Kinetic Theory Test

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Kinetic Theory Test - 76

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Weekly Quiz Competition

Question 1
1 / -0

In the figure, an ideal gas is expanded from volume $$V_0$$ to $$2V_0$$ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let $$\Delta U_1, \Delta U_2$$ and $$\Delta U_3$$ be the change in the internal energy of the gas is these three processes, Then:

A
$$\Delta U_1 > \Delta U_2 > \Delta U_3$$

B
$$\Delta U_1 < \Delta U_2 < \Delta U_3$$

C
$$\Delta U_2 < \Delta U_1 < \Delta U_3$$

D
$$\Delta U_2 < \Delta U_3 < \Delta U_1$$

Solution

According to first law of thermodynamic $$dQ=dU+P\triangle V,$$

Isobaric                                    Isothermal                                 Adiabatic
Pressure is constant.                $$\triangle T=0$$                    $$dQ=0$$
So, $$dQ-P\triangle V=\triangle V_1$$        $$\triangle U_2=0$$                                $$-P\triangle V=\triangle U_3$$

So, $$\triangle U_1>\triangle U_2>\triangle U_3$$
Hence, the answer is $$\triangle U_1>\triangle U_2>\triangle U_3.$$
Question 2
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Among the following, identify the substance in which molecules possess vibratory, rotatory and translatory motion, but their movements are not random.

A
Bromine

B
Iodine

C
Ammonia

D
Silicon dioxide

Solution
Question 3
1 / -0

Name $$A, B, C, D,E,$$ and $$F$$ in the following diagram showing change of state

A
A $$=$$ Melting, B$$=$$ Vaporisation, C $$=$$ Condensation, D $$=$$ Solidification, E $$=$$ Sublimation, F $$=$$ Solidification of gaseous state

B
A $$=$$ Sublimation, B $$=$$ Vaporisation, C $$=$$ Condensation, D $$=$$ Solidification, E $$=$$ Melting, F $$=$$ Solidification of gaseous state

C
A $$=$$ Condensation, B $$=$$ Vaporisation, C $$=$$ Melting, D $$=$$ Solidification, E $$=$$ Sublimation, F $$=$$ Solidification of gaseous state

D
A $$=$$ Vaporisation, B $$=$$ Melting, C $$=$$ Condensation, D $$=$$ Solidification, E $$=$$ Sublimation, F $$=$$ Solidification of gaseous state

Solution

A shows that solid is converted into liquid that is known as Melting.
B shows that liquid is converted into gas that is known as vaporization.
C shows that gas is converted into liquid that is known as condensation.
D shows that liquid is converted into solid that is known as solidification.
E shows that solid is converted into gas that is known as sublimation.
F shows that gas is converted into solid that is known as solidification of gaseous state.
Question 4
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If one mole of a monoatomic gas $$\left (\gamma = \dfrac {5}{3}\right )$$ is mixed with one mole of diatomic gas $$\left (\gamma = \dfrac {7}{5}\right )$$, the value of $$\gamma$$ for the mixture is :

A
$$1.40$$

B
$$1.50$$

C
$$1.53$$

D
$$3.07$$

Solution

For monoatomic gas :

$$\displaystyle C_v = \dfrac {3}{2}R$$

$$\displaystyle C_p = \dfrac {5}{2}R$$

For diatomic gas :

$$\displaystyle C_v = \dfrac {5}{2}R$$

$$\displaystyle C_p = \dfrac {7}{2}R$$

For mixture of 1 mole of each :

$$\displaystyle C_v = \dfrac {\dfrac {3}{2}R+\dfrac {5}{2}R}{2}=2R$$

$$\displaystyle C_p = \dfrac {\dfrac {5}{2}R+\dfrac {7}{2}R}{2}=3R$$

$$\displaystyle \gamma = \dfrac {C_p}{C_v}$$

$$\displaystyle \gamma = \dfrac {3R}{2R}$$

$$\displaystyle \gamma = 1.50$$

Hence, the correct option is B.
Question 5
1 / -0

According to the Kinetic Molecular Theory, temperature is directly proportional to:

A
average kinetic energy

B
lattice energy

C
ionization energy

D
free energy

E
average potential energy

Solution

The average kinetic energy of the particles in a gas is proportional to the temperature of the gas.
Reason - As the mass of these particles is constant, if the gas is confined in a container, the particles must move faster as the gas becomes warmer (to exert pressure proportional to the increasing temperature). This increase in velocity marks the increase in kinetic energy.
Question 6
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A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of $$30^\circ$$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall 2. ( $$F_2$$ ) provided to the molecule during collision satisfy :

