States of Matter Test

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States of Matter Test - 61

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Question 1
1 / -0

If $$K.E.$$ body is increased by $$100%$$. Then $$%$$ change $$'P'$$:

A
$$50\% $$

B
$$41\% $$

C
$$10\% $$

D
$$20\% $$

Solution

We have $$KE=\dfrac { 1 }{ 2 } { MV }^{ 2 }\quad \longrightarrow \left( 1 \right) $$
& linear momentum $$P=MV\quad \longrightarrow \left( 2 \right) $$
Equation $$(1)$$ & $$(2)$$ $$\Rightarrow KE=\dfrac { { P }^{ 2 } }{ 2M } $$
Given $$KE=2K$$
Let new momentum be $${ P }^{ \ast }$$
$$\therefore$$ $$\dfrac { { K }^{ \ast } }{ K } ={ \left( \dfrac { { P }^{ \ast } }{ P } \right) }^{ 2 }$$
$$\dfrac { 2K }{ K } ={ \left( \dfrac { { P }^{ \ast } }{ P } \right) }^{ 2 }\Rightarrow { 2P }^{ 2 }={ P }^{ { \ast }^{ 2 } }$$
or $${ P }^{ \ast }=\sqrt { 2 } P$$
The momentum increases by $$\sqrt { 2 } =1.41$$ times
$$=\left( \dfrac { 1.41-1 }{ 1 } \right) \times 100=41$$%
Question 2
1 / -0

Vapour pressure of a pure liquid $$X$$ is $$2$$ atm at $$300$$K. It is lowered to $$1$$ atm on dissolving $$1$$g of $$Y$$ in $$20$$g of liquid $$X$$. If molar mass of $$X$$ is $$200$$, what is the molar mass of $$Y$$?

A
$$20$$

B
$$10$$

C
$$100$$

D
$$30$$

Solution

Given= $$P^o=2atm$$

$$\Rightarrow P^o-P= 1 atm$$
$$W_2= 1$$
$$W_1= 20g$$
$$M_1=200$$
$$M_2=?$$

The expression for the lowering of the vapour pressure is given by,

$$\cfrac {P^o-P}{P^o}=x_2$$

$$\Rightarrow \cfrac {1}{2}= \cfrac {W_2 \times M_1}{M_2 \times W_1}$$

$$\Rightarrow \cfrac {1}{2}=\cfrac {1 \times 200}{M_2 \times 20}$$

$$\Rightarrow M_2= \cfrac {200 \times 2}{20}=20$$

Molar mass of $$Y$$ is $$20$$ .

Hence, the correct option is $$A$$
Question 3
1 / -0

When $${ CO }_{ 2 }$$ under high pressure is released from a fire extinguisher, particles of solid $$C$$ despite the low sublimation temperature $$\left( -{ 77 }^{ 0 }C \right) $$ of $${ CO }_{ 2 }$$ at 1.0 atm. It is:

A
the gas does work pushing back the atmosphere using KE of molecules and thus temperature

B
volume of the gas is decreased rapidly hence, temperature is lowered

C
both $$A$$ and $$B$$

D
none of the above
Solution
- Carbon dioxide ($$CO_2$$) extinguishers contain a mixture of liquid and gaseous carbon dioxide (a nonflammable gas).
- $$CO_2$$ is normally a gas at room temperature and pressure. It has to be stored under high pressure to make it a liquid. When you release the pressure, the gas expands enormously and makes a huge white jet.
- $$CO_2$$ attacks the fire triangle in two ways: it smothers the oxygen and, when it turns from a liquid back to a gas, it "sucks" in a massive amount of heat from its surroundings (the latent heat of vaporization), which cools whatever you spray it on by removing heat.
- $$CO_2$$ is designed for Class B and C (flammable liquid and electrical) fires only.
- When $$CO_2$$ under high pressure is released from a fire extinguisher, particles of solid C despite the low sublimation temperature it is the gas does work pushing back the atmosphere using KE of molecules and thus temperature.
- Hence option A is correct answer.
Question 4
1 / -0

What is the total pressure exerted by the mixture of 7.0 g of $$N_2$$, 2g of hydrogen and 8.0 g of sulphur dioxide gases in a vessel of 6.1L capacity that has been kept in a reservoir at $$27^oC$$?

