Question

At 273 k and 1.00 x 10^-2 atm, the density of a gas is 1.24 x 10^-5 g/cm3.

Answer 1

Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C$_{rms}$= √$\frac{3RT}{M}$=√$\frac{3PV}{M}$=√$\frac{3P}{d}$, where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.

Given,  T=273 K, P=1.00 x 10⁻² atm, d=1.24 x 10⁻⁵ g/cm³.

(a) Using the formula $C_{rms}$=√$\frac{3P}{d}$=√(3X1.00X10⁻²)/(1.24X10⁻⁵)=49.18

(b) Molar mass can be determined by using the formula $C_{rms}$=√{3RT}{M}

49.18=√$\frac{3X8.314X273}{M}$

49.18²=√(3X8.314X273)/M

M=$\frac{3X8.314X273}{49.18^{2} }$

M=1.67 ≅ 2

Molecular mass is 2.

(c) The gas is Helium (He) whose molecular mass is 2.

Answer 2

Answer:

a. $v_{rms}=0.495m/s$

b. $M=27.8g/mol$

c. Nitrogen.

Explanation:

Hello,

In this case, we consider the given conditions at which the gas is, in order to compute the required magnitudes:

b. We must start by this part as long as the Root-Mean-Square Velocity implies the usage of the molar mass, in such a way, by considering the given temperature, pressure and density, we use the ideal gas equation to solve for the molecular mass as follows:

$PV=nRT\\PV=\frac{m}{M} RT\\MPV=mRT\\M=\frac{mRT}{PV} \\M=\frac{\rho RT}{P}$

$M=\frac{1.24x10^{-5}\frac{g}{cm^3}*\frac{1000cm^3}{1L} *0.082 \frac{atm*L}{mol*K}*273K}{1.00x10^{-2}atm} \\\\M=27.8g/mol$

a. Now, the Root-Mean-Square Velocity is computed via:

$v_{rms}=\sqrt{\frac{3RT}{M}} =\sqrt{\frac{3*8.314 \frac{kg*m^2}{s^2*mol*K} * 273K}{27800kg/mol} } \\v_{rms}=0.495m/s$

c. Finally, considering the computed molar mass, the closest one matches with diatomic nitrogen, nevertheless there is a deviation of about 0.7% from the real molar mass which is 28g/mol indeed.

Best regards.