Kinetic Theory Test

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Kinetic Theory Test - 28

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Question 1
1 / -0

A vessel is filled with an ideal gas at a pressure of 10 atmospheres and temp 27$$^{0}$$C . Half of the mass of the gas is removed from the vessel & the temp. of the remaining gas is increased to 87$$^{0}$$C . Then the pressure of the gas in the vessel will be

A
5 atm

B
6 atm

C
7 atm

D
8 atm

Solution

We have Ideal gas Equation: $$PV=nRT$$
First the mass is halved, thus the moles of gas are halved.
$$P\propto n$$
So, $$P_{1}=\dfrac{P_0}{2}=5atm$$
Now the temperature is increased.
$$P\propto T$$
$$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}=\dfrac{360}{300}=\dfrac{6}{5}$$
$$\implies P_2=6atm$$
Question 2
1 / -0

2 grams of monoatomic gas occupies a volume of 2 litres at a pressure of $$8.3\times 10^{5}Pa$$ and 127$$^{0}$$C . Find the molecular weight of the gas. (R=8.3 joule/mole/K)

A
2 gram/mole

B
16 gram/mole

C
4 gram/mole

D
32 gram/mole

Solution

$$PV=\dfrac { m }{ M } RT\\ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 }=\dfrac { 2\times { 10 }^{ 3 } }{ M } \times 8.3\times 400\\ M=\dfrac { 2\times { 10 }^{ 3 }\times 8.3\times 400 }{ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 } } \\ M=4{ gram }/{ mole }$$
Question 3
1 / -0

From a disc of radius $$R$$ and mass $$M$$, a circular hole of diameter $$R$$, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

A
$$\dfrac{9MR^{2}}{32}$$

B
$$\dfrac{15MR^{2}}{32}$$

C
$$\dfrac{13MR^{2}}{32}$$

D
$$\dfrac{11MR^{2}}{32}$$

Solution

$$PV=\dfrac { m }{ M } RT\\ PV=\dfrac { 8 }{ 32 } RT\\ PV=\dfrac { RT }{ 4 } $$
Question 4
1 / -0

A reciever has a pressure of 144cm of Hg. After two strokes with an exhaust pump, the pressure is 36cm of Hg. After another two strokes the pressure will be.

A
9cm of Hg

B
2.4cm of Hg

C
6cm of Hg

D
3cm of Hg

Solution

$${ P }_{ n }=P{ x }^{ n }\\ For\quad two\quad strokes,\\ 36=144{ x }^{ 2 }\\ { x }^{ 2 }=\frac { 1 }{ 4 } \\ \therefore x=\frac { 1 }{ 2 } \\ { { P }_{ n } }^{ \backprime }=144{ x }^{ 4 }\\ { { P }_{ n } }^{ \backprime }=144{ \left( \frac { 1 }{ 2 } \right) }^{ 4 }=9cm\quad of\quad Hg$$
Question 5
1 / -0

Three flasks of identical volume are filled separately by
(a) 1 gram of $$H_{2}$$
(b) 1 gram of $$O_{2}$$
(c) 1 gram of $$CO_{2}$$
They are immersed in a tank of water so that all of them attain same temperature. The pressures $$P_{1},P_{2}$$ and $$P_{3}$$ have the on.

A
$$P_{1} > P_{2} > P_{3}$$

B
$$P_{1} = P_{2} = P_{3}$$

C
$$P_{1} < P_{2} < P_{3}$$

D
$$P_{3} > P_{1} > P_{2}$$

Solution

We know, $$PV=\dfrac { m }{ M } RT$$
Given the flasks have same volume, and mass m=1gm for all the gases, the temperature attained is also same.
Hence the equation can be reduced to,
PM=Constant, where M is the molecular weight.
since $${ M }_{ 1 }{ <M }_{ 2 }{ <M }_{ 3 }\\ { \therefore P }_{ 1 }{ >P }_{ 2 }{ >P }_{ 3 }$$
Question 6
1 / -0

A closed copper vessel contains water equal to half of its volume when the temperature. Of the vessel is raised to 447$$^{0}$$c the pressure of steam in the vessel is (Treat steam as an ideal gas, R=8310 J/k / mole, density of water =1000 kg/m$$^{3}$$ molecular weight of water =18 )

A
33.24 x 10$$^{7}$$ pa

B
16.62 x 10$$^{7}$$ pa

C
10.31 x 10$$^{7}$$ pa

D
8.31x 10$$^{7}$$ pa

Solution

Let V be the volume of the copper vessel.
Then considering steam as an ideal gas,
$$PV=nRT=\dfrac{m}{M}RT$$
Mass of the water in the vessel=$$m=\rho (\dfrac{V}{2})$$
$$\implies PV=\dfrac{\rho V}{2M}RT$$

$$\implies P=\dfrac{\rho RT}{2M}=16.62\times 10^7Pa$$
Question 7
1 / -0

A sample of $$O_{2}$$ gas and a sample of hydrogen gas both have the same no. of moles, the same volume and same pressure. Assuming them to be perfect gases, the ratio of temperature of oxygen gas to the temperature of hydrogen gas is :

