Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 27

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Question 1
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 2
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 3
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 4
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 5
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
$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } $$
$$\dfrac { 2.8{ \times V }_{ 1 } }{ 280 } =\dfrac { 1\times { V }_{ 2 } }{ 300 } $$
$${ V }_{ 2 }=3{ V }_{ 1 }$$
$$R\propto { V }^{ \dfrac { 1 }{ 3 } }$$
$$\therefore { R }_{ 2 }={ { 3 }^{ \dfrac { 1 }{ 3 } }R }_{ 1 }$$
Question 6
1 / -0

During an experiment an ideal gas is found to obey an additional law V$$^{2}$$P= constant. The gas is initially at a temperature T and volume V. When it expands to a volume 2V, the temperature becomes

A
T

B
2T

C
T$$\sqrt{2}$$

D
T/2

Solution

$${ V }^{ 2 }P=Constant\\ { V }^{ 2 }\left( \dfrac { RT }{ V } \right) =Constant\\ VT=Constant\\ \therefore \dfrac { { V }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { V }_{ 2 } }{ { T }_{ 1 } } \\ \\ { T }_{ 2 }=\dfrac { { T }_{ 1 }{ V }_{ 1 } }{ 2{ V }_{ 1 } } \\ \\ { T }_{ 2 }=\dfrac { { T }_{ 1 } }{ 2 } $$
Question 7
1 / -0

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

A
2P

B
P

C
P/2

D
4P

Solution

PV=nRT,
Since gas has same volume they contain same number of molecules and hence same number of moles.
Now since V and T remains same only n changes to 2n
Hence P=2P
Question 8
1 / -0

A vessel is filled with an ideal gas at a pressure of 200 atm and is at a temperature of 27$$^{0}$$ C. One half of the mass of the gas is removed from the vessel and the temperature of the remaining gas is increased to 87$$^{0}$$ C. At this temperature,the pressure of the gas will be

A
60 atm

B
120 atm

C
360 atm

D
240 atm

Solution

$${ P }_{ 1 }{ V }_{ 1 }=\dfrac { { m }_{ 1 } }{ M } R{ T }_{ 1 }\\ { P }_{ 2 }{ V }_{ 2 }=\dfrac { { m }_{ 2 } }{ M } R{ T }_{ 2 }\\ { V }_{ 1 }={ V }_{ 2 },{ \quad 2m }_{ 2 }={ m }_{ 1 }\\ \dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { m }_{ 1 }{ T }_{ 1 } }{ { { m }_{ 2 }T }_{ 2 } } \\ { P }_{ 2 }=\dfrac { 360\times 200 }{ 300\times 2 } \\ \\ { P }_{ 2 }=120atm$$
Question 9
1 / -0

A gas is heated through $$1^{o}\ C$$ in a closed vessel. Its pressure is increased by $$0.4$$%. The initial emperature of the gas is

A
$$250^{0}\ C$$

B
$$100^{0}\ C$$

C
$$-75^{0}\ C$$

D
$$-23^{0}\ C$$

Solution

Since it is a closed system the volume remains constant.
Hence,$$\dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { T }_{ 2 } } $$
$${ P }_{ 2 }=1.004{ P }_{ 1 }\\ \therefore { T }_{ 2 }=1.004{ T }_{ 1 }\\ but\quad { T }_{ 1 }+1={ T }_{ 2 }\\ \therefore { T }_{ 1 }=250K\\ { T }_{ 1 }=-{ 23 }^{ \circ }$$
Question 10
1 / -0

A given amount of a gas is heated till the volume and pressure both increase by $$2\%$$ each the percentage change in temperature of the gas is equal to nearly

A
$$2\%$$

B
$$3\%$$

C
$$4\%$$

D
$$1\%$$

Solution

$$\dfrac { dP }{ P } +\dfrac { dV }{ V } =\dfrac { dT }{ T } \\ Now\quad dP=0.02P,dV=0.02V\\ 0.02+0.02=\dfrac { dT }{ T } \\ \dfrac { dT }{ T } \times 100=4%=percentage\quad change\quad in\quad temperature$$

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Thermal Properties of Matter Test - 27

Result

Thermal Properties of Matter Test - 27

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SHARING IS CARING

Question 1

1

2

3

4

Solution

Question 2

110 cm of Hg

120 cm of Hg

150 cm of Hg

200 cm of Hg

Solution

Question 3

27$$^{0}$$C

127$$^{0}$$C

300$$^{0}$$C

400$$^{0}$$C

Solution

Question 4

120kPa

105kPa

150kPa

300kPa

Solution

Question 5

$$3^{1/3}$$

$$4^{1/3}$$

$$5^{1/3}$$

$$6^{1/3}$$

Solution

Question 6

T

2T

T$$\sqrt{2}$$

T/2

Solution

Question 7

2P

P

P/2

4P

Solution

Question 8

60 atm

120 atm

360 atm

240 atm

Solution

Question 9

$$250^{0}\ C$$

$$100^{0}\ C$$

$$-75^{0}\ C$$

$$-23^{0}\ C$$

Solution

Question 10

$$2\%$$

$$3\%$$

$$4\%$$

$$1\%$$

Solution

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