MEDIUMUPES B.Tech Admissions 2025
Ranked #42 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements | Last Date to Apply: 28th April
Pressure Of An Ideal Gas - Practice Questions & MCQ
Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
-
18 Questions around this concept.
Solve by difficulty
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 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)
A monoatomic ideal gas is expanded adiabatically to 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
JEE Main 2025: Session 2 Result Out; Direct Link
JEE Main 2025: College Predictor | Marks vs Percentile vs Rank
New: JEE Seat Matrix- IITs, NITs, IIITs and GFTI | NITs Cutoff
Latest: Meet B.Tech expert in your city to shortlist colleges
A gas at pressure 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 V versus temperature 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 is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure
will be equal to
When an ideal gas at pressure $P$, temperature $T$ and volume $V$ is isothermally compressed to a $V / n$, its pressure becomes $\mathrm{P}_{\mathrm{i}}$. If the gas is compressed adiabatically to $\mathrm{V} / \mathrm{n}$, its pressure becomes $\mathrm{P}_{\mathrm{a}}$. The ratio $\mathrm{P}_{\mathrm{i}} / \mathrm{P_a}$ is
An ideal gas is initially at temperature $T$ and volume V . Its volume is increased by $\Delta \mathrm{V}$ due to an increase in temperature $\Delta \mathrm{T}$, with 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 $V_1$ and $V_2$ and then compressed to original volume $V_1$ adiabatically. Initial pressure is $\mathrm{P}_1$ and final pressure is $\mathrm{P}_3$. The total work done is W Then:
JEE Main 2025 College Predictor
Know your college admission chances in NITs, IIITs and CFTIs, many States/ Institutes based on your JEE Main rank by using JEE Main 2025 College Predictor.
Use NowA 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
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
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
$
\begin{aligned}
& v_x=v_y=v_z \\
& A s v=\sqrt{v_x^2+v_y^2+v_z^2} \\
& \Rightarrow v=3 v_x^2=3 v_y^2=3 v_z^2
\end{aligned}
$
2. The time during a collision-Time between two successive collisions with the wall $A_1$
$
\begin{aligned}
& \text { I.e } \Delta t=\frac{\text { Distance travelled by molecule between two successive collision }}{\text { Velocity of molecule }} \\
& \text { or } \quad \Delta t=\frac{2 L}{v_x}
\end{aligned}
$
3. Collision frequency $(\mathbf{n})$ : It means the number of collisions per second.
$
\text { I.e } n=\frac{1}{\Delta t}=\frac{v_x}{2 L}
$
4. Change in momentum: This molecule collides with $A_1$ wall (A1) with velocity $v_x$ and rebounds with velocity ( $-\mathrm{v}_{\mathrm{x}}$ ) 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,
$
\begin{aligned}
& \Delta p_{\text {system }}=0 \\
& \Delta p_{\text {system }}=\Delta p_{\text {molecule }}+\Delta p_{\text {wall }}=0 \\
& \Delta p_{\text {wall }}=-\Delta p_{\text {molecule }}
\end{aligned}
$
the change in momentum of wall $\mathrm{A}_1$ is $\Delta p=2 m v_x$
5. Force on the wall: Force exerted by a single molecule on the $\mathrm{A}_1$ wall is equal to the rate at which the momentum is transferred to the wall by this molecule.
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 $\mathrm{A}_1$ 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 $\overline{v_x^2}=$ mean square of $x$ component of the velocity.
6. Pressure-As pressure is defined as force per unit area, hence the pressure on $\mathrm{A}_1$ wall
$
\begin{aligned}
& P_x=\frac{F_x}{A}=\frac{m N}{A L} \overline{v_x^2}=\frac{m N}{V} \overline{v_x^2} \\
& \text { As } \overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2} \\
& \text { 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{v^2}{3}
\end{aligned}
$
So Total pressure inside the container is given by
$
P=\frac{1}{3} \frac{m N}{V} \overline{v^2}=\frac{1}{3} \frac{m N}{V} v_{r m s}^2 \quad\left(\text { where } v_{r m s}=\sqrt{\overline{v^2}}\right)
$
Using total mass $=\mathrm{M}=\mathrm{mN}$
Pressure due to an ideal gas is given as
$
P=\frac{1}{3} \rho v_{r m s}^2=\frac{1}{3}\left(\frac{M}{V}\right) \cdot v_{r m s}^2
$
where
$m=$ mass of one molecule
$\mathrm{N}=$ Number of the molecule
$
v_{r m s}^2=\frac{v_1^2+v_2^2+\ldots \ldots \ldots}{n}
$
$v_{r m s}=\mathrm{RMS}$ velocity
Study it with Videos
Pressure of an ideal gas"Stay in the loop. Receive exam news, study resources, and expert advice!"
Books
Reference Books
Clear your Basics with NCERT
E-books & Sample Papers
Get Answer to all your questions
JEE Main Articles
Back to top