A
1:1

B
1:2

C
2:1

D
1:4

Solution

We know, PV=nRT
since P, V , n and R are same the ratio of temperatures is also same.
Hence the ratio is 1:1
Question 8
1 / -0

The density of a gas at N.T.P. is 1.5gm/lit. its density at a pressure of 152cm of Hg and temperature 27$$^{0}$$ C

A
$$\frac{273}{100}gm/ltr$$

B
$$\frac{150}{273}gm/ltr$$

C
$$\frac{1}{273}gm/ltr$$

D
1.5 $$gm/lit$$

Solution

Since volume is inversely proportional to density,
$$\dfrac { { P }_{ 1 } }{ { { d }_{ 1 }T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { { d }_{ 2 }T }_{ 2 } } \\ \dfrac { 760 }{ 1.5\times 273 } =\dfrac { 152\times { 10 } }{ { d }_{ 2 }\times 300 } \\ { d }_{ 2 }=\dfrac { 273 }{ 100 }gm/lit $$
Question 9
1 / -0

An electric bulb of $$250cc$$ was sealed off at a pressure $$10^{-3}$$ mm of Hg and temperature $$27^{0}$$ C. The number of molecules present in the gas is

A
$$8.02 \times 10^{15}$$

B
$$6.023 \times 10^{23}$$

C
$$8.021 \times 10^{23}$$

D
$$6 \times 10^{22}$$

Solution

$$PV=nRT\\ n=\dfrac { PV }{ RT } \\ n=\dfrac { 250\times { 10 }^{ -6 }\times 133.33\times { 10 }^{ -3 } }{ 8.314\times 300 } \\ n=1.336\times { 10 }^{ -8 }\\ N=n\times { N }_{ A }\\ N=1.336\times { 10 }^{ -8 }\times 6.022\times { 10 }^{ 23 }\\ N=8.02\times { 10 }^{ 15 }$$
Question 10
1 / -0

Two identical containers connected by a fine capillary tube contain air at N.T.P. if one of those containers is immersed in pure water, boilling under normal pressure then new pressure is

A
76 cm of Hg

B
152 cm of Hg

C
57 cm of Hg

D
87.76 cm of Hg

Solution

Since the number of moles of gas in the system remains constant:
Moles in first container earlier+Moles in second container earlier=Moles in first container later+Moles in second container later
$$\implies \dfrac{P_0V}{RT_{0}}+\dfrac{P_0V}{RT_0}=\dfrac{P_1V}{RT_{0}}+\dfrac{P_1V}{RT_1}$$

$$T_0=273\ K,T_1=373\ K,P_0=76\ cm\ of\ Hg$$

$$\implies P_1=87.76\ cm\ of\ Hg$$

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Kinetic Theory Test - 28

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Kinetic Theory Test - 28

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Question 1

5 atm

6 atm

7 atm

8 atm

Solution

Question 2

2 gram/mole

16 gram/mole

4 gram/mole

32 gram/mole

Solution

Question 3

$$\dfrac{9MR^{2}}{32}$$

$$\dfrac{15MR^{2}}{32}$$

$$\dfrac{13MR^{2}}{32}$$

$$\dfrac{11MR^{2}}{32}$$

Solution

Question 4

9cm of Hg

2.4cm of Hg

6cm of Hg

3cm of Hg

Solution

Question 5

$$P_{1} > P_{2} > P_{3}$$

$$P_{1} = P_{2} = P_{3}$$

$$P_{1} < P_{2} < P_{3}$$

$$P_{3} > P_{1} > P_{2}$$

Solution

Question 6

33.24 x 10$$^{7}$$ pa

16.62 x 10$$^{7}$$ pa

10.31 x 10$$^{7}$$ pa

8.31x 10$$^{7}$$ pa

Solution

Question 7

1:1

1:2

2:1

1:4

Solution

Question 8

$$\frac{273}{100}gm/ltr$$

$$\frac{150}{273}gm/ltr$$

$$\frac{1}{273}gm/ltr$$

1.5 $$gm/lit$$

Solution

Question 9

$$8.02 \times 10^{15}$$

$$6.023 \times 10^{23}$$

$$8.021 \times 10^{23}$$

$$6 \times 10^{22}$$

Solution

Question 10

76 cm of Hg

152 cm of Hg

57 cm of Hg

87.76 cm of Hg

Solution

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