Tag: physics

Questions Related to physics

A sample of gas is at $0^{\circ}C$. To what temperature must it be raised in order to double the rms speed of its molecules?

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $102^{\circ}C$

  2. $273^{\circ}C$

  3. $819^{\circ}C$

  4. $1092^{\circ}C$

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Correct Option: C

One mole of gas occupies 10 ml at 50 mm pressure. The volume of 3 moles of the gas at 100 mm pressure and same temperature is 

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. 15 ml

  2. 100 ml

  3. 200 ml

  4. 500 ml

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Correct Option: A

2 moles of an ideal monoatomic gas at temperature $T _0$ is mixed wth 4 moles of another ideal monoatomic gas at temperature $2T _0$ then  the temperature of the mixture is:

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $\frac{5}{3} T _0$

  2. $\frac{3}{2} T _0$

  3. $\frac{4}{3} T _0$

  4. $\frac{5}{4} T _0$

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Correct Option: A

In two vessels of the same volume, atomic hydrogen and helium with pressure 1 atm and 2 atm are filled. If temperature of both the same is the same, then the average speed of hydrogen atom $v _H$ will be related to helium $v _{He}$ as

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $v _{H}$ $= \sqrt{2}$ $v _{He}$ 

  2. $v _H$ $=$ $v _{He}$

  3. $v _H$ $=$ 2$v _{He}$

  4. $v _H$ $=$ $\dfrac{v _{He}}{2}$

Show answer
Correct Option: C
Explanation:

By Maxwell's speed distribution, $<v>\alpha \sqrt { \dfrac { RT }{ M }  } $. Since the temperature of two gases is same, hence


$<v>\alpha \sqrt { \dfrac { 1 }{ M }  } $

Also, ${ M } _{ He }=4{ M } _{ H }$

Hence, $<{ v } _{ H }>=2<{ v } _{ He }>$

Answer is option C.

The molecular weights of $O _2$ and $N _2$ are 32 and 28 respectively. At $15^0$C, the pressure of 1 gm will be the same as that of 1 gm in the same bottle at the temperature.

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $-21^0$C

  2. $13^0$C

  3. $15^0$C

  4. $56.4^0$C

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Correct Option: A

Average kinetic energy of a gas molecule is

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. Inversely proportional to the square of its absolute temperature

  2. Directly proportional to the square root of its absolute temperature

  3. Directly proportional to its absolute temperature

  4. Directly proportional to square of absolute temperature

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Correct Option: C

Maxwell's laws of distribution of velocities shows that

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. the number of molecules with most probable velocity is very large

  2. the number of molecules with most probable velocity is small

  3. the number of molecules with most probable velocity is zero

  4. the number of molecules with most probable velocity is exactly equal to 1

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Correct Option: A
Explanation:

The form of Maxwell's velocity distribution function is gaussian type. So the maximum of this function represents the speed at which most of the molecules travel. This speed is known as most probable speed.

The average velocity of the molecules in a gas in equilibrium is

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. proportional to $\sqrt{T}$

  2. proportional to T

  3. proportional to $T^{2}$

  4. equal to zero

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Correct Option: A
Explanation:

the average velocity of the gas molecules = $\sqrt{\frac{8RT}{\pi M}}$
so clearly, the average velocity $\alpha \sqrt{T}$
So, A is the correct answer.
Note that T is the temperature in Kelvins

The average kinetic energy of a gas molecule at ${27}^{o}C$ is $6.21\times {10}^{-21}J$, then its average kinetic energy at ${227}^{o}C$ is:

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $10.35\times {10}^{-21}J$

  2. ${11.35}\times {10}^{-21}J$

  3. $52.2\times {10}^{-21}J$

  4. $5.22\times {10}^{-21}J$

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Correct Option: A
Explanation:

Average kinetic energy of gas molecules $\propto$ Temperature (Absolute)
$\cfrac{K.E(at\quad {227}^{o}C)}{K.E (at\quad {27}^{o}C)}=\cfrac{273+227}{273+27}=\cfrac{500}{300}=\cfrac{5}{3}$
$K.E({227}^{o})=\cfrac{5}{3}\times 6.21\times {10}^{-21}J=10.35\times {10}^{-21}J$

For a given gas, which of the following relationships is correct at a given temp?

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $u _{rms} > u _{av} > u _{mp}$

  2. $u _{rms} < u _{av} < u _{mp}$

  3. $u _{rms} > u _{av} < u _{mp}$

  4. $u _{rms} < u _{av} > u _{mp}$

Show answer
Correct Option: A
Explanation:

$u _{rms} = $ Root mean square velocity
$u _{ar}= $Average velocity 
$u _{mp} $= Most probable velocity
$u _{mp}:u _{ar}:u _{rms}= 1: 1.128: 1224$
$\therefore u _{rms} > u _{ar} > u _{mp}$