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

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

  1. V/T $=$ k
  2. P/T $=$ k
  3. PV $=$ k
  4. $P _T=P _1+P _2+P _3$
  5. PT $=$ k
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
E Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

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: 

  1. 1 : 1 : 1 : 1

  2. 2 : 1 : 2 : 3

  3. 1 : 2 : 1 : 3

  4. 3 : 2 : 2 : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $0.224 l$
  2. $22.4 l$
  3. $2240 l$
  4. $2.24 l$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 22.4\times0.01$
                                            $= 0.224l$

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

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?

  1. $533$ mm
  2. $473$ mm
  3. $335$ mm
  4. $595$ mm
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

Chlorofluorocarbons such as $CCl _3F\ (M=137.5)$ and $CCl _2F _2\ (M=121)$ have been linked to ozone depletion in Antarctica. As of $2004$, these gases were found in $275$ and $605$ parts per trillion $(10^{12})$, ny volume. What are the concentrations of these gases under conditions typical of Antarctica stratosphere ($200\ K$ and $0.08\ atm$)? $(R=0.08\ l-atm/K-mol)$   

  1. $[CCl _3F]=1.375\times 10^{-12}\ mol\ l^{-1}$,
    $[CCl _2F _2]=3.025\times 10^{-12}\ mol\ l^{-1}$
  2. $[CCl _3F]=2.75\times 10^{-14}\ mol\ l^{-1}$,
    $[CCl _2F _2]=6.05\times 10^{-14}\ mol\ l^{-1}$
  3. $[CCl _3F]=2.75\times 10^{-10}\ mol\ l^{-1}$,
    $[CCl _2F _2]=6.05\times 10^{-10}\ mol\ l^{-1}$
  4. $[CCl _3F]=1.375\times 10^{-13}\ mol\ l^{-1}$,
    $[CCl _2F _2]=3.025\times 10^{-12}\ mol\ l^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$CCl _3F(mw)=133.5$
$CCl _2F _2(mw)=121$
By given data first we calculate density of solution
$\dfrac {P}{S}=\dfrac {RT}{Mw}\quad R=0.08\quad T=200$
$P=0.08$
for $CCl _3F-S(CCl _3F)=\dfrac {0.08\times 133.5}{0.08\times 200}$
$S=0.6875$
for $CCl _2F _2=\dfrac {0.08\times 121}{0.08\times 200}$
$=0.605$
given data in $(PPt)$
$10^{12}\ gm$ solution contains $=175\ gm$
$10^{3}\ gm$ solution contains $=\dfrac {275}{10^{12}}\times 10^3$
Mass $=275\times 10^{-9}\ gm$
find the volume $d=\dfrac {M}{v}$
$v=\dfrac {1000}{0.6785}=1454.5$
Molarity of $CCl _3F=\dfrac {275\times 10^{-9}}{137.5\times 1454.5}$
$=1.37\times 10^{-12}$
Similarly for $CCl _2F _2$
$10^{12}\ gm$ solution certain $=605$
$10^{3}\ gm$ solution certain $=\dfrac {605}{10^{12}}\times 10^3$
Mass $=605\times 10^{-9}$
$d=\dfrac {m}{v}\quad v=\dfrac {m}{v}=\dfrac {1000}{605}$
$v=1652.89$
molarity $=\dfrac {605\times 10^{-9}}{121\times 1652.89}$
$M=3.02\times 10^{-12}\ mol\ l^3$

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

A quantity of $4$g of oxygen occupies $10$ L at a particular pressure and temperature. If the pressure of gas is doubled and absolute temperature is halved, in order to maintain constant volume.

  1. $3$ g gas should be removed from container
  2. $3$ g gas should be added in the container
  3. $16$ g gas should be added in the container
  4. $12$ g gas should be added in the container
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,
$m _1=4g$ $V _1=10L$,$P _1=P$,$T _1=T$

$m _2=?$, $V _2=10L$,$P _2=2P$, $T _2=\dfrac{T}{2}$

We know that,

$PV=\dfrac{m}{M}RT$

At constant volume, V

$\dfrac{P _1}{P _2}=\dfrac{m _1RT _1}{m _2RT _2}$

$\dfrac{P}{2P}=\dfrac{4\times T}{m _2\times \dfrac{T}{2}}$


$\dfrac{1}{2}=\dfrac{8}{m _2}$

$m _2=16$g

Mass of gas added into container$=16-4=12$g.