Tag: gas laws

Quizzes (2) Questions (25)

Questions Related to gas laws

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

Which gas law should be used to solve a gas problem in which only the pressure and temperature of a gas change?

  1. Boyle's Law

  2. Ideal Gas Law

  3. Combined Gas Law

  4. Gay-Lussac's Law
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gay-Lussac's law states that for a given mass and constant volume of  an ideal gas, the pressure exerted on the sides of container is directly propotional to  absolute temperature.

So, for pressure-temperature graph, we use Gay-Lussac's law.

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

The ratio of volumes of $H _2$ and $O _2$ liberated on electrolysis of water is

  1. $1 : 2$
  2. $1 : 3$
  3. $2 : 1$
  4. $3 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$H _{2}O \rightarrow H^{+} +OH^{-}$
Cathode  :  4$OH^{-} \rightarrow 2H _{2}O +O _{2}+4e^{-} $
Anode :  4$H^{+} +4e^{-} \rightarrow 2H _{2} $
The ratio of volumes of H$ _{2}$ and O$ _{2}$ is $2 : 1$.

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

A glass container is sealed with a gas at $0.800$ atm pressure and at $25^o$C. The glass container sustain pressure of $2$ atm. Calculate the temperature to which gas can be heated before bursting the container.

  1. $432^o$C.
  2. $472^o$C.
  3. $372^o$C.
  4. $572^o$C.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have,

 $p _1=0.8$atm$,p _2=2$atm $T _1=298$k

Thus,

According to the Gas equation,

$\dfrac{p _1}{T _1}=\dfrac{p _2}{T _2}$

$\dfrac{0.8}{298}=\dfrac{2}{T _2}$

$T _2=745$k

$=472^0$C

Hence, the correct option is $\text{B}$
Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of ${ N } _{ 2 },{ H } _{ 2 }$ and $ { NH } _{ 3 }$ is $1:3:2$. This illustrates law of :

  1. definite proportions

  2. multiple proportions

  3. reciprocal proportions

  4. gaseous proportions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As per the Gay Lussac's law of combining volume of gases, the volumes of gaseous reactants and gaseous products bear a simple whole number ratio with each other if they are measured at same temperature and pressure.
In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of $N _2,  H _2$ and $ NH _3$ is $1:3:2$. This illustrates law of Gay Lussac's law of combining volumes of gases.

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

The volume of a gas is held constant while its temperature is raised. The pressure of the gas exerts on the walls of the container increases because:

  1. The masses of the molecules increases

  2. Each molecule loses more kinetic energy when it strikes the wall

  3. The molecules collide with the wall with relatively greater momentum

  4. The molecules strikes the walls more often
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Volume is kept constant and temperature increased.
We have relation, $PV=nRT$
$P\propto T$
Volume constant
So, $P\propto T$ pressure increases as temperature increases.
As we know the speed or average molecular speed depend on temperature.
As temperature increases molecules collide with wall with greater force or momentum.
Also temperature increases, molecular vibrations increases. So molecule will strike with wall more often.
So option C and D.
A is incorrect because there is no change in masses of molecule when we increase temperature.

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

A pre-weighed vessel was filled with oxygen at $NTP$ and weighed. It was then evacuated, filled with $SO {2}$ at the same temperature and pressure and again weighed. The weight of oxygen is _____________.

  1. the same as that of $SO _{2}$
  2. $\dfrac{1}{2}$ that of $SO _{2}$
  3. twice that of $SO _{2}$
  4. $\dfrac{1}{4}$ that of $SO _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant temperature and pressure, an equal volume of gas contains an equal number of moles.


$\therefore$ Moles of $O _{2}=$ Moles of $SO _{2}$
$1\ mole\ O _{2}=32\ g$
$1\ mole\ SO _{2}=64\ g$
$\therefore wt$ of $O _2=\dfrac{1}{2}$ that of $SO _{2}$
Option $B$ is correct.

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

A certain vessel $X$ has water and nitrogen gas at a total pressure of 2 $atm$ at 300 $K$. All the contents of the vessel are transferred to another vessel $Y$ having half the capacity of the vessel $X$.The pressure of ${N} _{2}$ in this vessel was 3.8 $atm$ at 300 $K$.The vessel $Y$ is heated to 320 $K$ and the total pressure observed was 4.32 $atm$. The pressure of ${N} _{2}$ at 320 $K$ is :

[Assume that the volume occupied by the gases in the vessel is equal to the volume of the vessel.]

  1. 4.0

  2. 4.05

  3. 5.05

  4. 1.05
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using PV = nRT, the moles of N2 are constant. Initial state in X: P_total = 2 atm. Since volume of Y is half of X, P_N2 in Y at 300K = 2 * P_N2_initial. Given P_N2 in Y = 3.8 atm, P_N2_initial = 1.9 atm. At 320K, P_N2_new = P_N2_old * (T2/T1) = 3.8 * (320/300) = 4.0533 atm.