is 6.21 $$\times$$ 10$$^{-21}$$ Joule. Its value at 227$$^o$$ C will be

A
$$12.35 \times 10^{-21} Joule$$

B
$$11.35 \times 10^{-21} Joule$$

C
$$10.35 \times 10^{-21} Joule$$

D
$$9.35 \times 10^{-21} Joule$$

Solution

Mean KE=$$\dfrac{3kT}{2}$$
so $$6.21\times 10^{-21}=3k\times 300/2$$
so $$k=1.38\times 10^{-23}$$
KE at 500 K ,$$KE=3\times 1.38\times 10^{-23}\times 500/2=10.35\times 10^{-21}\ Joule$$

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Kinetic Theory Test - 35

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Kinetic Theory Test - 35

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Question 1

$$\displaystyle \frac{3}{2} R$$

2R

$$\displaystyle \frac{5}{2} R$$

3R

Solution

Question 2

1.5

2.3

1.9

1.28

Solution

Question 3

Solution

Question 4

38 cm of Hg

76 cm of Hg

114 cm of Hg

152 cm of Hg

Solution

Question 5

1 mol of gas

1 g of gas

gas

1 litre of gas

Solution

Question 6

0$$^o$$C

-273$$^o$$C

100 K

100$$^o$$C

Solution

Question 7

four times

three times

double

half

Solution

Question 8

10$$^o$$C

10$$^1$$K

10$$^2$$K

10$$^3$$J

Solution

Question 9

$$6 \times 10^{23} cm^{-3}$$

$$2.68 \times 10^{19} cm^{-3}$$

$$2.5 \times 10^{6} cm^{-3}$$

$$2400 cm^{-3}$$

Solution

Question 10

$$12.35 \times 10^{-21} Joule$$

$$11.35 \times 10^{-21} Joule$$

$$10.35 \times 10^{-21} Joule$$

$$9.35 \times 10^{-21} Joule$$

Solution

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