A
$$F_1\, >\, F_2$$

B
$$F_1\, <\, F_2$$

C
$$F_1\, =\,F_2$$, bothnon-zero

D
$$F_1\, =\, F_2\, =\, 0$$

Solution

Work done by the wall during a collision is equal to the change in momentum of the collided particle per unit time of collision.
Let the collision time in both cases be $$t$$, the mass of the particle be $$m$$ and the speed be $$u$$.
Change in momentum for the collision at Wall1 is $$\Delta p_1 = mu-(-mu) = 2mu$$
The final momentum of the gas molecule after its collision with wall 2 makes $$120^o$$ angle with the momentum just before its collision with wall 2.
Change in momentum for the collision at Wall2 is $$\Delta p_2 = \sqrt{(mu)^2+(mu)^2-2(mu)(mu)\cos(120^\circ)} = \sqrt{3}mu$$
Force is given by $$F = \dfrac{p}{t}$$
Thus, $$F_1 = 2mu/t$$ and $$F_2 = \sqrt{3}mu/t$$.
Hence, $$F_1>F_2$$
Question 7
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A gaseous mixture enclosed in a vessel consists of one g mole of a gas A with $$\displaystyle \gamma =\left ( \frac{5}{3} \right ) $$ and some amount of gas B with $$\displaystyle \gamma = \frac{7}{5} $$ at a temperature The gasses A and B do not react with each other and are assumed to be ideal Find the number of g moles of the gas B if $$\displaystyle \gamma $$ for the gaseous mixture is $$\displaystyle \left ( \frac{19}{13} \right ) $$

A
2

B
5

C
4

D
3

Solution

Change in internal energy of the first gas + change in internal energy of the second gas = change in internal energy of the mixture

or $$\mu_1 (C_{v})_{1}\Delta T+\mu _{2}(C_{v})_{2} \Delta T=(\mu _{1}+\mu _{2})(C_{v})_{mix}\Delta T$$

or $$\mu _{1}\dfrac{R}{\gamma _{1}-1}+\mu _{2}\dfrac{R}{\gamma _{2}-1}=(\mu _{1}+\mu _{2})\dfrac{R}{\gamma _{mix}-1}; \, \therefore\, \gamma_{mix}=\dfrac{\mu _{1}\gamma _{1}(\gamma _{2}-1)+\mu _{2}\gamma _{2}(\gamma _{1}-1)}{\mu _{1}(\gamma _{2}-1)+\mu _{2}(\gamma _{1}-1)}$$

or $$\dfrac{19}{13}=\dfrac{1\times (5/3)(7/5-1)+\mu _{2}\times (7/5)(5/3-1)}{1(7/5-1)+\mu _{2}(5/3-1)}$$ or $$\mu _{2}=2$$
Question 8
1 / -0

If the radii of two copper spheres are in the ratio $$1 : 3$$ and increase in their temperatures are in the ratio $$9 : 1$$ then the ratio of the increase in their internal energy will be

A
$$1 : 4$$

B
$$2 : 1$$

C
$$4 : 1$$

D
$$1 : 3$$

Solution

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Question 9
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Relation between pressure ($$P$$) and energy density ($$E$$) of an ideal gas is-

A
$$P=2/3E$$

B
$$P=3/2E$$

C
$$P=3/5E$$

D
$$P=E$$

Solution

Kinetic energy $$=\dfrac{1}{2}{ MV }_{ rms }$$
$$\Rightarrow \dfrac{1}{2}M\left( \dfrac { 3RT }{ M } \right) $$        $$[M=$$ molar mass,$$ { V }_{ rms }=\sqrt { \dfrac { 3KT }{ { m } } } =\sqrt { \dfrac { 3RT }{ M } } ]$$
$$=\dfrac{3}{2}RT$$
$$\Rightarrow K.E=\dfrac{3}{2}PV$$          $$[PV=RT]$$
$$\Rightarrow \dfrac{K.E}{V}=\dfrac{3}{2}P$$
$$\Rightarrow E=\dfrac{3P}{2}$$        $$E=$$ Energy density.
Hence, the answer is $$P=\dfrac{2}{3}E.$$
Question 10
1 / -0

An insulated container is divided into two equal portions. One portion contains an ideal monoatomic gas at pressure $$P$$ and temperature $$T$$, while the other portion is a perfect vacuum. If a hole is opened between the two portions, the change in internal energy of the gas is

A
Zero

B
Equal to work done by the gas

C
Equal to work done on the gas

D
$$3RT/2$$

Solution

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