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

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?

  1. $102^{\circ}C$

  2. $273^{\circ}C$

  3. $819^{\circ}C$

  4. $1092^{\circ}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rms speed C is proportional to sqrt(T). If the speed doubles, the temperature must increase by a factor of 2^2 = 4. Initial temperature = 273K. Final temperature = 4 * 273K = 1092K. In Celsius, this is 1092 - 273 = 819C.

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

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 

  1. 15 ml

  2. 100 ml

  3. 200 ml

  4. 500 ml
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law PV = nRT, since T is constant, P1V1 / n1 = P2V2 / n2. Given P1=50, V1=10, n1=1; P2=100, n2=3. Solving for V2: (50 * 10) / 1 = (100 * V2) / 3. 500 = (100 * V2) / 3, so V2 = 1500 / 100 = 15 ml.

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

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:

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

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

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

  4. $\frac{5}{4} T _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a mixture of gases, the final temperature T_mix = (n1*T1 + n2*T2) / (n1 + n2). Here, n1=2, T1=T0, n2=4, T2=2T0. T_mix = (2*T0 + 4*2T0) / (2 + 4) = (2T0 + 8T0) / 6 = 10T0 / 6 = 5/3 * T0.

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

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

  1. $v _{H}$ $= \sqrt{2}$ $v _{He}$ 

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

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

  4. $v _H$ $=$ $\dfrac{v _{He}}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  1. $-21^0$C

  2. $13^0$C

  3. $15^0$C

  4. $56.4^0$C
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a fixed mass of gas in a constant volume, P is proportional to T/M. P1 = P2 implies T1/M1 = T2/M2. T1 = 15C = 288K. M1 = 32 (O2). M2 = 28 (N2). 288/32 = T2/28. 9 = T2/28. T2 = 252K. In Celsius, 252 - 273 = -21C.

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

Average kinetic energy of a gas molecule is

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The average kinetic energy of a gas molecule is given by (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature. Thus, it is directly proportional to the absolute temperature.

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

Maxwell's laws of distribution of velocities shows that

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. proportional to $\sqrt{T}$

  2. proportional to T

  3. proportional to $T^{2}$

  4. equal to zero
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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:

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

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

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

  4. $5.22\times {10}^{-21}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $u _{rms} &gt; u _{av} &gt; u _{mp}$

  2. $u _{rms} &lt; u _{av} &lt; u _{mp}$

  3. $u _{rms} &gt; u _{av} &lt; u _{mp}$

  4. $u _{rms} &lt; u _{av} &gt; u _{mp}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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