Kinetic Theory Test

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Kinetic Theory Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The volume of a gas at $$27^o$$C and 2 atmospheric pressure is 2 litres. If the pressure is doubled and absolute temperature is made half, the new volume of gas is

A
50 ml

B
500 ml

C
1000 ml

D
2000 ml

Solution

from ideal gas equation, PV = nRT
So, this gives $$V = \frac{nRT}{P}$$
so, V is proportional to absolute temperature and inversely proportional to the pressure of the gas
Now temperature is made half and pressure is made 2 times
So, clearly the volume becomes one-fourth of it's original value.
So, B is the correct answer.
Question 2
1 / -0

A closed vessesl contains 8 gms of oxygen and 7gm of Nitrogen. Total pressure at a certain temperature is 10 atm. When all the oxygen is removed from the system without chage in temperature then the pressure will be

A
10 x 7/15atm

B
10 x 8/15atm

C
10 x 8/16 atm

D
10 x 8/32 atm

Solution

$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { m }_{ 1 }{ M }_{ 2 } }{ { m }_{ 2 }{ M }_{ 1 } } \\ \\ \dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { 8\times 14 }{ 7\times 32 } \\ { P }_{ 1 }=10\times \dfrac { 8 }{ 16 } $$
Question 3
1 / -0

The volume occupied by 8 gm of oxygen at S.T.P. is

A
11.2 lit

B
22.4 lit

C
2.8 lit

D
5.6 lit

Solution

32 gm of oxygen = 1 mole = 22.4lit (standard result according to gas governing equation)
8 gm of oxygen = 1/4 mole = 22.4/4 = 5.6lit
Question 4
1 / -0

An air bubble rises from bottom to top of a liquid of density 1.5 g/cm$$^{3}$$. If its volume is doubled, the depth of liquids is

A
689 cm

B
398 cm

C
760 cm

D
152 cm

Solution

In this case, $$(P + \rho g h) V$$ $$=$$ $$2V P$$
where, $$P$$ $$=$$ atm pressure
$$\rho$$ $$=$$ density of the liquid
$$h$$ $$=$$ height of the lake
$$V$$ $$=$$ volume of the bubble
Solving we get,
$$h$$ $$=$$ $$\dfrac{P}{\rho g}$$
here, $$P$$ $$=$$ 76 X 13.6 X 9.8
$$\rho$$ $$=$$ 1.5 $$gm/cm^{3}$$ $$=$$ 1500 $$kg/m^{3}$$
Putting these values we get, $$h$$ $$=$$ 689 cm
Question 5
1 / -0

If pressure of an ideal gas contains in a closed vessel is increased by 0.5% the increase in temperature is 20k the initial temperature of the gas is

A
27$$^{0}$$C

B
127$$^{0}$$C

C
300$$^{0}$$C

D
400$$^{0}$$C

Solution

In this case,
$$P_1 = P$$, $$P_2= 1.05P$$
$${T_1} = T$$, $$T_2 = T+20$$
Putting these values in $$\dfrac{P_1}{T_1}= \dfrac{P_2} {T_2}$$ we get, T = 400 or T = $$127^o$$C
Question 6
1 / -0

Two closed vessels of the equal volume contain air at 105kPa at 300K and are connected through a narrow tube. If one of the vessels is now maintained at 300K and the other at 400K then the pressure becomes.

A
120kPa

B
105kPa

C
150kPa

D
300kPa

Solution

In this case, let the initial pressure = P and final pressure = P'. Since (P/T) = constant, considering the system of two spheres of equal volume, we get,
$$\dfrac{P}{300} + \dfrac{P}{300} = \dfrac{P'}{300} + \dfrac{P'}{400}$$
here, P = 105k Pa
Solving we get, $$P' = 120 kPa$$
Question 7
1 / -0

At the top of a mountain a thermo meter read 7$$^{0}$$ C and barometer reads 70 cm of Hg. At the bottom of the mountain the barometer reads 76cm of Hg and thermometer reads 27$$^{0}$$ C. The density of air at the top of mountains is ______ times the density at the bottom.

