Kinetic Theory Test

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Kinetic Theory Test - 43

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Question 1
1 / -0

The variation of the product of pressure and volume $$PV$$ with volume $$V$$ of a fixed mass of an ideal gas at constant temperature is graphically represented by the curve

A

B

C

D

Solution
Question 2
1 / -0

The temperature to which a gas must be cooled before it can't be liquefied by pressure alone is called its

A
Critical temperature

B
Saturation point

C
Boiling point

D
Freezing point

Solution

$$Answer:-$$ A
In this P-V curve the isotherm having $$T_c$$ is that critical temperature and corresponding to that there is critical pressure $$P_c$$ .

Critical temperature is the temperature of a gas in its critical state, above which it cannot be liquefied by pressure alone.
Critical pressure is the pressure of a gas or vapour in its critical state.
Question 3
1 / -0

The molecular kinetic energy of a liquid is:

A
more than molecular kinetic energy of solids

B
more than molecular kinetic energy of gases

C
less than molecular kinetic energy of solids

D
dependent on the matter

Solution

The inter-molecular attractions in liquids are lesser compared to solids. As, liquids can move freely due to less inter molecular attraction they have more molecular kinetic energy compared to solids.
Question 4
1 / -0

When we cool a gas below its condensation point, the K.E. of its molecules:

A
increases

B
decreases

C
remains the same

D
first decreases then increases

Solution

As we know kinetic energy of molecules of a gas is a function of Temperature
So, As we decrease its temperature or say cool it, its kinetic energy gradually decrease. At a certain point which decreasing kinetic energy it starts bonding up whit molecules and thus condensation starts.
Hence, the answer is decreases.
Question 5
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Two gases $$A$$ and $$B$$ having the same temperature $$T$$, same pressure $$P$$ and same volume $$V$$ are mixed. If the mixture is at the same volume $$V$$, the pressure of the mixture is

A
$$P$$

B
$${P}/{2}$$

C
$$2P$$

D
$$4P$$

Solution

From Ideal Gas Law,
$$PV=nRT$$
$$\implies$$ Moles of a gas=$$n=\dfrac{PV}{RT}$$
When the mixture is created, the total number of moles is equal to the sum of that of individual gases.
Hence $$N=n_1+n_2$$
$$\implies \dfrac{P'V}{RT}=\dfrac{PV}{RT}+\dfrac{PV}{RT}$$
$$\implies P'=2P$$
Question 6
1 / -0

Real gases obey ideal gas laws more closely at:

A
low pressure and low temperature

B
low pressure and high temperature

C
high pressure and low temperature

D
high pressure and high temperature

Solution

Real gases obey ideal gas laws at low pressure and high temperature because at low pressure the number of molecules per unit volume is less so attractive force between them is negligible. At high temperature the speed of the molecules is very high so collisions becomes elastic.
Question 7
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Which of the following is an assumption of Kinetic theory of matter?

A
Molecules are in a state of continuous motion and possess kinetic energy.

B
The kinetic energy of molecules increases with increase in temperature.

C
The molecules of matter always attract each other due to forces of cohesion and adhesion.

D
All of the above

Solution

The following are the assumptions made regarding the motion of molecules in matter:
1. Molecules are in a state of continuous motion and possess kinetic energy.
2. The kinetic energy of molecules increases with an increase in temperature and decreases with a decreases in temperature.
3. The molecules of matter always attract each other due to the intermolecular force of attraction.
4. Molecules of matter have space in between them.
5. If the intermolecular space is small, intermolecular force of attraction increases.
Question 8
1 / -0

To what temperature should the hydrogen at $$327^oC$$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?

A
$$-123^oC$$

B
$$123^oC$$

C
$$-100^oC$$

D
$$0^oC$$

Solution

$$v_{rms}\propto \sqrt{\displaystyle\frac{3RT}{M}}$$
$$\Rightarrow\displaystyle T\propto v^2_{rms}$$
$$\Rightarrow \displaystyle\frac{T_2}{T_1}=\left[\displaystyle\frac{v_2}{v_1}\right]^2=\frac{1}{4}$$
$$\Rightarrow T_2=\displaystyle\frac{T_1}{4}=\frac{(273+327)}{4}$$
$$=150K=-123^oC$$
Question 9
1 / -0

At a constant volume the specific heat of a gas is $$0.075$$ and its molecular weight is $$40$$. The gas is :

A
monoatomic

B
diatomic

C
triatomic

D
none of the above

Solution

We know that,
Molar heat capacity at constant volume,

$$C_{v} =$$ Specific heat at constant volume $$\times $$ Mol. wt.

$$=0.075 \times 40 = 3.0\ cal$$

$$\because C_{p} - C_{v} = R$$

or $$C_{p} = R + C_{v} = 2 + 3 = 5$$

Now, $$\dfrac {C_{p}}{C_{v}} = \gamma; \therefore \gamma = \dfrac {5}{3} = 1.66$$

This value shows that the gas is monoatomic.

Hence, option $$A$$ is correct
Question 10
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The average kinetic energy of a gas molecule at $${27}^{o}C$$ is $$6.21\times {10}^{-21}J$$, then its average kinetic energy at $${227}^{o}C$$ is:

A
$$10.35\times {10}^{-21}J$$

B
$${11.35}\times {10}^{-21}J$$

C
$$52.2\times {10}^{-21}J$$

D
$$5.22\times {10}^{-21}J$$

Solution

Average kinetic energy of gas molecules $$\propto$$ Temperature (Absolute)
$$\cfrac{K.E(at\quad {227}^{o}C)}{K.E (at\quad {27}^{o}C)}=\cfrac{273+227}{273+27}=\cfrac{500}{300}=\cfrac{5}{3}$$
$$K.E({227}^{o})=\cfrac{5}{3}\times 6.21\times {10}^{-21}J=10.35\times {10}^{-21}J$$