A
2,5 bar

B
4.5 bar

C
10 atm

D
5.7 bar

Solution

Total moles of gases $$=\left(\dfrac{7}{28}+\dfrac{2}{2}+\dfrac{8}{64}\right)=\left(\dfrac{1}{4}+\dfrac{1}{8}+1\right)=\dfrac{11}{8}$$ moles.
volume of vessel $$=6.1 \mathrm{~L}$$
and, temperature

$$=27^{\circ}\mathrm{C}=300\mathrm{k}$$.
By ideal gas equation, $$\quad P V=n R T$$.
$$\begin{aligned}\Rightarrow P=\dfrac{n R T}{V} &=\dfrac{11 \times 0.0831 . x 300}{6.1} \mathrm{bar} .\\& \approx 5.7 \mathrm{bar} .\end{aligned}$$
Thus, total pressure exerted by the mixture $$=5.7$$ bar.
so, option (D) is correct.
Question 5
1 / -0

A $$4.40\ g$$ piece of solid $$CO_{2}$$ is allowed to sublime in a balloon. The final volume of the balloon is $$1.00L$$ at $$300K$$. What is the pressure of the gas?

A
$$0.122$$

B
$$122$$

C
$$2.46$$

D
$$24.6$$

Solution

moles of $$C{ O }_{ 2 }=\dfrac { 4.40 }{ 44 } =0.1$$

$$P=0.1\times 0.0821\times \dfrac { 300 }{ 1.00 } =2.46\quad atm$$
Question 6
1 / -0

$${N}_{2}+3{H}_{2}\rightleftharpoons 2{NH}_{3}$$. Starting with one mole of nitrogen and $$3$$ moles of hydrogen, at equilibrium $$50$$% of each had reacted. If the equilibrium pressure is $$P$$, the partial pressure of hydrogen at equilibrium would be:

A
$$P/2$$

B
$$P/3$$

C
$$P/4$$

D
$$P/6$$

Solution

Initial mole ratio N2:H2 is 1:3
At equilibrium, 50% of N2, H2 has reacted and equilibrium pressure is P
$${ N }_{ 2 }+3{ H }_{ 2 }\rightarrow 2N{ H }_{ 3 }\\ y\quad \quad \quad 3y\quad \quad \quad \quad 2y$$
At equilibrium, 50% of each reactant reacted, the number of moles will become half.
The mole of N2=$$y-\frac { y }{ 2 } $$
The moles of H2=$$3y-\frac { 3y }{ 2 } $$
THE moles of NH3=$$\frac { 2y }{ 2 } $$
Total moles at equilibrium=$$\frac { y }{ 2 } +\frac { 3y }{ 2 } +\frac { 2y }{ 2 } =3y$$
Partial pressure of H2=$$\frac { Moles\quad of\quad H2 }{ Total\quad moles } \times \quad Total\quad pressure$$=$$\frac { 3y }{ 2 } \times \frac { 1 }{ 3y } \times \quad P$$=$$\frac { P }{ 2 } $$
Question 7
1 / -0

Two liquids X and Y form an ideal solution.
The mixture has a vapour pressure of 400mm at 300K, when mixed in the molar ratio of 1:1 and a vapour pressure of 350mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are?

A
250mm, 550mm

B
350mm, 450mm

C
350mm, 700mm

D
550mm, 250mm

Solution

Total pressure
$$P=pA^0\dfrac{2}{A}+pB^0\dfrac{2}{B}$$
for case (i)
$$400p=pA^0\dfrac{1}{2}+pB^0\dfrac{1}{2}$$ $$\{molar\, ratio\theta=1:1\}$$
$$pA^0+pB^0=800$$
for case (ii)
$$350=pA^0\dfrac{1}{3}+pB^0\dfrac{2}{3}$$ $$\{molar\,ration\theta=1:2\}$$
$$pA^0+2pB^0=1050$$.....(2)
from (1) and (2)
$$pA^0=550mm$$ $$pB^0=250\ mm$$
Question 8
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A volume $$V$$ of a gas at a temperature $$T_{1}$$ and a pressure $$p$$ is enclosed in a sphere. It is connected to another sphere of volume $$V/2$$ by a tube and stopcock is opened the temperature of the gas in the second sphere becomes $$T_{2}$$. The first sphere is maintained at a temperature $$Y_{1}$$. What is the final pressure $$p_{1}$$ within the apparatus?

