States of Matter Test

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States of Matter Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

In atmosphere which gas have high partial pressure?

A
$${ N }_{ 2 }$$

B
$${ O }_{ 2 }$$

C
$${ H }_{ 2 }O$$

D
$${ CO }_{ 2 }$$

Solution

Since mole fraction of nitrogen is the highest in atmosphere & partial pressure is directly proportional to mole fraction, Nitrogen will have the highest partial pressure.
Partial pressure= P $$\times$$ mole fraction.
Question 2
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0.5 mole of each of $$H, \ SO_2$$ and $$CH_4$$, are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:

A
$$P(CH_{4}) > P(SO_{2}) > P(H_{2})$$

B
$$P(H_{2}) > P(CH_{4}) > P(SO_{2})$$

C
$$P(SO_{2}) > P(CH_{4}) > P(H_{2})$$

D
$$P(H_{2}) > P(SO_{2}) > P(CH_{4})$$

Solution

$$SO_2$$ Molar mass= $$32+32=64g/mol$$
$$CH_4$$ Molar mass= $$12+4=16 g/mol$$
$$H_2$$ Molar mass= $$2 \times 1= 2 g/mol$$
Rate of diffusion is inversely related to molecular weight. Lighter the gas, more is the diffusion. The rate of diffusion is given by,
$$r_{H_2} >r_{CH_4} >r_{SO_2}$$
Hence the order of partial pressure will be
$$P_{SO_2}>P_{CH_4}>P_{H_2}$$
Question 3
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The molecular weight of $$O_2$$ and $$SO_2$$ are $$32$$ and $$64$$ respectively. If one litre of $$O_2$$ at $$15^0$$C and $$759$$mm pressure contains N molecules, the number of molecules in two litre of $$SO_2$$ under the same conditions of temperature and pressure will be:

A
$$N/2$$

B
$$N$$

C
$$2N$$

D
$$4N$$

Solution

Solution:- (C) $$2 N$$
From ideal gas equation-
$$PV = nRT$$
$$\Rightarrow n = \cfrac{PV}{RT}$$
For $${O}_{2}$$-
$$V = 1 \; L \; \left( \text{Given} \right)$$
No. of moles of $${O}_{2} = \cfrac{P}{RT}$$
Therefore,
No. of molecules of $${O}_{2} = n \times {N}_{a} = N \; \left( \text{Given} \right)$$
$$\Rightarrow \cfrac{P}{RT} \times {N}_{a} = N ..... \left( 1 \right)$$
For $$S{O}_{2}$$-
$$V = 2 \; L \; \left( \text{Given} \right)$$
No. of moles of $$S{O}_{2} = \cfrac{2P}{RT}$$
Therefore,
No. of molecules of $$S{O}_{2} = n \times {N}_{a}$$
$$\Rightarrow$$ No. of moles of $$S{O}_{2} = \cfrac{2P}{RT} \times {N}_{a}$$
$$\Rightarrow$$ No. of molecules of $$S{O}_{2} = 2 N \; \left( \text{From } \left( 1 \right) \right)$$
Hence the no. of molecules of $$S{O}_{2}$$ is $$2 N$$.
Question 4
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Suppose that we change $$U_{RMS}$$ of gas in a closed container from $$5 \times 10^{2}$$ cm/sec to $$10 \times 10^{2}$$ cm/sec, which one of the following might correctly explain how this change was accomplished?

A
By heating the gas w double the temperature

B
By removing 75% of the gas at constant volume we decrease the pressure to one quarter of its original value

C
By heating the gas we quadruple the pressure

D
By pumping in more gas at constant temperature we quadruple the pressure.

Solution

We know that $$U_{RMS}=\sqrt{\cfrac{SRT}{M}}\Longrightarrow \cfrac{U_2}{U_1}=\sqrt{\cfrac{T_2}{T_1}}$$
$$\Longrightarrow \cfrac{10\times 10^2\text{ }cm/sec}{5\times 10^2\text{ }cm/sec}=\sqrt{\cfrac{T_2}{T_1}}$$
$$\Longrightarrow \cfrac{T_2}{T_1}=\cfrac{4}{1}$$
again, $$C_{RMS}=\sqrt{\cfrac{3P}{P}}$$
$$\Longrightarrow \cfrac{P_2}{P_1}=\cfrac{4}{1}$$
By heating the gas we quadruple the pressure .
So, the correct answer is option $$C.$$
Question 5
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According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.

A
increases, number of moles; P & T

B
decreases, number of moles; P & T

C
increases; T & P; number of moles

D
decreases; P & T; number of moles

Solution

According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.
OR
For a given mass of ideal gas,
Volume$$\propto$$Number of moles of the gas (if temperature and pressure are constant)
So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.
Question 6
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Vapours of a liquid can exist only at ?

