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 speed of a longitudinal wave in a mixture containing 4 moles of He and 1 mole of Ne at 300 K will be

  1. 930 m/s

  2. 541 m/s

  3. 498 m/s

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The mixture contains $n _1=4\ mol$ of $He$ and $n _2=4\ mole$ of $Ne$ act. $T=300\ K$.
Forth $He$ and $Ne$ are monoatomic, so we take $r$ mix $=5/3\quad \left [r=\dfrac {cp}{cv}\right]$ 
$M'=\dfrac {M _1n _1+M _2n _2}{n _1+n _2}$
$=\dfrac {4\times 4+20\times 1}{4+1}=\dfrac {24}{5}=4.8\ g/mol$
$=4.8\times 10^{-3}\ kg/mol$
$\therefore \ $ Speed of sound through mixture
$v=\sqrt {\dfrac {vRT}{M'}}=\sqrt {\dfrac {5}{3}\times \dfrac {8.3\times 300}{4.8\times 10^{-3}}}\sim 929.83\ m/s$


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

$2$ grams of mono atomic gas occupies a volume of $2$ litres at a pressure of $8.3 \times 10^5$ Pa and $127^0C$. Find the molecular weight of the gas.

  1. $2$ grams/mole

  2. $16$ grams/mole

  3. $4$ grams/mole

  4. $32$ grams/mole
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using PV = (m/M)RT, where m=2g, P=8.3e5 Pa, V=2e-3 m^3, T=400K, R=8.314. M = mRT/PV = (2 * 8.314 * 400) / (8.3e5 * 2e-3) = 6651.2 / 1660 = 4.006 g/mol.

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

Non-linear triatomic gas has f=6. If 50% dissociates into 3 atoms, the new mixture has 0.5 moles of triatomic (f=6) and 0.5 * 3 = 1.5 moles of monatomic (f=3). Average f = (0.5*6 + 1.5*3) / (0.5 + 1.5) = (3 + 4.5) / 2 = 7.5 / 2 = 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

When an ideal monoatomic  gas is heated zt constant pressure , which of the following may be true

  1. $\dfrac {dU}{dQ} = \frac {3}{5}$

  2. $\dfrac {dW}{dQ} = \frac {2]}{5}$

  3. $\dfrac {dU}{dQ} = \frac {4}{5}$

  4. $dW + dU = dQ $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From First Law: dQ = dU + dW. This is always true for any process. For monatomic gas at constant pressure, dU = (3/2)nRdT and dW = PdV = nRdT, so dQ = (5/2)nRdT. Then dU/dQ = 3/5 and dW/dQ = 2/5, not 4/5. Option D is the fundamental First Law statement.

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

If $\gamma $ be the ration of specific heats of a perfect gas, the number of degree of freedom of a molecule of the gas is:

  1. $\dfrac{{25}}{2}\left( {\gamma - 1} \right)$

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

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

  4. $\dfrac{9}{2}(\gamma - 1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The relationship between the adiabatic index gamma and degrees of freedom f is gamma = 1 + 2/f. Solving for f gives f = 2 / (gamma - 1).

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 magnetic monment of a diamagnetic atom is 

  1. Much greater than one

  2. 1

  3. Between zero and one

  4. Equal to zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Diamagnetic materials have no permanent magnetic moment; the induced magnetic moment opposes the external magnetic field, resulting in a net magnetic moment of zero in the absence of an external field.

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

On increasing temperature of the reacting system by $10$ degrees the rate of reaction almost doubles. The most appropriate reason for this is

  1. collision frequency increases

  2. activation energy decreases by increases in temperatuer

  3. the fraction of molecules having energy equal to threshold energy or more increase

  4. the value of threshold energy decreases
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

An ant is walking on the horizontal surface. The number of degree of freedom of ant will be

  1. $1$

  2. $2$

  3. $3$

  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An ant walking on a horizontal surface is constrained to move in a 2D plane (x and y coordinates). However, if the ant is considered a rigid body capable of rotation about its center of mass, it possesses 3 degrees of freedom (2 translational and 1 rotational).

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 monoatomic gas undergoes an isothermal expansion at temperature T during which its volume becomes 4 times. The work done on the gas and change in internal energy of the gas respectively is

  1. n RT Ln 4,0
    • n RT Ln 4,0

  3. n RT Ln 4 $\frac { 3 n R T } { 2 }$
    • n RT Ln 4, $\frac { 3 n R T } { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} w=nRT\, \, \, \ln { \left( { \frac { { 4v } }{ v }  } \right)  }  \ =nRT\, \, \ln { 4 } \, \, \, \, \, \, & \, \, \, \Delta u=0 \end{array}$

$\therefore$ Option $A$ is correct.