Kinetic Theory Test

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Kinetic Theory Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

An ideal gas is that which

A
Cannot be liquified

B
Can be easily liquified

C
Has strong inter molecular forces

D
Has a large size of molecules.

Solution

The concept of liquefaction of gases works on the principle of intermolecular forces of attraction between the gas molecules. Gases liquefy when these forces increase between the molecules, binding them together. But an ideal gas is one where the molecules do not influence each other, the explanation comes from Joule Thompson effect.
Question 2
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A box contains x molecules of a gas. How will the pressure of the gas be affected if the number of molecules is made 2x?

A
Pressure will decrease.

B
Pressure will remain unchanged.

C
Pressure will be doubled.

D
Pressure will become three times.

Solution

we know that from ideal gas equation, PV = nRT
now, n = number of moles of gas = $$\dfrac{N}{N_{A}}$$, where N = number of molecules of the gas and $$N_{A}$$ is the Avogadro's number
So, putting the value of n in ideal gas equation, we get
$$PV = \dfrac{N}{N_{A}}RT$$
this means that pressure of the gas is proportional to the number of molecules of the gas at constant temperature and volume
So, if we double the number of molecules of the gas (2x), then pressure will also double
So, C is the correct answer.
Question 3
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In a gas equation, PV = RT, V refers to the volume of:

A
any amount of a gas

B
one gram mass of a gas

C
one gram mole of gas

D
one litre of gas

Solution

Given PV = RT
Comparing with the ideal gas equation, PV = nRT, we
find that here n = 1
So, number of moles = 1
Hence we must refer to molar volume
So, C is the correct answer.
Question 4
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The relation between volume V, pressure P and absolute temperature T of an ideal gas is PV = xT, where x is a constant. The value of x depend upon

A
the mass of the gas molecule

B
the average kinetic energy of the gas molecules

C
P, V and T

D
the number of gas molecules in volume V.

Solution

From ideal gas equation we know that PV = nRT
We are given that PV = xT
Comparing the 2, we get that x = nR
Now, n = number of moles of the gas = $$\frac{N}{N_{A}}$$, where N is the number of molecules of the gas and $$N_{A}$$ is the avogadro's number
So, x = $$\frac{N}{N_{A}}R$$
Clearly $$N_{A}$$ and R are universal constants
So, x depends on N = number of molecules of the gas.
So, D is the correct answer.
Question 5
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The density of an ideal gas

A
is directly proportional to its pressure and absolute temperature

B
is directly proportional to its pressure and inversely proportional to its absolute temperature

C
is inversely proportional to its pressure and directly proportional to its absolute temperature

D
is inversely proportional to both its pressure and absolute temperature of the gas

Solution

Let us derive the expression for the density of an ideal gas
density = $$\frac{Mass}{Volume} = \frac{w}{V}$$
Now from ideal gas equation, PV = nRT
and n = $$\frac{w}{M}$$ where w is the mass of the gas and M is the molar mass of the gas
Putting this in ideal gas equation, we get
$$PV = \frac{w}{M}RT$$
now substitute the value of V in the density expression to get
$$d = \frac{PM}{RT}$$
so, density of an ideal gas is proportional to the pressure and inversely proportional to the absolute temperature
So, B is the correct answer.
Question 6
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The average velocity of the molecules in a gas in equilibrium is

A
proportional to $$\sqrt{T}$$

B
proportional to T

C
proportional to $$T^{2}$$

D
equal to zero

Solution

the average velocity of the gas molecules = $$\sqrt{\frac{8RT}{\pi M}}$$
so clearly, the average velocity $$\alpha \sqrt{T}$$
So, A is the correct answer.
Note that T is the temperature in Kelvins
Question 7
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Choose the only correct statement from the following

A
The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.

B
The product of pressure and volume of a gas is always constant.

C
The average kinetic energy of molecules of a gas is proportional to its absolute temperature.

D
The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.