A
below boiling point

B
below critical temperature

C
below inversion temperature

D
above critical temperature

Solution

At critical temperature, there will be an equilibrium between the vapors of liquid and gaseous state. The liquid starts changing to gaseous state above critical temperature. Hence, vapor of liquid exist only below critical temperature.
Question 7
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Air contains 79% $${ N }_{ 2 }$$ and 21% $${ O }_{ 2 }$$ by volume. If the pressure is $$750$$ of $$Hg$$, the partial pressure of $${ O }_{ 2 }$$ is:

A
$$157.5$$ of $$Hg$$

B
$$175.5$$ of $$Hg$$

C
$$315.0$$ of $$Hg$$

D
$$257.5$$ of $$Hg$$

Solution

Solution:- (A) $$157.5 \text{ mm of Hg}$$
Given:-
Total pressure $$\left( P \right) = 750 \text{ mm of Hg}$$
Percentage of $${O}_{2}$$ in air $$= 21 \%$$
Therefore,
$${P}_{{O}_{2}} = \cfrac{\% \text{ of } {O}_{2} \text{ in the air}}{100} \times P$$
$$\Rightarrow {P}_{{O}_{2}} = \cfrac{21}{100} \times 750 = 157.5 \text{ mm of Hg}$$
Hence the partial pressure of $${O}_{2}$$ is $$157.5 \text{ mm of Hg}$$.
Question 8
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For a fixed amount of perfect gas, which of these statements must be true?

A
$$E$$ and $$H$$ each depend only on $$T$$

B
$$C_p$$ is constant

C
$$PdV = nRdT$$ for every infinitesimal process

D
Both $$(A)$$ and $$(C)$$ are true

Solution

For an ideal gas$$ =, PV=nRT$$
$$E$$ and $$H$$ are dependent only on temperature where $$H$$ is enthalpy and $$E$$ is energy.
$$\cfrac{C_p}{C_v}$$ remains constant for a real gas.
Question 9
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A weather balloon is inflated with helium. The balloon has a volume of $$100 m^3 $$ and it must be inflated to a pressure of $$ 0.10 atm$$. If $$50 L$$ gas cylinders of helium at a pressure of $$100 atm $$ are used, how many cylinders are needed? Assume that temperature is constant.

A
$$ 2 $$

B
$$ 3 $$

C
$$ 4 $$

D
$$ 1 $$

Solution

If the cylinders were at 1.00 atm, they would contain 50.00 L each, so at 100.0 atm they contain 5000. L each.
$$1m^3=1000L$$,
So $$100.0 m^3= 1.000 \times 10^5 L$$.
Since the pressure in the balloon is only 0.10 atm, though, the gas volume equals only $$1.000 \times 10^4 L $$at 1 atm.
(1.000 × 10⁴ L) / (5000. L/cylinder) = 2.00 cylinders.
$$\dfrac{1.000\times 10^4 L}{5000L}$$ /cylinder $$=2$$ cylinders
Question 10
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A gaseous mixture containing $$ 0.35_{g} $$ of $$ N_{2} $$ and 5600 ml of $$ O_{2} $$ at STP has kept in a 5 liters at 300K. The total temperature of the gaseous mixture is?

A
1.293 atm

B
1.2315 atm

C
12.315 atm

D
0.616 atm

Solution

Moles of $$O_2=\cfrac{5600}{22400}=0.25=n_{O_2}$$
Moles of $$N_2=\cfrac{0.35}{28}=0.0125=n_{N_2}$$
Initial pressure of $$N_2=\cfrac{n_{N_2}RT}{V}=\cfrac{0.0125\times 0.0821\times 300}{5}=0.061\text{ }atm$$
Potential pressure of $$O_2=\cfrac{n_{O_2}RT}{V}=\cfrac{0.25\times 0.0821\times 300}{5}=1.231\text{ }atm$$
$$\therefore$$ Total pressure $$=1.231+0.061=1.292\text{ }atm$$

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States of Matter Test - 40

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States of Matter Test - 40

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Question 1

$${ N }_{ 2 }$$

$${ O }_{ 2 }$$

$${ H }_{ 2 }O$$

$${ CO }_{ 2 }$$

Solution

Question 2

$$P(CH_{4}) > P(SO_{2}) > P(H_{2})$$

$$P(H_{2}) > P(CH_{4}) > P(SO_{2})$$

$$P(SO_{2}) > P(CH_{4}) > P(H_{2})$$

$$P(H_{2}) > P(SO_{2}) > P(CH_{4})$$

Solution

Question 3

$$N/2$$

$$N$$

$$2N$$

$$4N$$

Solution

Question 4

By heating the gas w double the temperature

By removing 75% of the gas at constant volume we decrease the pressure to one quarter of its original value

By heating the gas we quadruple the pressure

By pumping in more gas at constant temperature we quadruple the pressure.

Solution

Question 5

increases, number of moles; P & T

decreases, number of moles; P & T

increases; T & P; number of moles

decreases; P & T; number of moles

Solution

Question 6

below boiling point

below critical temperature

below inversion temperature

above critical temperature

Solution

Question 7

$$157.5$$ of $$Hg$$

$$175.5$$ of $$Hg$$

$$315.0$$ of $$Hg$$

$$257.5$$ of $$Hg$$

Solution

Question 8

$$E$$ and $$H$$ each depend only on $$T$$

$$C_p$$ is constant

$$PdV = nRdT$$ for every infinitesimal process

Both $$(A)$$ and $$(C)$$ are true

Solution

Question 9

$$ 2 $$

$$ 3 $$

$$ 4 $$

$$ 1 $$

Solution

Question 10

1.293 atm

1.2315 atm

12.315 atm

0.616 atm

Solution

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