Kinetic Theory Test

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Kinetic Theory Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Water is falling from 160m height. Assuming that half the K.E. of falling water gets converted into heat, the rise in temperature of water is approximately

A
$$0.1^{0}C$$

B
$$0.2^{0}C$$

C
$$0.3^{0}C$$

D
$$0.4^{0}C$$

Solution

As the water falls from 160m height, its potential energy is converted into kinetic energy.
Given that half of the KE is converted to heat energy, essentially means half of PE is converted.
$$\displaystyle \frac {1}{2}mgh=mc\Delta T$$
$$\Rightarrow \Delta T=\dfrac {0.5(gh)}{c}=\dfrac {0.5(10)(160)}{4200}=0.19^oC$$
Question 2
1 / -0

The temperature of a gas is due to

A
P.E. of its molecules

B
K.E. of its molecules

C
Attractive forces between molecules

D
Repulsive forces between molecules

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, B is the correct answer.
Question 3
1 / -0

When the volume of a gas is decreased at constant temperature the pressure increases because the molecules

A
strike unit area of the walls of the container more often

B
strike the unit area of the walls of the container with higher speed

C
strike the unit area of the wall of the container with lesser speed

D
move with more kinetic energy

Solution

We first note that the speed and hence the kinetic energy of the molecules of an ideal gas depends solely on temperature and since here temperature remains constant, the pressure cannot increase due to factors mentioned in options B,C and D
Now A is correct and we explain why.
Consider a small area on the wall of the container. Previously when volume was high, it had a lower probability of being hit by a molecule but now space is less, so it will be hit by more molecules so, more pressure will be generated on this area.
Hence pressure increases
Question 4
1 / -0

The graph between temperature and pressure of a perfect gas is

A
hyperbola.

B
parabola.

C
a straight line parallel to pressure axes at $$73^{0}C$$.

D
a straight line intercepting the temperature axes at $$273^{0}C$$.

Solution

From ideal gas equation, PV = nRT
so P-T curve at constant volume is a straight line.
If the temperature is written in Kelvin, it turns out that the intercept is at 273
Question 5
1 / -0

The graph drawn between pressure and temperature at constant volume for a given mass of different molecular weights $$M_{1}$$ and $$M_{2}$$ are the straight lines as shown in the figure then

A
$$M_{2} > M_{1}$$

B
$$M_{1} > M_{2}$$

C
$$M_{1} = M_{2}$$

D
$$M_{1}^{3} = M_{2}$$

Solution

From ideal gas equation, PV = nRT = $$\frac{w}{M}RT$$
Slope of P-T curve = $$\frac{wR}{MV}$$
Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is molecular weight.
So, $$M_{1} > M_{2}$$
Question 6
1 / -0

At constant pressure, density of a gas is :

A
directly proportional to absolute temperature

B
inversely proportional to absolute temperature

C
independent of temperature

D
directly proportional to square root of absolute temperature
Solution
Hint: Density of a substance is given by $$\rho = \dfrac{m}{V}$$ , where $$m$$is the mass of the substance and $$V$$ is its volume.
Correction Option: B
Explanation for correct option:
- By ideal gas equation: $$PV = nRT$$, where $$P$$,$$V$$, $$T$$ and $$n$$ are the pressure, volume, temperature and moles of the gas and $$R$$ is the universal gas constant.
$$ \Rightarrow PV \alpha nRT$$
$$ \Rightarrow PV = \dfrac{m}{M} RT$$, where $$m$$ and $$M$$ are weight and molecular weight of the gas.
$$ \Rightarrow P = \dfrac{m}{V}\dfrac{{RT}}{M}$$
Here, $$\rho = \dfrac{m}{V}$$ and $$R' = \dfrac{R}{M}$$, where $$R'$$ is the specific gas constant for the gas.
$$ \Rightarrow P = \rho R'T$$
$$ \Rightarrow \rho = \dfrac{P}{{R'T}}$$
$$\therefore \rho \alpha \dfrac{1}{T}$$
Hence, at constant pressure, density of a gas is inversely proportional to absolute temperature.
Question 7
1 / -0

Universal gas constant per molecule is called

A
Rydberg constant

B
Kelvin constant

C
Boltzman constant

D
Stefan's constant
Question 8
1 / -0

All gasses deviate from gas laws at

A
high pressure and low temperature.

B
low pressure and high temperature.

C
high pressure and high temperature.

D
low pressure and low temperature.

Solution

At high pressure and low temperature, the molecular interaction terms become significant.
So, we cannot neglect these terms.
Also, the total size of molecules becomes significant at low temperatures.
So, at low temperature and high pressure, there is a significant deviation from ideal gas equation.
Question 9
1 / -0

Statement - I : The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume because.
Statement - II : The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

A
1) Statement - I is true, Statement - II is true; statement - II is a correct explanation for statement - I

B
2) Statement - I is true, statement - II is true; statement - II is NOT a correct explanation for statement - I

C
3) statement - I is true, statement - II is false

D
4) statement - I is false, statement - II is true

Solution

The rms velocity is given as $$v_{rms}=\sqrt{\displaystyle\frac{3RT}{M}}$$, where M is the molecular weight of the given gas. Thus the kinetic energy is given as $$\displaystyle\frac{1}{2}Mv^2_{rms}=\frac{3}{2}RT$$. Using gas equation PV=RT we get the kinetic energy as $$\displaystyle\frac{3}{2}PV$$.
Both the statement are correct but 2 is not the correct explanation of 1.
Question 10
1 / -0

If the slope of P-T graph for a given mass of a gas increases, then the volume of the gas

A
Increases

B
Decreases

C
Does not change

D
May increase or decrease

Solution

From ideal gas equation, PV = nRT = $$\frac{w}{M}RT$$
slope of P-T curve = $$\frac{wR}{MV}$$
since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is volume.
So as slope increases, volume decreases

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Kinetic Theory Test - 22

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Kinetic Theory Test - 22

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Question 1

$$0.1^{0}C$$

$$0.2^{0}C$$

$$0.3^{0}C$$

$$0.4^{0}C$$

Solution

Question 2

P.E. of its molecules

K.E. of its molecules

Attractive forces between molecules

Repulsive forces between molecules

Solution

Question 3

strike unit area of the walls of the container more often

strike the unit area of the walls of the container with higher speed

strike the unit area of the wall of the container with lesser speed

move with more kinetic energy

Solution

Question 4

hyperbola.

parabola.

a straight line parallel to pressure axes at $$73^{0}C$$.

a straight line intercepting the temperature axes at $$273^{0}C$$.

Solution

Question 5

$$M_{2} > M_{1}$$

$$M_{1} > M_{2}$$

$$M_{1} = M_{2}$$

$$M_{1}^{3} = M_{2}$$

Solution

Question 6

directly proportional to absolute temperature

inversely proportional to absolute temperature

independent of temperature

directly proportional to square root of absolute temperature

Solution

Question 7

Rydberg constant

Kelvin constant

Boltzman constant

Stefan's constant

Question 8

high pressure and low temperature.

low pressure and high temperature.

high pressure and high temperature.

low pressure and low temperature.

Solution

Question 9

1) Statement - I is true, Statement - II is true; statement - II is a correct explanation for statement - I

2) Statement - I is true, statement - II is true; statement - II is NOT a correct explanation for statement - I

3) statement - I is true, statement - II is false

4) statement - I is false, statement - II is true

Solution

Question 10

Increases

Decreases

Does not change

May increase or decrease

Solution

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