Tag: states of matter: gaseous and liquid states

Questions Related to states of matter: gaseous and liquid states

Law of definite proportions when expressed in terms of volumes becomes:

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. Dalton's Law

  2. Berzelius hypothesis

  3. Gay-Lussac's Law

  4. Avogadro's Law

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Correct Option: C
Explanation:

Law of definite proportions: It states that a given chemical substance (compound) always contains the same elements combined in a fixed proportion by weight. It is also called the law of constant composition.
The law of combining volumes was given by Gay Lussac in 1808. It is applicable only to a gaseous reaction. According to this law, gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure. So, the law of definite proportions when expressed in terms of volumes becomes 'Gay-Lussac's Law'. 
Avogadro's law states that 'equal volumes of all gases, at the same temperature and pressure, have the same number of molecules'.
According to Berzelius hypothesis, equal volumes of all gases contain an equal number of atoms under similar conditions of temperature and pressure.

For which of the following reactions, is Gay Lussac's law not applicable?

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. Formation of $HI$ from its constituents

  2. Formation of $NH _3$ from its constituents

  3. Formation of $CO _2$ from its constituents

  4. Formation of $SO _3$ from $SO _2$ and $O _2$

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Correct Option: C
Explanation:

Gay Lussac's law not applicable to formation of $CO _{2}$ from its constituents as carbon and oxygen exist in different physical states.

Which one is correct about  Gay-Lussac's law?

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. V/T $=$ k

  2. P/T $=$ k

  3. PV $=$ k

  4. $P _T=P _1+P _2+P _3$

  5. PT $=$ k

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Correct Option: B
Explanation:

According to Gay-Lussac's Law :


$\displaystyle \dfrac {P}{T}= k $ 

At constant volume, the pressure of a given mass of a gas varies directly with the temperature.

$\displaystyle P \propto T $ (V and n constant)

Hence, the correct option is B.

Which one proposes a relationship between the combining volumes of gases with respect to the reactants and gaseous products.

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. Avogadro's number

  2. $P _1V _1=P _2V _2$

  3. $V _1T _2=V _2T _1$

  4. Dalton's Theory

  5. Gay-Lussac's Law

Show answer
Correct Option: E
Explanation:

Gay-Lussac's Law proposes a relationship between the combining volumes of gases with respect to the reactants and gaseous products. According to this law, when gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure.

Gay Lussac's Law of combining volumes is applicable for combustion of carbon.

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. True

  2. False

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Correct Option: B
Explanation:
Gay Lussac's Law of Combining Volumes. Gay Lussac's Law of Combining Volumes states that whengases react, they do so in volumeswhich bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

The specific heat of a bivalent metal is 0.16. The approximate equivalent mass of the metal will be:

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. 40

  2. 20

  3. 80

  4. 10

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Correct Option: B
Explanation:

atomic weight = $\dfrac{6.4}{specific heat}$


atomic weight = 40
equivalent weight = $\dfrac{40}{2}$ (bivalent)

equivalent mass = 20

Equal volumes of different gases at any definite temperature and pressure have:

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. equal weights

  2. equal masses

  3. equal densities

  4. equal number of moles

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Correct Option: D
Explanation:

At equal volume of different gases at any definite temperature and pressure have equal no. of particles.

hence, equal number of moles.

Four flasks of 1 litre capacity each arc separately filled with gases $H _2, He, O _2$ and $O _3$. At the same temperature and pressure the ratio of the number of atoms of these gases present in different flasks would be: 

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. 1 : 1 : 1 : 1

  2. 2 : 1 : 2 : 3

  3. 1 : 2 : 1 : 3

  4. 3 : 2 : 2 : 1

Show answer
Correct Option: A
Explanation:

Ans ; A

Here in this question 4 different  types of gasses are filled in same volume of flasks i.e. all 4 types of gasses have same number of molecules.
THe ratio of number of atoms of these gasses present in different flasks would be = 1:1:1:1

The volume occupied by 0.01 moles of helium gas at STP is:

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. $0.224  l$

  2. $22.4  l$

  3. $2240  l$

  4. $2.24  l$

Show answer
Correct Option: A
Explanation:

As 1 mole of a gaseous substance occupies 22.4 litre ($22.4 l =$ gram molar volume)
Volume occupied $=GMV \times 0.01$( gram molar volume )

Hence, volume occupied $= 22.4\times0.01$
                                            $= 0.224l$

A mixture of CO and $CO _2$ has a density of $1.5 g/l $ at $27^o$C and $760$ mm pressure. If $1\ l$ of the mixture is exposed to alkali, what would be the pressure of the remaining gas at the same volume and temperature?

chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

  1. $533$ mm

  2. $473$ mm

  3. $335$ mm

  4. $595$ mm

Show answer
Correct Option: C
Explanation:

The given density$=d=\dfrac{m}{volume}$ …….$(1)$


so the given mass is a mixture of CO & $CO _2$

$m=n _{CO}\times 28+n _{CO _2}\times 44$

${$ using no. of moles$=\dfrac{mass}{mol. mass}}$

Also using $PV=nRT$
$V=\dfrac{(n _{CO}+n _{CO _2})RT}{P}$

Putting in $(1)$
$1.5=\dfrac{n _{CO}\times 28+44\times n _{CO _2}}{{(n _{CO _2}+n _{CO})\times 0.082\times 300}}$ $[{760$mm$=1$atm$}, P=1atm]$

$1.5=\dfrac{28n _{CO}+44n _{CO _2}}{24.6(n _{CO}+n _{CO _2})}$


$\Rightarrow 36.9(n _{CO}+n _{CO _2})=28n _{CO}+44n _{CO _2}$

$9n _{CO}=7n _{CO _2}$

After the reaction with alkali, all $CO _2$ will be used so the remaining pressure will be of $CO$.

$P _{CO _2}=\dfrac{n _{CO}}{n _{CO}+n _{CO _2}}\times P _{Total}$

$=\dfrac{n _{CO}}{n _{CO}+\dfrac{9}{7}n _{CO}}$

$=\dfrac{7}{16}P _{Total}=\dfrac{7}{16}\times 760$

$ \approx 335$ mm
Option C.