A
$$\dfrac {2pT_{2}}{2T_{2}+T_{1}}$$

B
$$\dfrac {2pT_{2}}{T_{2}+2T_{1}}$$

C
$$\dfrac {pT_{2}}{2T_{2}+T_{1}}$$

D
$$\dfrac {2pT_{2}}{T_{1}+T_{2}}$$

Solution

We have the ideal gas equation as follow:-
$$PV={ nRT }$$
The number of moles of gases can be written as:-
$$\\ { n }=\frac { PV }{ RT } $$
Number of moles present in 1st sphere before stopcock open is
$$\\ { n }=\frac { pV }{ R{ T }_{ 1 } }$$
Number of moles present in 1st sphere after stopcock open is
$$ \\ { n }_{ 1 }=\frac { { p }_{ 1 }V }{ { RT }_{ 1 } } $$
Number of moles present in 2st sphere before stopcock open is
$$\\ { n }_{ 1 }=\frac { { { p }_{ 2 } }\frac { V }{ 2 } }{ R{ T }_{ 2 } } $$
Now we have,
$$\\ { n }{ ={ n }_{ 1 } }{ +{ n }_{ 2 } }$$

$$\\ \frac { pV }{ { RT }_{ 1 } } =\frac { p_{ 1 }V }{ { RT }_{ 1 } } +\frac { { p }_{ 1 }V }{ { 2RT }_{ 2 } } $$
After solving above expression we get final pressure as
$$\\ { p }_{ 1 }=\frac { { 2pT }_{ 2 } }{ { 2T }_{ 2 }+{ T }_{ 1 } } $$
Correct option is A
Question 9
1 / -0

A $$2.24 L$$ cylinder of oxygen at $$1atm$$ and $$273 K$$ is found to develop a leakage. When the leakage was plugged the pressure dropped to $$570 mm$$ of Hg. The number of moles of gas that escaped will be:

A
$$0.025$$

B
$$0.050$$

C
$$0.075$$

D
$$0.09$$

Solution

Original pressure, $$\left( {P}_{i}\right) = 1 \text{ atm} = 760 \text{ mmHg}$$
Final pressure, $$\left( {P}_{f} \right) = 570 \text{ mmHg}$$

$$\therefore$$ Drop in pressure, $$\left( P \right) = {P}_{i} - {P}_{f} = \left( 760 - 570 \right) \text{ mmHg} = 190 \text{ mmHg} = \cfrac{190}{760} \text{ atm}$$
Volume, $$\left( V \right) = 2.24 L$$
$$R = 0.0821 \text{ L atm } {K}^{-1} {mol}^{-1}$$
Temperature, $$\left( T \right) = 273 K$$

According to ideal gas equation,
$$PV = nRT$$
$$\Rightarrow \; n = \cfrac{PV}{RT}$$
$$\Rightarrow \; n = \cfrac{\left( \cfrac{190}{760} \right) \times 2.24}{0.0821 \times 273} = 0.02498 = 0.025 \text{ moles}$$

Hence the number of moles of gas that escaped will be $$0.025 \text{ moles}$$.
Question 10
1 / -0

Equal weights of methane and hydrogen are mixed in an empty container at $${25}^{o}C$$. The fraction of the total pressure exerted by hydrogen is:

A
$$\cfrac{1}{2}$$

B
$$\cfrac{8}{9}$$

C
$$\cfrac{1}{9}$$

D
$$\cfrac{16}{17}$$

Solution

Molar ratio of $$C{ H }_{ 4 }$$ : $${ H }_{ 2 }$$ = 16 : 2 = 8 : 1 (Mixed in equal quantities)
Pressure is directly proportional to a number of moles.
Methane is 8/9 of total no. of moles.
So,
Fraction of pressure by $$C{ H }_{ 4 }$$ = 8/9 .
Hence $$\dfrac { 8 }{ 9 } $$ is the answer.

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 61

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States of Matter Test - 61

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SHARING IS CARING

Question 1

$$50\% $$

$$41\% $$

$$10\% $$

$$20\% $$

Solution

Question 2

$$20$$

$$10$$

$$100$$

$$30$$

Solution

Question 3

the gas does work pushing back the atmosphere using KE of molecules and thus temperature

volume of the gas is decreased rapidly hence, temperature is lowered

both $$A$$ and $$B$$

none of the above

Solution

Question 4

2,5 bar

4.5 bar

10 atm

5.7 bar

Solution

Question 5

$$0.122$$

$$122$$

$$2.46$$

$$24.6$$

Solution

Question 6

$$P/2$$

$$P/3$$

$$P/4$$

$$P/6$$

Solution

Question 7

250mm, 550mm

350mm, 450mm

350mm, 700mm

550mm, 250mm

Solution

Question 8

$$\dfrac {2pT_{2}}{2T_{2}+T_{1}}$$

$$\dfrac {2pT_{2}}{T_{2}+2T_{1}}$$

$$\dfrac {pT_{2}}{2T_{2}+T_{1}}$$

$$\dfrac {2pT_{2}}{T_{1}+T_{2}}$$

Solution

Question 9

$$0.025$$

$$0.050$$

$$0.075$$

$$0.09$$

Solution

Question 10

$$\cfrac{1}{2}$$

$$\cfrac{8}{9}$$

$$\cfrac{1}{9}$$

$$\cfrac{16}{17}$$

Solution

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