States of Matter Test

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States of Matter Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The volume of a gas is ______ when the pressure is reduced to half and the absolute temperature is doubled.

A
doubled

B
quadrupled

C
tripled

D
none of these

Solution

$$\cfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } =\cfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \Longrightarrow \cfrac { { V }_{ 2 } }{ V_{ 1 } } =\cfrac { { P }_{ 1 } }{ P_{ 2 } } \cdot \cfrac { { T }_{ 1 } }{ T_{ 2 } } =2\cdot 2=4$$
Therefore, the volume of the gas is quadrupled.
Question 2
1 / -0

If the velocity of an electron in the second orbit of the hydrogen atom is x, the velocity in fourth orbit will be:

A
$$\dfrac{x }{ 16}$$

B
$$0.5 x$$

C
$$\dfrac{x }{ 4}$$

D
$$\dfrac{x }{ 3}$$

Solution

$$mv{ r }_{ n }=\cfrac { nh }{ 2\pi } and\quad { r }_{ n }=\cfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }m{ e }^{ 2 } } $$ (since Z=1 for hydrogen atom)
$$\therefore v=\cfrac { 1 }{ { r }_{ n } } \cdot \cfrac { nh }{ 2\pi m } =\cfrac { 4{ \pi }^{ 2 }m{ e }^{ 2 } }{ { n }^{ 2 }{ h }^{ 2 } } \cdot \cfrac { nh }{ 2\pi m } =\cfrac { 1 }{ n } \cdot \cfrac { 2\pi { e }^{ 2 } }{ h } \\ \therefore v\propto \cfrac { 1 }{ n } \\ { v }_{ n=2 }=x\quad and\quad { v }_{ n=4 }=?$$
Now, $${ v }_{ n=2 }\times 2={ v }_{ n=4 }\times 4\Longrightarrow 2x={ v }_{ n=4 }\times 4\Longrightarrow { v }_{ n=4 }=0.5x$$
Question 3
1 / -0

Which of the following pair of gases cannot be separated by diffusion method?

A
$$SO_2$$ and $$H_2$$

B
$$CO_2$$ and $$N_2O$$

C
$$NH_3$$ and $$N_2$$

D
$$CO_2$$ and $$H_2$$

Solution

Previously we told that, $$D\propto \cfrac { 1 }{ \sqrt { M } } $$
Now, $${ M }_{ C{ O }_{ 2 } }$$ $$=2\times 16+12=44$$ g/mol
$${ M }_{ { N }_{ 2 }O }$$ $$=2\times 14+16=44$$ g/mol
$$\therefore { D }_{ C{ O }_{ 2 } }={ D }_{ { N }_{ 2 }O }$$, as diffusion coefficient of $$CO_{2}$$ and $$N_{2}O$$ is equal, $$CO_{2}$$ and $$N_{2}O$$ cannot be separated by diffusion method.
Question 4
1 / -0

Statement I
The volume of a gas at 100 deg C and 600 mmHg will be lower at STP
Because
Statement II
Decreasing temperature and increasing pressure will cause the volume of a gas to decrease

A
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

B
Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

C
Statement 1 is correct but Statement 2 is not correct.

D
Statement 1 is not correct but Statement 2 is correct.

E
Both the Statement 1 and Statement 2 are not correct.

Solution

Answer(D)

Using ideal gas law $$PV$$$$=$$$$nRT$$
for given amount of a gas volume $$V$$$$=$$$$kT/P$$ ,where $$k$$ is a constant.
so $$V$$$$=$$$$K*373/600$=$$$$0.621k$$
at $$STP$$ volume $$V$$$$=$$$$k*273/760$$$$=$$$$=$$$$.35k$$
so The volume of a gas at 100 deg C and 600 mmHg will be higher than at STP,so this statement is not true
Decreasing temperature and increasing pressure will cause the volume of a gas to decrease,this is correct
Question 5
1 / -0

The value of the Avogadro constant is:

A
$$6.022 \times 10^{13}$$

B
$$6.022 \times 10^{22}$$

C
$$6.022 \times 10^{23}$$

D
$$6.022 \times 10^{24}$$

Solution

Correct Answer: Option C
Explanation:

The value of the Avogadro constant is $$6.022 \times 10^{23}$$.

It is a number of atoms or molecules or ions or particles present in one mole of a substance.

