Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 28

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Question 1
1 / -0

Two sample of Hydrogen and Oxygen of same mass possess same pressure and volume. The ratio of their temperature is

A
$$1:8$$

B
$$1:16$$

C
$$8:1$$

D
$$16:1$$

Solution

$${ P }_{ 1 }{ V }_{ 1 }=\dfrac { { m }_{ 1 } }{ { M }_{ 1 } } R{ T }_{ 1 }\\ { P }_{ 2 }{ V }_{ 2 }=\dfrac { { m }_{ 2 } }{ { M }_{ 2 } } R{ T }_{ 2 }\\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } \\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { 1 }{ 16 } \left( { M }_{ 1 }=1gm,{ \ M }_{ 2 }=16gm \right) $$
Question 2
1 / -0

Two gases A&B having same pressure P, volume V and absolute temperature T are mixed. If the mixture has volume and temperature as V & T respectively then the pressure of mixture is

A
2P

B
P

C
P/2

D
4P

Solution

Moles of gas A $$=\dfrac{PV}{RT}$$

Moles of gas B $$=\dfrac{PV}{RT}$$
Moles of the gas mixture $$=\dfrac{P'V}{RT}$$
Since the number of moles(mass) remains same before and after mixing,
$$\dfrac{PV}{RT}+\dfrac{PV}{RT}=\dfrac{P'V}{RT}$$
$$\implies P'=2P$$
Question 3
1 / -0

One litre of Helium gas at a pressure of 76 cm - Hg and temperature 27$$^{0}$$C is heated till its pressure and volume are doubled. The final temperature attained by the gas is

A
900$$^{0}$$C

B
927$$^{0}$$C

C
627$$^{0}$$C

D
327$$^{0}$$C

Solution

$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { PV }{ 300 } =\dfrac { 2P\times 2V }{ { T }_{ 2 } } \\ { T }_{ 2 }=1200K\\ { T }_{ 2 }={ 927 }^{ \circ }C$$
Question 4
1 / -0

The mass of oxygen gas occupy a volume of 11.2 lit at a temperature 27$$^{0}$$C and a pressure of 76 cm of mercury in kilogram is (molecular weight of oxygen =32)

A
0.001456

B
0.01456

C
0.1456

D
1.1456

Solution

$$PV=\dfrac { m }{ M } RT\\ 11.2\times { 10 }^{ -3 }\times 1.01\times { 10 }^{ 5 }=\dfrac { m }{ 32\times { 10 }^{ -3 } } \times 8.314\times 300\\ m=0.01456$$
Question 5
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 6
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
$$\implies \dfrac{P_0V}{RT_{0}}+\dfrac{P_0V}{RT_0}=\dfrac{P_1V}{RT_{0}}+\dfrac{P_1V}{RT_1}$$

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

$$\implies P_1=87.76\ cm\ of\ Hg$$
Question 7
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 8
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 9
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 10
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 $${ M }_{ 1 }{ <M }_{ 2 }{ <M }_{ 3 }\\ { \therefore P }_{ 1 }{ >P }_{ 2 }{ >P }_{ 3 }$$

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Thermal Properties of Matter Test - 28

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Thermal Properties of Matter Test - 28

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Question 1

$$1:8$$

$$1:16$$

$$8:1$$

$$16:1$$

Solution

Question 2

2P

P

P/2

4P

Solution

Question 3

900$$^{0}$$C

927$$^{0}$$C

627$$^{0}$$C

327$$^{0}$$C

Solution

Question 4

0.001456

0.01456

0.1456

1.1456

Solution

Question 5

5 atm

6 atm

7 atm

8 atm

Solution

Question 6

76 cm of Hg

152 cm of Hg

57 cm of Hg

87.76 cm of Hg

Solution

Question 7

2 gram/mole

16 gram/mole

4 gram/mole

32 gram/mole

Solution

Question 8

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

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

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

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

Solution

Question 9

9cm of Hg

2.4cm of Hg

6cm of Hg

3cm of Hg

Solution

Question 10

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

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

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

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

Solution

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