Kinetic Theory Test

Result

Kinetic Theory Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
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 2
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 3
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 4
1 / -0

If $$\rho $$ is the density, m is the mass of 1 molecule and K is the Boltzman constant for a gas then the pressure of the gas is:

A
$$P=\dfrac{\rho KT}{m}$$

B
$$P=\dfrac{mKT}{\rho }$$

C
$$P=\dfrac{\rho mT}{K}$$

D
$$P=\rho KmT$$

Solution

$$PV=\dfrac { m }{ M } RT\\ P\dfrac { m }{ \rho } =nkT\\ P=\dfrac { \rho nkT }{ m } ,\quad \\ if\quad n=1,\quad P=\dfrac { \rho kT }{ m } $$
Question 5
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 6
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

$$\dfrac { dP }{ P } +\dfrac { dV }{ V } =\dfrac { dT }{ T } \\ Now\quad dP=0.02P,dV=0.02V\\ 0.02+0.02=\dfrac { dT }{ T } \\ \dfrac { dT }{ T } \times 100=4%=percentage\quad change\quad in\quad temperature$$
Question 7
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 8
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 9
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 10
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

$$PV=\dfrac { m }{ M } RT\\ 11.2\times { 10 }^{ -3 }\times 1.01\times { 10 }^{ 5 }=\dfrac { m }{ 32\times { 10 }^{ -3 } } \times 8.314\times 300\\ m=0.01456$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Kinetic Theory Test - 27

Result

Kinetic Theory Test - 27

Score

-

Rank

-

SHARING IS CARING

Question 1

T

2T

T$$\sqrt{2}$$

T/2

Solution

Question 2

2P

P

P/2

4P

Solution

Question 3

60 atm

120 atm

360 atm

240 atm

Solution

Question 4

$$P=\dfrac{\rho KT}{m}$$

$$P=\dfrac{mKT}{\rho }$$

$$P=\dfrac{\rho mT}{K}$$

$$P=\rho KmT$$

Solution

Question 5

$$250^{0}\ C$$

$$100^{0}\ C$$

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

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

Solution

Question 6

$$2\%$$

$$3\%$$

$$4\%$$

$$1\%$$

Solution

Question 7

$$1:8$$

$$1:16$$

$$8:1$$

$$16:1$$

Solution

Question 8

2P

P

P/2

4P

Solution

Question 9

900$$^{0}$$C

927$$^{0}$$C

627$$^{0}$$C

327$$^{0}$$C

Solution

Question 10

0.001456

0.01456

0.1456

1.1456

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.