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The Gas Laws - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Gas laws(I) is considered one of the most asked concept.

  • 21 Questions around this concept.

Solve by difficulty

A think tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm containing mercury and two equal ends containing air at standard atmospheric pressure. if the tube is now turned to a vertical position, by what amount will the mercury be displaced? 

(Given: cross-section of the tube can be assumed to be uniform)

For a perfect gas, two pressures $P_1$  and $P_2$ are shown in figure. The graph shows :

For an ideal gas following the isochoric process if $T_1=127^{\circ} \mathrm{C}_{\&} P_2=5$ bar then what is the value of $\left(\frac{P_1 T_2}{P_2 T_1}\right)$

  An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross-section 8.0 x 10-3 m2. Initially, the gas is at 300K and occupies a volume of 2.4 x 10-3 m3 and the spring is in its relaxed state as shown in the figure. The gas is heated by a small heater until the piston moves out slowly by 0.1 m. The force constant of the spring is 8000 N/m and the atmospheric pressure is 1.0 x 105 N/m2. The cylinder and the piston are thermally insulated.  The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :

(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

Choose the correct option 

a) Boyle's law                               1)    

 

 

b) Charle's law                           2) 

 

 

 

c) Gag Lussac's Law                 3)  

A thermodynamic cycle $x y z x$ is shown on a V - T diagram.

The P - V diagram that best describe this cycle is : (Diagrams are schematic and not to scale)

 

A cylindrical tube of uniform cross-sectional area $A$ is fitted with two air-tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the pressure of the gas is $P_0$ and temperature is $T_0$, atmospheric pressure is also $P_0$.Now, the temperature of the gas is increased to $2T_0$, the tension of the wire will be

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"At constant volume, the pressure P of a given mass of a gas is directly proportional to its absolute temperature T " 

This statement  is called 

If initially 1 mole of ideal gas in a container has a volume $V_1$. Then if we add same gas in the container keeping its pressure and tempreture constant, till its volume become trice of initial volume. Then how many no. of moles we have added

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A gas expands obeying the relation as shown in the \mathrm{P / V} diagram. The maximum temperature in this process is equal to


 

Concepts Covered - 3

Gas laws(I)

BOYLE'S LAW
Boyle's law : It states that, for a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure.

$
\begin{aligned}
V & \propto \frac{1}{P} \\
\text { or, } \quad P . V & =\text { constant } \\
\Rightarrow P_1 V_1 & =P_2 V_2
\end{aligned}
$


We can also write the above equation as,

$
\begin{aligned}
P V & =P\left(\frac{m}{\rho}\right)=\text { constant } \\
\Rightarrow \quad \frac{P}{\rho} & =\text { constant or } \frac{P_1}{\rho_1}=\frac{P_2}{\rho_2}
\end{aligned}
$


We can represent the Boyle's law through the various graph, which is shown as -

                                                  

 

CHARLE'S LAW - 

Charle's law : It states that, if the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

From the above statement we can conclude the following equations - 

                                                                           

$
\begin{aligned}
\boldsymbol{V} & \propto \boldsymbol{T} \\
\frac{V}{T} & =\text { Constant } \\
\frac{V_1}{T_1} & =\frac{V_2}{T_2}
\end{aligned}
$


This equation can also be written in terms of density and temperature as -

$
\frac{V}{T}=\frac{m}{\rho T}=\text { constant }\left(\text { As volume } V=\frac{m}{\rho}\right)
$

or, $\quad \rho T=$ constant $\Rightarrow \rho_1 \mathbf{T}_1=\rho_2 \mathbf{T}_2$

We can represent the Charle's law through the various graph, which is shown as -

 

                         

 

 

 

Gas laws(II)

Gay-Lussac’s law  - 

Gay-Lussac’s law or pressure law : If the volume remains constant, then the pressure of a given mass of a gas is directly proportional to its absolute temperature.

So, We can conclude the above statement in the following equation - 

 

                                                 $P \propto T$ or $\frac{P}{T}=$ constant $\Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2}$

The graphical representation of Gay-Lussac's law is - 

 

                                               

 

AVAGADRO'S LAW - 

Avogadro’s law : Equal volume of all the gases under similar conditions of temperature and pressure contain equal number of molecules. It implies that - 

                                                               $N_1=N_2$
N = Number of molecules in a particular gas.

Gas laws(III)

GRAHAM’S LAW OF DIFFUSION 

Graham’s law of diffusion: It states that when any two gases at the same pressure and temperature are allowed to diffuse into each other, then the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas.

So we can say that, 

                                                               

$$
r \propto \frac{1}{\sqrt{\rho}} \propto \frac{1}{\sqrt{M}} \alpha V_{r m s}
$$


Where, $r=$ rate of diffusion of gas

$$
\begin{aligned}
& \rho=\text { Density of the gas } \\
& \mathrm{M}=\text { Molecular weight of the gas } \\
& V_{r m s}=\text { Root mean square velocity }
\end{aligned}
$$


Now, from the above equation we can write,

$$
\frac{r_1}{r_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt{\frac{M_2}{M_1}}
$$


DALTON'S LAW OF PARTIAL PRESSURE -
Dalton's law of partial pressure :It states that the total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gases exert if it alone occupied the same volume at a given temperature.

Now, let us have a mixture of ' $n$ ' gases, so from the above statement we can conclude that -

$$
\text { For } n \text { gases } P=P_1+P_2+P_3+\ldots \ldots P_n
$$


Here, $\mathrm{P}=$ Pressure exerted by the mixture of gases

$$
P_1, P_2 \ldots \ldots P_n=\text { Partial pressure of the component gases. }
$$
 

 

 

 

 

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Gas laws(I)
Gas laws(II)
Gas laws(III)

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Books

Reference Books

Gas laws(I)

Physics Part II Textbook for Class XI

Page No. : 321

Line : 31

Gas laws(II)

Physics Part II Textbook for Class XI

Page No. : 319

Line : 44

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