Kinetic Theory Test

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Kinetic Theory Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

In a given species of tobbaco there is 0.1 mg of virus per c.c. The mass of virus is $$4 \times 10^{7}$$ Kg per kilomol. The number of the molecules of virus present in 1 c.c. will be

A
$$10^{13}$$

B
$$10^{11}$$

C
$$1.5 \times 10^{12}$$

D
$$10^{14}$$

Solution

$${ m }_{ v }=\dfrac { M }{ { N }_{ A } } =\dfrac { 4\times { 10 }^{ 7 } }{ 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } } \\ mass\quad of\quad virus\quad per\quad cc=0.1\times { 10 }^{ -6 }kg\\ $$
Number of molecules in cc
$$ n=\dfrac { { m }_{ v } }{ { N }_{ A } } =\dfrac { 0.1\times { 10 }^{ -6 }\times 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } }{ 4\times { 10 }^{ 7 } } \\ n=1.5\times { 10 }^{ 12 }$$
Question 2
1 / -0

A vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of a vessel. The length of vessel is 90 cm and the piston divides the cylinder in the ratio 2 : 1. Each of the two parts contain 0.1 mole of an ideal gas. The temperature of the gas is $$27^{o}$$C in each part. The mass of the piston is

A
41.5 kg

B
8.97 kg

C
9.57 kg

D
11.57 kg

Solution

$$PV=nRT$$
$$P = nRT/V$$
$$PA=nRT/l$$
$$PA=0.1\times 8.31\times 300/0.6$$
$$PA=41.5 N$$
Therefore option(A) is correct.
Question 3
1 / -0

At what temperature the linear kinetic energy of a gas molecule will be equal to that of an electron accelerated through a potential difference of 10 volt?

A
77.3 x 10$$^{3}$$ K

B
38.65 x 10$$^{3}$$ K

C
19.x10$$^{3}$$ K

D
273 K

Solution

Energy of an electron accelerated through a potential V: $$E=eV$$
Given,
$$E=\dfrac{3}{2}kT$$
$$ \Rightarrow eV = \dfrac{3}{2}kT$$
$$ \Rightarrow T =\dfrac{2eV}{3k} =\dfrac{2e \times 10}{3 \times 1.38 \times 10^{-23}}= \dfrac{20\: eV}{3 \times 1.38 \times 10^{-23}}=\dfrac{20 \times 1.6 \times 10^{-19}J}{3 \times 1.38 \times 10^{-23}}=77.3 \times 10^3\:K$$
Question 4
1 / -0

One gram mol of helium at 27$$^{0}$$ C is mixed with three gram mols of oxygen at 127$$^{0}$$ C at constant pressure. If there is no exchange of heat with the atmosphere then the final temperature will be

A
375 K

B
175K

C
475 K

D
575K

Solution

In this case, N1 = 1, N2 = 3, T1 = 300 and T2 = 400.
Putting these values in the equation T(N1 + N2) = N1T1 + N2T2
we get final temperature T = 375K
Question 5
1 / -0

The mean molecular energy of gas at 300 K will be

A
6.2 x 10$$^{20}$$ Joule

B
6.2 x 10$$^{-20}$$Joule

C
6.2 x 10$$^{-21}$$Joule

D
2.6 x 10$$^{-20}$$Joule

Solution

Average kinetic energy: $$(KE)_{avg}= \frac{3}{2}kT=\frac{3}{2}\times 1.38 \times 10^{-23} \times 300=6.2 \times 10^{-21}J$$ where k is
Boltzman's constant and $$T=(30 + 273)^{\circ}K= 300^{\circ}K$$
Question 6
1 / -0

The degrees of freedom of a diatomic gas at normal temperature is

A
3

B
4

C
5

D
6

Solution

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp, vibration is not possible. Hence, the total no of degrees of freedom is $$f= 3+ 2=5$$
Question 7
1 / -0

Equal masses of $$N_{2}$$ and $$O_{2}$$ gases are filled in vessel A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

