Question

If the limiting slope of density vs. pressure plot for air at $0^{o}C$ is $1.27\times {10}^{-5}kg{m}^{-3}P{a}^{-1}$, the molar mass of air would be:

• A. $14.42\times {10}^{-3}kgmo{l}^{-1}$
• B. $23.89\times {10}^{-3}kgmo{l}^{-1}$
• C. $28.82\times {10}^{-3}kgmo{l}^{-1}$
• D. $32.89\times {10}^{-3}kgmo{l}^{-1}$

Answer 1

The correct option is C $28.82\times {10}^{-3}kgmo{l}^{-1}$

As we know,
${\left[\frac{d}{P}\right]}_{P\to 0}=\frac{M}{RT}$
$M=RT\ast 1.27\times {10}^{-5}$
$M=1.27\times {10}^{-5}\times 8.314\times 273$
$=28.82\times {10}^{-3}kgmo{l}^{-1}$