Question

At$\mathbf{273}\mathit{K}\mathbf{}$and$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$, the density of a gas is$\mathbf{1}\mathbf{.}\mathbf{24}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$.

1. Find${\mathit{v}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$for the gas molecules.
2. Find the molar mass of the gas
3. Identify the gas.

Short answer

1. The rms value of velocity ${\text{v}}_{rms}$is$494\text{\hspace{0.17em}m}/\text{s}$ .
2. Molar mass of gas is$27.9\text{\hspace{0.17em}g}/\text{mole}$.
3. The gas is ${\text{N}}_2$.

Step-by-step solution

 Step 1: Given data

• Temperature is$T=273\text{\hspace{0.17em}K}$
• Pressure is$P=1.00\times {10}^{-2}\text{atm}$
• Density is$\rho =1.24\times {10}^{-5}\text{g}/{\text{cm}}^3$

Understanding the concept

The expression for the RMS speed is given by,

$v=\sqrt{\frac{3RT}{M}}$

Here V is the RMS speed, R is the universal gravitational constant, T is the temperature of the gas, M is the molar mass of the gas.

The relation between mass and molar mass is given by,

$n=\frac{W}{M}$

Here n is the number of moles, w is the mass of the gas,M is the molar mass of the gas.

(a) Find vrms for the gas molecules

Since,

$$\begin{gathered} \rho =\frac{Pm}{RT} \\ RT=\frac{Pm}{\rho } \end{gathered}$$

Now, rms velocity is given by

$V_{rms}=\sqrt{\frac{3RT}{m}}$

Substitute$\frac{Pm}{\rho }$ for RT into the above equation,

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{3Pm}{m\rho }}\\ v_{rms}=\sqrt{\frac{3P}{\rho }}\end{array}$

Substitute$1.01\times {10}^3\text{Pa}$for P and $1.24\times {10}^{-2}\text{kg}/{\text{m}}^3$for P into the above equation,

$\begin{array}{c}{V}_{\text{rms}}=\sqrt{\frac{3\times 1.01\times {10}^3}{1.24\times {10}^{-2}}}\\ V_{\text{rms}}=494\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore the RMS speed of the gas molecule is$494\text{\hspace{0.17em}m}/\text{s}$ .

(b) Find the molar mass of the gas

Now we have to find the molar mass.

Since

$\rho =\frac{Pm}{RT}$

Substitute$1.24\times {10}^{-2}\text{kg}/{\text{m}}^3$ for$\rho$ ,$1.01\times {10}^3\text{Pa}$ for P, $8.314J/mol\cdot K$for R ,$273K$ for T into the above equation.

$\begin{array}{c}1.24\times {10}^{-2}=\frac{1.01\times {10}^3\times m}{8.31\times 273}\\ \text{m}=27.9g/mol\end{array}$

(c) Identify the gas

The molar mass of the gas is around $28g/mol$, so the gas molecule is $N_2$.