Solution

We know that the kinetic energy of gas molecules and the temperature are related as
KE $$\alpha$$ kT
So, the temperature of gas molecules is due to the kinetic energy of molecules.
So, C is the correct answer.
Question 8
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If pressure and temperature of an ideal gas are doubled and volume is halved, the number of molecules of the gas

A
becomes half

B
becomes two times

C
becomes 4 times

D
remains constant

Solution

From ideal gas equation, PV = nRT
This means $$n = \frac{PV}{RT}$$
Now, pressure and temperature both are doubled and volume is halved
So, clearly the number of moles will also become half
So, A is the correct answer.
Question 9
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The absolute temperature T of a gas is plotted against its pressure P for two different constant volumes V$$_{1}$$ and V$$_{2}$$ where V$$_{1}$$ > V$$_{2}$$. T is plotted along x-axis and P along y-axis.

A
Slope for curve corresponding to volume V$$_{1}$$ is greater than that corresponding to volume V$$_{2}$$.

B
Slope for curve corresponding to volume V$$_{2}$$ is greater than that corresponding to volume V$$_{1}$$.

C
Slope for both curves are equal.

D
Slope for both curves are unequal such that they intersect at T = 0.

Solution

From ideal gas equation, PV = nRT
This means that $$P = \frac{nR}{V}T$$
So, the slope of P-T curve = $$\frac{nR}{V} \alpha \frac{1}{V}$$
this means that more slope => less volume
Since we are given that $$V_{2} < V_{1}$$, so we get that
Slope for curve corresponding to volume $$V_2$$ is greater than that corresponding to volume $$V_1$$.
So, B is the correct answer.
Question 10
1 / -0

The universal gas constant has the units

A
dyne/$$^{0}C$$

B
erg/mole/k

C
erg-cm/k

D
watt/k

Solution

from ideal gas equation, PV = nRT
now, the dimensions of PV is same as dimension of energy (or dimension of work)
So, dimension of PV is ergs
and the dimension of n is mol, dimension of T is Kelvin
So, dimension of R = $$\dfrac{dimension\ of\ PV}{dimension\ of\ n \times dimension\ of\ Temperature}$$
$$ = \dfrac{erg}{mol-K}$$
so, B is the correct answer.

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Kinetic Theory Test - 23

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Kinetic Theory Test - 23

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Question 1

Cannot be liquified

Can be easily liquified

Has strong inter molecular forces

Has a large size of molecules.

Solution

Question 2

Pressure will decrease.

Pressure will remain unchanged.

Pressure will be doubled.

Pressure will become three times.

Solution

Question 3

any amount of a gas

one gram mass of a gas

one gram mole of gas

one litre of gas

Solution

Question 4

the mass of the gas molecule

the average kinetic energy of the gas molecules

P, V and T

the number of gas molecules in volume V.

Solution

Question 5

is directly proportional to its pressure and absolute temperature

is directly proportional to its pressure and inversely proportional to its absolute temperature

is inversely proportional to its pressure and directly proportional to its absolute temperature

is inversely proportional to both its pressure and absolute temperature of the gas

Solution

Question 6

proportional to $$\sqrt{T}$$

proportional to T

proportional to $$T^{2}$$

equal to zero

Solution

Question 7

The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.

The product of pressure and volume of a gas is always constant.

The average kinetic energy of molecules of a gas is proportional to its absolute temperature.

The average kinetic energy of molecules of a gas is proportional to the square root of its absolute temperature.

Solution

Question 8

becomes half

becomes two times

becomes 4 times

remains constant

Solution

Question 9

Slope for curve corresponding to volume V$$_{1}$$ is greater than that corresponding to volume V$$_{2}$$.

Slope for curve corresponding to volume V$$_{2}$$ is greater than that corresponding to volume V$$_{1}$$.

Slope for both curves are equal.

Slope for both curves are unequal such that they intersect at T = 0.

Solution

Question 10

dyne/$$^{0}C$$

erg/mole/k

erg-cm/k

watt/k

Solution

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