Tag: behaviour of perfect gas and kinetic theory of gases

Quizzes (2) Questions (105)

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

A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

  1. 5

  2. 6

  3. 4

  4. 3.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant pressure, Q = n*Cp*deltaT and W = n*R*deltaT. Given Q = 4Q (this seems to be a typo in the prompt, likely Q_heat = 4*W). If Q_heat = 4*W, then Cp*deltaT = 4*R*deltaT, so Cp = 4R. Since Cp = (f/2 + 1)R, then f/2 + 1 = 4, f/2 = 3, f = 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

N moles of an ideal diatomic gas is contained in a cylinder at temperature $T.$ On supplying some heat to cylinder, $N/3$ moles of gas disassociated into atoms while temperature remains constant. Heat supplied to the gas is

  1. $\dfrac {NRT}{3}$

  2. $\dfrac {5NRT}{2}$

  3. $\dfrac {8NRT}{3}$

  4. $\dfrac {NRT}{6}$
Reveal answer Fill a bubble to check yourself
B 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

The heat capacity at constant volume of a sample of a monoatomic gas is $35\ J/K$. Find the number of moles.

  1. $12.81 \ \ mol  $

  2. <span>$21.81 \ \ mol &nbsp;$</span>

  3. <span>$4.81 \ \ mol &nbsp;$</span>

  4. <span>$2.81 \ \ mol &nbsp;$</span>
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For monoatomic gas, degrees of freedom is 3. 


Since ${ C } _{ V }=\dfrac { f }{ 2 } nR$

Hence, $35=\dfrac { 3 }{ 2 } n(8.314)$

$n=\dfrac { 70 }{ 3\times 8.314 } =2.81mol$

Answer is $2.81mol.$

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

Relation between pressure ($P$) and energy density ($E$) of an ideal gas is-

  1. $P=2/3E$

  2. $P=3/2E$

  3. $P=3/5E$

  4. $P=E$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Kinetic energy $=\dfrac{1}{2}{ MV } _{ rms }$
$\Rightarrow \dfrac{1}{2}M\left( \dfrac { 3RT }{ M }  \right) $        $[M=$ molar mass,$ { V } _{ rms }=\sqrt { \dfrac { 3KT }{ { m } }  } =\sqrt { \dfrac { 3RT }{ M }  } ]$
$=\dfrac{3}{2}RT$
$\Rightarrow K.E=\dfrac{3}{2}PV$          $[PV=RT]$
$\Rightarrow \dfrac{K.E}{V}=\dfrac{3}{2}P$
$\Rightarrow E=\dfrac{3P}{2}$        $E=$ Energy density.
Hence, the answer is $P=\dfrac{2}{3}E.$
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 of volume $0.3 \ { { m }^{ 3 } }$ contains Helium at $20.0$. The average kinetic energy per molecule for the gas is:

  1. $6.07\times { 10 }^{ -21 }J$

  2. $7.3\times { 10 }^{ 3 }J$

  3. $14.6\times { 10 }^{ 3 }J$

  4. $12.14\times { 10 }^{ -21 }J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given temperature of gas $=20°C$
                                            $=293K$
$\Rightarrow$ Average translational kinetic energy $=\dfrac{3}{2}KT$
                                                                   $=\dfrac{3}{2}\times1.38\times10^{-23}\times293$
                                                                   $=6.07\times10^{-21}J$
Hence, the answer is $6.07\times10^{-21}J.$