Gases and Gas Laws

Using the formula lambda = 1 / (sqrt(2) * pi * n * d^2) with n = 2.7 * 10^25 m^-3 and d = 2 * 10^-10 m, the result is approximately 2.81 * 10^-7 m.

V, n, T $\rightarrow  same$(25) so $P\rightarrow $ also same ( P  5  25)
$\displaystyle \sigma \rightarrow same (25)$
given

$\displaystyle (v {rms})\times

x=\dfrac{1}{\sqrt{6}}(v _{avg.}) _{CH _{4}}$ &

$v _{rms}=\sqrt{\dfrac{3\pi }{8}}(v _{avg.})$ so
$\displaystyle \sqrt{\dfrac{3\pi }{8}}(v _{avg.}) _{CH _{4}}$
$\displaystyle \dfrac{(v _{avg.})x}{(v _{avg.}) _CH _{4}}=\sqrt{\dfrac{8}{3\pi }}.\frac{1}{\sqrt{6}}=\dfrac{2}{3\sqrt{\pi }}$
For X (9< ) : $\displaystyle Z _{1}=\sqrt{2}\pi \sigma ^{2}(v _{avg.}) _{x}N^{\ast }$
For CH{4} (9< ) : $\displaystyle Z _{1}=\pi \sigma ^{2}(v _{avg.}) _{CH _{4}}N^{\ast }$
Since T, P, v, n are same, $N\ast $ will also be same.
$\displaystyle



\frac{Z _{1}X}{Z _{1}(CH _{4})}=\sqrt{2}\frac{(v _{avg.}) _{x}}{(v _{avg.}) _{CH _{4}}}=\sqrt{2}.\frac{2}{3\sqrt{\pi

}}$
$\displaystyle Z _{1}(X)=Z _{1}(CH _{4}).\frac{2\sqrt{2}}{3\sqrt{\pi }}$

Weight is calculated as mass times gravity (W = mg). Using g = 10 m/s^2, 14.2 kg * 10 m/s^2 = 142 N.

CGS unit of density is $g cm^{-3}$

Molar mass of mixture $=\cfrac{total mass}{total mole}$

At STP, 1 mole of any gas occupies 22.4 liters. 2.24 liters of O2 is 0.1 moles. The molar mass of O2 is 32 g/mol. Mass = 0.1 moles * 32 g/mol = 3.2 grams. The closest option is 3.4 grams, likely due to a typo in the question or options.

The reaction is C(s) + CO2(g) -> 2CO(g). According to the stoichiometry, 1 volume of CO2 produces 2 volumes of CO. Therefore, 26cc of CO2 will produce 26 * 2 = 52cc of CO.

Density is calculated as mass divided by volume. The total mass of the nitrogen gas remains constant throughout the compression process. By finding the initial mass from the initial volume and density, and dividing it by the final volume of 3.92 ml, the resulting density is obtained.

According to Henry's Law, the solubility of a gas is proportional to its partial pressure. Using the standard solubility of CO2 in water at 25 degrees Celsius and 1 atm, one can scale the value to 2.4 atm.

A mole is equal to the relative molecular mass of a gas. As given, the number of moles of gas X and hydrogen are same in the two flasks.
Therefore no. of moles in 2g of hydrogen gas = 1
No. of moles in 1.01g of hydrogen gas = 1/2
Now the weight of gas X that contains 1/2 moles = 16.2g
Weight of gas X that contains 1mole = $\frac{16.2}{1/2}$ = 32g

$At\quad STP,\quad volume\quad of\quad 1\quad mole\quad gas\quad =\quad 22.4\quad L\ Mass\quad of\quad 1L\quad gas\quad =\quad 2g\ Mass\quad of\quad 22.4L\quad gas\quad =\quad 2\times 22.4\quad =\quad 44.8\ So,\quad molecular\quad weight\quad of\quad gas\quad =\quad 44.8\ So,\quad correct\quad answer\quad is\quad option\quad B.$

$N _2O _4\longrightarrow 2NO _2$