Question

An ideal gas has a density of $\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$atand $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$. Identify the gas.

Short answer

The ideal gas is Oxygen.

Step-by-step solution

Step 1

The density of gas$\rho =1.33\times {10}^{-6}g/c{m}^3$at the pressure$P=1.00\times {10}^{-3}atm$at$20°C$

Therefore, pressure is given at standard atmosphere(atm) it is changed to pascal (Pa) $\rho =1.00\times {10}^{-3}\times 1.013\times {10}^{5}Pa$and temperature is given at Celsius (C) is converted to Kelvin (K) T=293K.

Step 2

For the identification of gas, Molar Mass (M) is calculated

$p=\frac{P.M}{RT}$

Here, R is gas constant and the value of R is 8.315 J/mol

Putting all the values, molar mass (M) is

$\begin{array}{r}M=\frac{\rho \times R\times T}{P}\\ =\frac{1.33\times 8.315\times 293}{1\times {10}^{-3}\times 1.013\times {10}^5}\\ =31.98\mathrm{g}/\mathrm{mol}\end{array}$

As Oxygen mole is 32 which is approximately equivalent to 31.98

Hence, thus the given ideal gas is oxygen.