A
0.99

B
0.9

C
0.89

D
0.95

Solution

$$\dfrac{P_1}{d_1T_1}=\dfrac{P_2}{d_2T_2}$$ (Standard equation of governance)
Dirctly putting up the value,
$$ \therefore \dfrac{70}{d_1\times ( 7+273)} = \dfrac{76}{d_2\times(27+ 273)}$$
$$\dfrac{d_1}{d_2} = \cfrac{75}{76}=0.987$$
Question 8
1 / -0

At a given temperature and pressure 64 gm of Oxygen and X gm of H$$_{2}$$ occupy the same volume. Then x= ......gm

A
1

B
2

C
3

D
4

Solution

Since they occupy same volume, the two gases should contain same number of molecules, i.e. they have same number of moles.
Now 1 mole of oxygen = 32gm.
Hence 64 gm = 2 mole of oxygen.
$$\therefore $$ number of moles of hydrogen is 2.
1 mole of hydrogen= 2gm
Hence X= 4gm
Question 9
1 / -0

A one litre sphere and a two litre sphere are connected with a capillary tube of negligible volume. They contain an ideal gas at 27$$^{0}$$C at a pressure of 100cm of Hg. Keeping the temperature of one litre sphere constant at 27$$^{0}$$C, if temperature of two litre sphere is increased to 127$$^{0}$$C, then the final pressure is

A
110 cm of Hg

B
120 cm of Hg

C
150 cm of Hg

D
200 cm of Hg

Solution

Moles of gas in 1L sphere before increasing temperature $$=\dfrac{P_0 V}{RT}$$

Moles of gas in 2L sphere before increasing temperature $$=\dfrac{P_0 (2V)}{RT}$$

Moles of gas in 1L sphere after increasing temperature $$=\dfrac{PV}{RT}$$
Moles of gas in 2L sphere after increasing temperature $$=\dfrac{P (2V)}{R(T+100)}$$
Since the total number of moles remain the same,
$$\dfrac{P_0V}{RT}+\dfrac{2P_0 V}{RT}=\dfrac{PV}{RT}+\dfrac{2PV}{R(T+100)}$$
Hence, $$P=P_0 (\dfrac{3(T+100)}{3T+100})$$
$$T=300K$$, $$P_0=100cm Hg$$
$$\implies P=120cm Hg$$
Question 10
1 / -0

At the bottom of a lake where the temperature is $${7}^{0}\,C$$ the pressure is $$2.8$$ atmosphere. An air bubble of radius $$1\, cm$$ at the bottom rises to the surface. Where the temperature is $${27}^{0}\, C$$. The radius of the air bubble at the surface is

A
$$3^{1/3}$$

B
$$4^{1/3}$$

C
$$5^{1/3}$$

D
$$6^{1/3}$$

Solution

We know that, by gas law
$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } $$
$$\dfrac { 2.8{ \times V }_{ 1 } }{ 280 } =\dfrac { 1\times { V }_{ 2 } }{ 300 } $$
$${ V }_{ 2 }=3{ V }_{ 1 }$$
$$R\propto { V }^{ \dfrac { 1 }{ 3 } }$$
$$\therefore { R }_{ 2 }={ { 3 }^{ \dfrac { 1 }{ 3 } }R }_{ 1 }$$

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Kinetic Theory Test - 26

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Kinetic Theory Test - 26

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Question 1

50 ml

500 ml

1000 ml

2000 ml

Solution

Question 2

10 x 7/15atm

10 x 8/15atm

10 x 8/16 atm

10 x 8/32 atm

Solution

Question 3

11.2 lit

22.4 lit

2.8 lit

5.6 lit

Solution

Question 4

689 cm

398 cm

760 cm

152 cm

Solution

Question 5

27$$^{0}$$C

127$$^{0}$$C

300$$^{0}$$C

400$$^{0}$$C

Solution

Question 6

120kPa

105kPa

150kPa

300kPa

Solution

Question 7

0.99

0.9

0.89

0.95

Solution

Question 8

1

2

3

4

Solution

Question 9

110 cm of Hg

120 cm of Hg

150 cm of Hg

200 cm of Hg

Solution

Question 10

$$3^{1/3}$$

$$4^{1/3}$$

$$5^{1/3}$$

$$6^{1/3}$$

Solution

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