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Pressure Of An Ideal Gas - Practice Questions & MCQ

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

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  • 17 Questions around this concept.

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For Brownian motion of particle, match columns I and II.

               Factor                                                                    Effect

     (i) Decrease in size of Brownian particle            (P) Increase of Brownian motion

      (ii) Decrease in density of medium                    (Q) Decrease of Brownian motion

       (iii) Increase in temperature of medium             (R) Unaffected

       (iv) Increase in viscosity of the medium  

  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 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)

A monoatomic ideal gas is expanded adiabatically to \mathrm{n} times its initial volume, The ratio of the final rate of collision of molecules with a unit area of container walls to the initial rate will be

A gas at pressure \mathrm{P_{0}} is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

 The volume \mathrm{V}versus temperature \mathrm{T} graphs for a certain amount of a perfect gas at two pressures \mathrm{P_{1}} and \mathrm{P_{2}} are shown in the figure. Here

Gas at pressure \mathrm{P_{0}}  is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure \mathrm{P} will be equal to

When an ideal gas at pressure P, temperature T and volume V is isothermally compressed to a \mathrm{V / n}, its pressure becomes \mathrm{P_{i}}.If the gas is compressed adiabatically to \mathrm{V} / \mathrm{n},  its pressure becomes \mathrm{P}_{\mathrm{a}}.The ratio \mathrm{P}_{\mathrm{i}} / \mathrm{Pa} is

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An ideal gas is initially at temperature \mathrm{T} and volume \mathrm{V}. It volume is increased by \Delta \mathrm{V} due to an increase in temperature \Delta \mathrm{T}, pressure remaining constant. The quantity \delta=\Delta \mathrm{V} / \mathrm{V} \Delta \mathrm{T} varies with temperature as :

An ideal gas expands isothermally from a volume \mathrm{V}_{1} and \mathrm{V}_{2} and then compressed to original volume \mathrm{V}_{1} adiabatically. Initial pressure is \mathrm{P}_{1} and final pressure is \mathrm{P}_{3}.The total work done is \mathrm{W}Then:

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A cylindrical tube of uniform cross-sectional area \mathrm{A} is fitted with two air tight frictionless pistons. The piston are connected to each other by a metallic wire. Initially, the pressure of the gas is \mathrm{p_{0}} and temperature is \mathrm{T_{0}}, atmospheric pressure is also \mathrm{p_{0}}.Now, the temperature of the gas is increased to \mathrm{2 T_{0}}, the tension of the wire will be

Concepts Covered - 1

Pressure of an ideal gas

Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L as shown in the below figure.

1. Instantaneous velocity- 

Any molecule of gas moves with velocity \vec{v} in any direction

\text { where } \quad \vec{v}=v_{x} \hat{i}+v_{y} \hat{j}+v_{z} \hat{k}

And Due to the random motion of the molecule

v_{x}=v_{y}=v_{z}

  \\ As \ v=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}\\ \Rightarrow v=3v_x^2=3v_y^2=3v_z^2

2. The time during a collision-Time between two successive collisions with the wall A1

     \\ I.e \ \ \Delta t=\frac{\text { Distance travelled by molecule between two successive collision }}{\text { Velocity of molecule }}\\ or \ \ \ \ \ \ \ \ \Delta t=\frac{2L}{v_x}

3.  Collision frequency (n): It means the number of collisions per second.

         I.e \ \ n=\frac{1}{\Delta t}=\frac{v_{x}}{2 L}

4.  Change in momentum: This molecule collides with A1  wall (A1) with velocity vx and rebounds with velocity (-vx)
  The change in momentum of the molecule is given by 

        \Delta p=\left(-m v_{x}\right)-\left(m v_{x}\right)=-2 m v_{x}

As the momentum remains conserved in a collision,

 \Delta p_{system}=0\\ \Delta p_{system}=\Delta p_{molecule}+\Delta p_{wall}=0\\ \Delta p_{wall}=-\Delta p_{molecule}

the change in momentum of wall A1 is \Delta p= 2 m v_{x}

5. Force on the wall: Force exerted by a single molecule on the A1 wall is equal to the rate at which the momentum is transferred to the wall by this molecule.

 \text { i.e. } F_{\text {Single molecule }}=\frac{\Delta p}{\Delta t}=\frac{2 m v_{x}}{\left(2 L / v_{x}\right)}=\frac{m v_{x}^{2}}{L}

The total force on the wall A1 due to N molecules

F_{x}=\frac{m}{L} \sum v_{x}^{2}=\frac{m}{L}\left(v_{x_{1}}^{2}+v_{x_{2}}^{2}+v_{x_{3}}^{2}+\ldots\right)=\frac{m N}{L} \overline {{v_{x}^{2}}}

where  \overline{v_{x}^{2}}=\text { mean square of } x \text { component of the velocity. }

6. Pressure-As pressure is defined as force per unit area, hence the pressure on  A1 wall

P_{x}=\frac{F_{x}}{A}=\frac{m N}{A L} \overline{v_{x}^{2}}=\frac{m N}{V} \overline{v_{x}^{2}}

\\ As \ \overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}} \\ So \ \ \overline{v^{2}}=\overline{v_{x}^{2}}+\overline{v_{y}^{2}}+v_{z}^{2}\\ \Rightarrow \overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}}=\frac{\overline{v^{2}}}{3}

So Total pressure inside the container is given by 

\left.P=\frac{1}{3} \frac{m N}{V}\overline {{v}^{2}}=\frac{1}{3} \frac{m N}{V} v_{rm s}^{2} \quad \text { (where } v_{rm s}=\sqrt{\overline{ v^{2}}}\right)

Using total mass= M=mN

 Pressure due to an ideal gas is given as

P= \frac{1}{3}\rho \: v_{rms}^{2}  = \frac{1}{3}\left ( \frac{M}{V} \right ) \cdot \: v_{rms}^{2}

where 

m= mass of one molecule

N = Number of the molecule

v_{rms}^{2}= \frac{v_{1}^{2}+v_{2}^{2}+..........}{n}

 v_{rms} = RMS velocity

 

 

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Pressure of an ideal gas

Physics Part II Textbook for Class XI

Page No. : 331

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