Kinetic Theory Test

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Kinetic Theory Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Select the correct formula :
(where k=Boltzmann's constant, R= gas constant, n= moles, r = density, M= molecular weight, p= pressure, T= kelvin temperature, V= volume)
a) k=RN$$_{av}$$
b)$$r=\dfrac{nM}{V}$$
c)$$\dfrac{p}{r}=\dfrac{RT}{M}$$
d) R=kN$$_{av}$$

A
a,b,c

B
a,b,d

C
b,c,d

D
a,c,d

Solution

(a) $$k= R/N_{av}$$
(b) r is density it is clearly satisfying the relation. nM gives toatal mass and when divided by volume it gives density of the substance
(c)We know that , PV=nRT
PV=wRT/M
PV/w=RT/M
P/r=RT/M
(d)$$k=R/N_{av}$$ therefore, $$R=kN_{av}$$
hence (b) (c) (d) are correct
Hence option(C)
Question 2
1 / -0

In the equation PV=constant, the numerical value of constant depends upon
a) temperature b) mass of the gas
c) system of units used d) nature of the gas

A
a & b

B
b & c

C
c & d

D
all

Solution

Since PV=nRT
Therefore value of PV depends on temperature, mass of gas , nature of gas and also system of units used
Question 3
1 / -0

Two containers of equal volume containing the same gas at pressure $$P_{1}$$ and $$P_{2}$$ and absolute temperature $$T_{1}$$ and $$T_{2}$$ respectively were connected with narrow capillary tube. The gas reaches a common pressure P and a common temperature T. The ratio P/T is equal to :

A
$$\dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}$$

B
$$\dfrac{1}{2}\left ( \dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}\right )$$

C
$$\dfrac{P_{1}T_{2}+P_{2}T_{1}}{T_{1}+T_{2}}$$

D
$$\dfrac{P_{1}T_{2}-P_{2}T_{1}}{T_{1}-T_{2}}$$

Solution

Since the mass remains constant before and after opening the tube, moles of the gases remain same.
Thus $$\dfrac{P_1V}{RT_1}+\dfrac{P_2V}{RT_2}=\dfrac{P(2V)}{RT}$$
$$\implies \dfrac{P}{T}=\dfrac{1}{2}(\dfrac{P_1}{T_1}+\dfrac{P_2}{T_2})$$
Question 4
1 / -0

The value of $$\gamma$$ for gas X is 1.66, then x is :

A
Ne

B
O$$_3$$

C
N$$_2$$

D
H$$_2$$

Solution

Given that value of gamma is 1.66 i.e $$\dfrac{5}{3}$$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.
Question 5
1 / -0

According to the Boltzmann's law of equipartition of energy, the energy per degree of freedom and at a temperature T K is :

A
(3/2) KT

B
(2/3) KT

C
KT

D
1/2 KT

Solution

According to the Boltzmann's law of equipartition of energy, the energy per degree of freedom and at a temperature T (in kelvin ) is $$\dfrac{1}{2}$$ KT

where K=Boltzmann's constant and
T= temperature (in kelvin )
Question 6
1 / -0

When the pressure of a gas is changed, then:

A
The density of the gas also changes.

B
The ratio of the pressure to the density remains unaffected.

C
The velocity of the sound remains unaffected.

D
The value of $$\gamma$$ changes.

Solution

When the pressure of a gas is changed, then the ratio of the pressure to the density remains unaffected as $$\displaystyle = \frac{P}{\rho} = \frac{RT}{M}$$
Question 7
1 / -0

A system consists of N particles, which have independent K relations among one another. The number of degrees of freedom of the system is given by :

A
3 NK

B
N/3K

C
3 N/K

D
3N - K

Solution

For a system of N particles having K independent relations among them, the degrees of freedom of the system is given by 3N-K. 3N is due to three degrees of freedom associated with each particle if all the particles are independent of each other (i.e K=0) and due to K relation among them, degrees of freedom reduces to 3N-K
Hence, Option D is correct.
Question 8
1 / -0

The value of C$$_v$$ for 1 mol of polyatomic gas is (F $$=$$ number of degrees of freedom) :

A
$$\displaystyle \frac{fR}{2T}$$

B
$$\displaystyle \frac{fR}{2}$$

C
$$\displaystyle \frac{fRT}{2}$$

D
$$2fRT$$

Solution

The value of C$$_{v}$$ for 1 mol of polyatomic gas is $$\dfrac{fR}{2}$$
where f is no of degrees of freedom (in a gas for every degree of freedom we have energy distribution as $$\dfrac{kT}{2}$$)
Question 9
1 / -0

A man is climbing up a spiral type staircase. His degrees of freedom are :

A
1

B
2

C
3

D
more than 3

Solution

There will be three degrees of freedom. Two are along x-direction and y-directions due to translation and the last degree of freedom due to angular rotation as the man climbs up.
Hence, Option C is correct.
Question 10
1 / -0

The law of equipartition of energy was given by :

A
Claussius

B
Maxwell

C
Boltzmann

D
Carnot

Solution

The law is given by Claussius which states that for any dynamical system in a thermal equilibrium, the total energy is equally divided among the degrees of freedom.

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Kinetic Theory Test - 13

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Kinetic Theory Test - 13

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Question 1

a,b,c

a,b,d

b,c,d

a,c,d

Solution

Question 2

a & b

b & c

c & d

all

Solution

Question 3

$$\dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}$$

$$\dfrac{1}{2}\left ( \dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}\right )$$

$$\dfrac{P_{1}T_{2}+P_{2}T_{1}}{T_{1}+T_{2}}$$

$$\dfrac{P_{1}T_{2}-P_{2}T_{1}}{T_{1}-T_{2}}$$

Solution

Question 4

Ne

O$$_3$$

N$$_2$$

H$$_2$$

Solution

Question 5

(3/2) KT

(2/3) KT

KT

1/2 KT

Solution

Question 6

The density of the gas also changes.

The ratio of the pressure to the density remains unaffected.

The velocity of the sound remains unaffected.

The value of $$\gamma$$ changes.

Solution

Question 7

3 NK

N/3K

3 N/K

3N - K

Solution

Question 8

$$\displaystyle \frac{fR}{2T}$$

$$\displaystyle \frac{fR}{2}$$

$$\displaystyle \frac{fRT}{2}$$

$$2fRT$$

Solution

Question 9

1

2

3

more than 3

Solution

Question 10

Claussius

Maxwell

Boltzmann

Carnot

Solution

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