A
16:7

B
16:14

C
32:7

D
32:28

Solution

The mathematical form of the Ideal Gas Law is:
PV = nRT and n = m/MW
Where:P - pressure
V - volume
n - number of moles
T - temperature
m - mass
MW - Molecular Weight
R - ideal gas constant
So we can write,
$$\dfrac{P_a}{
P_b}$$ =$$\dfrac{M_b \times V_b}{ M_a \times V_a}$$
here, $$V_b = 2 V_a $$and$$ M_a$$ = atomic wt of $$O_2$$ = 8 and $$M_b$$ = atomic wt of $$N_2$$ = 7
Putting these values we get,
$$P_a/ P_b$$ = 16/7
Question 8
1 / -0

A partially inflated balloon contains 500 m$$^{3}$$ of helium at 27$$^{0}$$ C and 1 atm pressure. The volume of the helium at an altitude of 6000 m, where the pressure is 0.5atm and the temperature is -3$$^{0}$$C is

A
22.4 lit

B
11.2 lit

C
900 m$$^{3}$$

D
50 m$$^{3}$$

Solution

Assuming helium to be an ideal gas, and using relation:
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

$$V_2=V_1\dfrac{P_1T_2}{P_2T_1}=900m^3$$
Question 9
1 / -0

The temperature of gas is increased from 27$$^{o}$$C to 127$$^{o}$$C. The ratio of its mean kinetic energies will be

A
$$\dfrac{3}{4}$$

B
$$\dfrac{4}{3}$$

C
$$\dfrac{9}{16}$$

D
$$\dfrac{16}{9}$$

Solution

We know that , K.E $$\alpha$$ T ; where, T is absolute temperature
So, ratio of kinetic energies = ratio of temperatures (in Kelvin)
= $$\dfrac{300K}{400K} = \dfrac{3}{4}$$
Question 10
1 / -0

Figure shows a cylindrical tube of cross-sectional area A filled with two frictionless pistons. The pistons are connected through wire. The tension in the wire if the temperature rises from T$$_{0}$$ to 2T$$_{0}$$ is (Initial pressure is P$$_{0}$$, atmospheric pressure)

A
$$P_{0}A$$

B
$$\frac{P_{0}A}{2}$$

C
$$2P_{0}A$$

D
$$\frac{2P_{0}A}{3}$$

Solution

Since the wired would be taut, there would be no change in the volume of enclosed gas, so $$\dfrac{P}{T}=constant$$
$$\implies \dfrac{P}{2T_0}=\dfrac{P_0}{T_0}$$
$$\implies P=2P_0$$
Thus, an equilibrium to exist, the difference in inside and outside forces must be balanced by the tension in the wire.
Thus, $$T=2P_0A-P_0A=P_0A$$

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Kinetic Theory Test - 30

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Kinetic Theory Test - 30

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SHARING IS CARING

Question 1

$$10^{13}$$

$$10^{11}$$

$$1.5 \times 10^{12}$$

$$10^{14}$$

Solution

Question 2

41.5 kg

8.97 kg

9.57 kg

11.57 kg

Solution

Question 3

77.3 x 10$$^{3}$$ K

38.65 x 10$$^{3}$$ K

19.x10$$^{3}$$ K

273 K

Solution

Question 4

375 K

175K

475 K

575K

Solution

Question 5

6.2 x 10$$^{20}$$ Joule

6.2 x 10$$^{-20}$$Joule

6.2 x 10$$^{-21}$$Joule

2.6 x 10$$^{-20}$$Joule

Solution

Question 6

3

4

5

6

Solution

Question 7

16:7

16:14

32:7

32:28

Solution

Question 8

22.4 lit

11.2 lit

900 m$$^{3}$$

50 m$$^{3}$$

Solution

Question 9

$$\dfrac{3}{4}$$

$$\dfrac{4}{3}$$

$$\dfrac{9}{16}$$

$$\dfrac{16}{9}$$

Solution

Question 10

$$P_{0}A$$

$$\frac{P_{0}A}{2}$$

$$2P_{0}A$$

$$\frac{2P_{0}A}{3}$$

Solution

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