Maxwell-boltzmann speed distribution function - class-XI

For a diatomic gas like hydrogen, the molar specific heat at constant volume depends on the temperature. At low temperatures, only translational degrees of freedom are active (3R/2). At moderate temperatures, rotational degrees of freedom are active (5R/2). At very high temperatures, vibrational degrees of freedom may also contribute (7R/2).

The area under the Maxwell-Boltzmann speed distribution curve and the speed axis represents the total fraction of gas molecules, which must always equal 1 regardless of the temperature. Therefore, the areas A_1 and A_2 at temperatures T_1 and T_2 are equal.

The rms speed C is given by sqrt(3RT/M). The ratio C_A / C_B = sqrt((T_A / M_A) / (T_B / M_B)). Given T_A / M_A = 4 * (T_B / M_B), the ratio is sqrt(4) = 2.

$\begin{array}{l} \frac { { rx } }{ { ry } } =\frac { { { w _{ x } } } }{ { { n _{ y } } } } \sqrt { \frac { { { M _{ y } } } }{ { { M _{ x } } } }  }  \ =\frac { 3 }{ 2 } \sqrt { \frac { { 80 } }{ { 20 } }  } =\frac { 3 }{ 1 } =3:1 \ \therefore \frac { { dis\tan  ce, , travelled, , by, , gas, , X } }{ { dis\tan  ce, , travelled, , by, , gas, , Y } } =3:1 \ \therefore dis\tan  ce, , of, , A, , from, , Q=\frac { { 150 } }{ 3 } =50, , cms \end{array}$

Use the ideal gas law PV = NkT, where N/V = P / (kT). Convert pressure to Pascals (1 mm Hg = 133.322 Pa) and temperature to Kelvin (300K). Calculate the number density N/V.

From the ideal gas law PV = NkT, the number of molecules per unit volume n = N/V = P / (kT). Given P = 10^6 * kT, substituting this gives n = (10^6 * kT) / (kT) = 10^6.

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.

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.

For a mixture of gases, the final temperature T_mix = (n1T1 + n2T2) / (n1 + n2). Here, n1=2, T1=T0, n2=4, T2=2T0. T_mix = (2T0 + 42T0) / (2 + 4) = (2T0 + 8T0) / 6 = 10T0 / 6 = 5/3 * T0.

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

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.

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.

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 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

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

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

Let the volume occupied $=V$

The Maxwell-Boltzmann speed distribution function
$\left(N _v=\dfrac{dN}{dv}\right)$ depends on the mass of the gas
molecule. [Here, dN is the number of molecules with speeds between v and
$(v+dv)$]. The masses of hydrogen and oxygen molecules are different.