Hence, the correct answer is option $$C$$.
Question 6
1 / -0

Which statement is NOT compatible with the ideal gas law?

A
Gas particles have no definite volume

B
Gas temperature depends on particles mass and velocity

C
Gas particles are in constant, random motion

D
Gas particles have no mass

Solution

(A) True: Gas particles have no definite volume.
(B) True: Gas temperature depends on particles mass and velocity.
(C) True: Gas particles are in constant, random motion.
(D) False: Gas particle has mass.

The correct option is (D).
Question 7
1 / -0

Which gas law uses pressure, number of moles, temperature, volume and the gas constant (R)?

A
Boyle's Law

B
Ideal Gas Law

C
Combined Gas Law

D
Charles' Law

Solution

The ideal gas low equation uses all the terms,
$$PV=nRT$$
where $$P$$ is the pressure
$$V$$ is the volume
$$T$$ is the temperature
$$n$$ is the mole of gas.
Question 8
1 / -0

Statement I: In the kinetic theory of gases, collisions between gas particles and the walls of the container are considered elastic.
Statement II: Gas molecules are considered pointlike, volumeless particles with no intermolecular forces and in constant, random motion.

A
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

B
Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1

C
Statement 1 is correct but Statement 2 is not correct

D
Statement 1 is not correct but Statement 2 is correct

E
Both the Statement 1 and Statement 2 are not correct

Solution

Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions.Temperature remains the same, so the average kinetic energy and the rms speed should remain the same.
In Kinetic theory Gas molecules are considered pointlike, volumeless particles with no intermolecular forces and in constant, random motion.
Question 9
1 / -0

Two gases A and B are taken in same volume containers under similar conditions of temperature and pressure. In container A, there are '2N' molecules of gas A. How many number molecules does container B have?

A
2N

B
4N

C
N

D
8N

Solution

Avogadro's law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Thus ans is 2N.
Question 10
1 / -0

Directions For Questions

In the above experiment, the vapor pressure of isopropanol is determined. Throughout the experiment, the temperature was held constant at $$\displaystyle { 25 }^{ \circ }C$$ and the atmospheric pressure remained at $$760\ mm\ of\ \displaystyle Hg$$.

...view full instructions

The initial pressure inside the flask is:

A
near zero (vacuum)

B
$$27\ mm\ $$$$\displaystyle Hg$$

C
$$760\ mm\ $$$$\displaystyle Hg$$

D
$$(760-27)\ mm\ Hg $$

Solution

Since here in the figure given the mercury is at the same height in both the arms containing the sample and the arm is atom to the atmosphere, the pressure inside the flask initially is $$760mHg$$.
Option (C) is correct.

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States of Matter Test - 19

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States of Matter Test - 19

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Question 1

doubled

quadrupled

tripled

none of these

Solution

Question 2

$$\dfrac{x }{ 16}$$

$$0.5 x$$

$$\dfrac{x }{ 4}$$

$$\dfrac{x }{ 3}$$

Solution

Question 3

$$SO_2$$ and $$H_2$$

$$CO_2$$ and $$N_2O$$

$$NH_3$$ and $$N_2$$

$$CO_2$$ and $$H_2$$

Solution

Question 4

Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

Statement 1 is correct but Statement 2 is not correct.

Statement 1 is not correct but Statement 2 is correct.

Both the Statement 1 and Statement 2 are not correct.

Solution

Question 5

$$6.022 \times 10^{13}$$

$$6.022 \times 10^{22}$$

$$6.022 \times 10^{23}$$

$$6.022 \times 10^{24}$$

Solution

Question 6

Gas particles have no definite volume

Gas temperature depends on particles mass and velocity

Gas particles are in constant, random motion

Gas particles have no mass

Solution

Question 7

Boyle's Law

Ideal Gas Law

Combined Gas Law

Charles' Law

Solution

Question 8

Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1

Statement 1 is correct but Statement 2 is not correct

Statement 1 is not correct but Statement 2 is correct

Both the Statement 1 and Statement 2 are not correct

Solution

Question 9

2N

4N

N

8N

Solution

Question 10

near zero (vacuum)

$$27\ mm\ $$$$\displaystyle Hg$$

$$760\ mm\ $$$$\displaystyle Hg$$

$$(760-27)\ mm\ Hg $$

Solution

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