A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226g of N₂. Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. If the root-mean-square speed of the gas molecules is 182 m/s, what is the pressure of the gas?

Assumption;

The pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy

Given data;

The volume of the initially evacuated container is V = 1.64L

Mass of the N₂ gas is $m_{N_2}=0.226g$

Root-mean-square speed of the gas molecules is $v=182m/s$

Boltzmann constant R = 8.314J/mol*K

Formulas;

The ideal gas equation

PV = nRT ............... (1)

Root-mean-square speed of the gas molecules

$v_{rms}=\sqrt{\frac{3RT}{M}}$ ................ (2)

Number of moles

role="math" localid="1665123094731" $n=\frac{m}{M}$ .............. (3)

From equation (2), put the value and get T

$$\begin{gathered} T=\frac{{v^2}_{rms}*M}{3R} \\ =\frac{{\left(182\right)}^2*28.014*{10}^{-3}kg/mol}{3*8.314J/mol*K} \\ =37.20K \end{gathered}$$

From equation (3) put the value and get

$$\begin{gathered} n=\frac{0.226*{10}^{-3}kg}{28.014*{10}^{-3}kg/mol} \\ =8.067*{10}^{-3}mol \end{gathered}$$

From equation (1)

$$\begin{gathered} P=\frac{nRt}{V} \\ =\frac{8.067*{10}^{-3}mol*8.314J/molK*37.20K}{1.64L} \\ =1.67*{10}^{3}Pa \end{gathered}$$

Hence, the pressure of the gas is$1.67*{10}^{3}Pa$