Kinetic Theory Test

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Kinetic Theory Test - 37

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Question 1
1 / -0

One mole of a gas at a pressure 2 Pa and temperature 27$$^o$$C is heated till both pressure and volume are doubled. What is the temperature of the gas?

A
1200 K

B
900 K

C
600 K

D
300 K

Solution

We have the relation PV=nRT (ideal gas equation)

T=$$\dfrac{PV}{nR}$$

Given, that pressure and volume are doubled i.e, $$P_1=2P$$,$$V_1=2V$$.

$$T_1=\dfrac{P_1V_1}{nR}$$

$$=\dfrac{\left( 2P \times 2V\right)}{nR}$$

$$=4T$$

Given that T=27$$^o$$C i.e, 300K

$$T_1=4\times 300$$

$$=1200K$$
Question 2
1 / -0

What is number of degrees of freedom of an ideal diatomic molecule at ordinary temperature?

A
7

B
6

C
5

D
3

Solution

The degrees of freedom of an ideal diatomic molecule at ordinary temperature is 5.
Question 3
1 / -0

The internal energy of air in a room of volume 50 m$$^3$$ at atmospheric pressure will be

A
2.5 X 10$$^7$$ erg

B
2.5 X 10$$^7$$ Joule

C
5.255 X 10$$^7$$ Joule

D
1.25 X 10$$^7$$ Joule

Solution

The internal energy of the room is given by the molecular motion and is called as the kinetic internal energy given by $$\displaystyle\frac{f}{2}nRT$$. Using the gas law we have $$PV=nRT$$ and as the gas is mainly comprised of nitrogen and oxygen both being diatomic we take the degree of freedom as 5. Thus we calculate internal energy as $$\displaystyle\frac{5}{2}PV=\frac{5}{2}\times 1.013\times 10^5\times 50=1.25\times 10^7 J$$
Question 4
1 / -0

If the value of R = 2/5 C$$_v$$ for a gas, then the atomicity of the gas will be:

A
monoatomic

B
diatomic

C
polyatomic

D
any of these

Solution

We have $$C_v=\displaystyle\frac{5}{2}R$$ which is the $$C_v$$ value for diatomic molecules.
Question 5
1 / -0

Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?

A
Kinetic energy

B
Momentum

C
Density

D
Speed

Solution

In case of ideal gases the average velocity is always zero. Hence the average momentum is zero.
Whereas average speed is non- zero so the kinetic energy also non-zero, as these two are scalar quantities.
Question 6
1 / -0

The total kinetic energy of 8 litres of helium molecules at 5 atmosphere pressure will be

A
$$6078 \ erg$$

B
$$6078 \ Joule$$

C
$$607 \ erg$$

D
$$607 \ Joule$$
Question 7
1 / -0

The Avogadro's number when to tends to infinity, the phenomenon of Brownian motion would

A
remain completely unaffected.

B
become more vigorous, than that observed with present finite value of Avogadro's number, for all sizes of the Brownian particles.

C
becomes more vigorous, than that observed with the present finite value of Avogadro's number, only for relatively large Brownian particle.

D
become practically unobservable as the molecular impact would tend to balance one another, for practically.

Solution

Brownian motion is caused due to the scattering of particles resulting from the collisions between different particles in a fluid and hence has no relation to the Avogadro's number.

Hence, answer is option A.
Question 8
1 / -0

A product from a chemical industry passes through three states - gas, liquid, solid. The product is initially formed in gaseous state which is then liquefied and finally solidified. For same mass (say 5g) which state has maximum internal energy?

A
solid

B
liquid

C
gas

D
all have equal internal energy

Solution

Internal energy is dependent on the freedom of movement of particles. Since in gaseous state, particles are most free to move, this state has maximum internal energy.
Question 9
1 / -0

The temperature of an ideal gas is increased from 27$$^o$$C to 927$$^o$$C. Then, the mean kinetic energy of gas molecules :

A
becomes $$\displaystyle \frac{927}{27}$$ times

B
becomes $$\displaystyle \sqrt{\frac{927}{27}}$$ times

C
become double

D
become four times

Solution

$$T_{1}=27^{0}C=300K$$
$$T_{2}=927^{0}C=1200K$$

$$V_{rms}$$( root mean square velocity )$$=\sqrt\frac{3RT}{M}$$

$$KE_{1}=\dfrac{1}{2}\times{m}\times{V_{rms}^{2}}$$

$$KE_{2}=\dfrac{1}{2}\times{m}\times{ \dfrac{3R\times{1200}}{m} }$$

$$=4KE_{1}$$
Question 10
1 / -0

In two vessels of the same volume, atomic hydrogen and helium with pressure 1 atm and 2 atm are filled. If temperature of both the same is the same, then the average speed of hydrogen atom $$v_H$$ will be related to helium $$v_{He}$$ as

A
$$v_{H}$$ $$= \sqrt{2}$$ $$v_{He}$$

B
$$v_H$$ $$=$$ $$v_{He}$$

C
$$v_H$$ $$=$$ 2$$v_{He}$$

D
$$v_H$$ $$=$$ $$\dfrac{v_{He}}{2}$$

Solution

By Maxwell's speed distribution, $$<v>\alpha \sqrt { \dfrac { RT }{ M } } $$. Since the temperature of two gases is same, hence

$$<v>\alpha \sqrt { \dfrac { 1 }{ M } } $$

Also, $${ M }_{ He }=4{ M }_{ H }$$

Hence, $$<{ v }_{ H }>=2<{ v }_{ He }>$$

Answer is option C.

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Kinetic Theory Test - 37

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Kinetic Theory Test - 37

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Question 1

1200 K

900 K

600 K

300 K

Solution

Question 2

7

6

5

3

Solution

Question 3

2.5 X 10$$^7$$ erg

2.5 X 10$$^7$$ Joule

5.255 X 10$$^7$$ Joule

1.25 X 10$$^7$$ Joule

Solution

Question 4

monoatomic

diatomic

polyatomic

any of these

Solution

Question 5

Kinetic energy

Momentum

Density

Speed

Solution

Question 6

$$6078 \ erg$$

$$6078 \ Joule$$

$$607 \ erg$$

$$607 \ Joule$$

Question 7

remain completely unaffected.

become more vigorous, than that observed with present finite value of Avogadro's number, for all sizes of the Brownian particles.

becomes more vigorous, than that observed with the present finite value of Avogadro's number, only for relatively large Brownian particle.

become practically unobservable as the molecular impact would tend to balance one another, for practically.

Solution

Question 8

solid

liquid

gas

all have equal internal energy

Solution

Question 9

becomes $$\displaystyle \frac{927}{27}$$ times

becomes $$\displaystyle \sqrt{\frac{927}{27}}$$ times

become double

become four times

Solution

Question 10

$$v_{H}$$ $$= \sqrt{2}$$ $$v_{He}$$

$$v_H$$ $$=$$ $$v_{He}$$

$$v_H$$ $$=$$ 2$$v_{He}$$

$$v_H$$ $$=$$ $$\dfrac{v_{He}}{2}$$

Solution

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