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Pressure Of An Ideal Gas MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 29 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  

Which graph shows the correct Relationship ( P= pressure, \rho = density)

Two non-reactive monoatomic ideal gases have partial pressure when enclosed in a cylinder at a constant temperature is  3: 4 and thin atomic masses is  3: 2 . what is the ratio of these Densities?

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In the process  P V =  constant, pressure (P) various density (\rho) graph of an ideal  gas is
 

( P= Pressure      T= Temperature)

(i)

(ii)

(iii)


choose the wrong answer.

From the P-T graph, what conclusion can be drawn?

16 gm of oxygen, 14 gm nitrogen and 11 gm of carbon dioxide are mixed in an enclosed volume of 5L and temp 27^{0}C. The pressure exerted by the mixture is - 

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Variation of internal energy with the density of one mole of monoatomic gas is depicted in the figure given below, corresponding variation of pressure with volume can be depicted as,

Pressure versus temperature graph of an ideal gas is as shown in figure. Density of the gas at point A is P0 . Density at point B will be

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The mass of hydrogen molecule is 3.32 × 10-27 . If 1023 hydrogen molecules strike per sound, a fixed wall of area 2 cm2 at an angle of 45° from normal and rebound back with the speed of 1000 m/sec, then the pressure exerted on the wall is

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}

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

As the momentum remains conserved in a collision, the change in momentum of the 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_{m s}^{2} \quad \text { (where } v_{m 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

Line : 15

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