Tag: behaviour of perfect gas and kinetic theory of gases

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Questions Related to behaviour of perfect gas and kinetic theory of gases

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The total Kinetic energy of $1\ mole$ of ${N}^{} _{2}$ at $27^{o} _{}{C}$ will be approximately :-

  1. $1500\ J$

  2. $15633\ cal$

  3. $1500\ kcal$

  4. $1500\ erg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $n$ mole of any gas the total  kinetic energy is given as $E=\dfrac{3}{2}nRT$

Where $R$ is gas constant having value $8.31J/mole-K$ or $8.31\times 4.18 cal /mole-K=34.74\text{Cal per mole per Kelvin}$
$T$ is temperature in Kelvin which is $T=27+273=300K$
So putting all values we get $E=1.5\times 1 \times 34.74\times 300=15633Calorie$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes $5.66$V while its temperature falls to $T/2$. How many degrees of freedom do the gas molecules have?

  1. 7

  2. $5$.

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Adiabatic equation of perfect gas is given as $TV^{r-1}=$ constant
$m=T _{1}V _{1}^{(r-1)}=T _{2}V _{2}^{(r-1)}$
$T _{1}=T _{1}V _{1}=V _{1}V _{2}=5.66\ V$
and $T _{2}=\dfrac{T}{2}$
$TV^{r-1}=\dfrac{T}{2}(5.66\ V)^{r-1}$
$2=5.66^{r-1}$
Taking $\log$ on both sides
$(r-1)\log 5.66=\log 2(r-1)$
$r=\dfrac{\log 2}{\log 5.66}=1+0.3010/0.75$
$r=1+0.4$
$r=1.4$ for $r=1.4$ Agree of freedom $=5$
Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The total kinetic energy of $1$ mole of $N _2$ at $27$C will be approximately

  1. <span><span><span class="mrow"><span class="mn">3739.662 J</span></span></span></span>&nbsp;

  2. 1500 calorie

  3. 1500 kilo calorie

  4. 1500 erg.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The kinetic enrgy of one mole is given by:

KE=$\dfrac{3}{2} K _BT$
The kinetic enrgy of 1 mole of $N _2$ atoms is:
KE=$\dfrac{3}{2}K _B T$ where $N$ is Avogadro's number,$K _B$ is Boltzmann's constant and $T$ is temperature
KE=$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$
$=3739.662 J$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The de-Broglie wavelength of a particle accelerated with $150\ volt$ potential is $10^{-10}\ m$. If it accelerated by $600\ volts$ p.d. its wavelength will be

  1. $0.25\ A^{o}$

  2. $0. 5\ A^{o}$

  3. $1.5\ A^{o}$

  4. $2\ A^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$

$\lambda \ \alpha \ \dfrac{1}{V}$

${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$

$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$

Divide (2) by (1)

$ \dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4} $

$ \Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}} $ 

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

Three particles are situated on a light and rigid rod along Y-axis as shown in the figure. If the system is rotating with angular velocity of $2 rad/sec$ about X axis, then the total kinetic energy of the system is :

  1. $92 J $

  2. $184 J $

  3. $ 276 J $

  4. $46 J $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

A gas has molar heat capacity $C = 4.5\ R$ in the process $PT = constant$. Find the number of degrees of freedom (n) of molecules in the gas.

  1. $n = 7$

  2. $n = 3$

  3. $n = 5$

  4. $n = 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a process PT = constant, P * (PV/nR) = constant, so P^2 * V = constant. This is a polytropic process PV^x = constant with x = 1/2. Molar heat capacity C = R/(gamma-1) + R/(1-x). 4.5R = R/(2/f) + R/(1-0.5) = fR/2 + 2R. 4.5 = f/2 + 2, so f/2 = 2.5, f = 5.

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The degrees of freedom of a triatomic gas is? (consider moderate temperature)

  1. $6$

  2. $4$

  3. $2$

  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general epression for degree of freedom is $DOF=3N-n$

here, DOF means degree of freedom, N is number of particle, and n is the number of holonomic constraints.
for a triatomic molecule, the number of particle is 3 and since the separation between three atoms are fixed so, the number of constraints is 3.
hence, $DOF=(3\times 3)-3$
$DOF=9-3$
$DOF=6$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

A vessel contains a non-linear triatomic gas. If $50$% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation 

  1. $2.15$

  2. $3.75$

  3. $5.25$

  4. $6.35$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let's assume we have $1$ mole of triatomic gas

$\therefore 3Na$ is present
So, $0.5$ moles= $1.5 Na$ atoms
$1$ part of $0.5$ moles remains untouched
Degree of dissociation= $0.5 \times 6=3$
Degree of freedom for $0.5Na= 1.5 \times 0.5=0.75$
Total=$3+0.75=3.75$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

For gas, if the ratio of specific heats at constants pressure $P$ and constant volume $V$ is $\gamma $, then the value of degree of freedom is:

  1. $\dfrac{\gamma +1}{\gamma -1}$

  2. $\dfrac{\gamma -1}{\gamma +1}$

  3. $\dfrac{1}{2}(\gamma-1)$

  4. $\dfrac{2}{\gamma-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer