# A 0.275*L volume of an unspecified gas exerts a pressure of 732.6*mm*Hg at a temperature of -28.2 ""^@C. What is the molar quantity of this gas?

Apr 3, 2017

Approx. $0.013 \cdot m o l$

#### Explanation:

We need to know that $1 \cdot a t m$ will support a column of mercury $760 \cdot m m$ high, and thus we can use the length of a mercury column to express pressure.

And thus P=(732.6*mm*Hg)/(760*mm*Hg*atm^-1)=??*atm

And we use the $\text{Ideal Gas Law}$, $n = \frac{P V}{R T}$

$n = \frac{P V}{R T} = \frac{\frac{732.6 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 0.275 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 245 \cdot K} =$

$1.32 \times {10}^{-} 2 \cdot m o l .$

Why did I change the temperature to $\text{degrees Kelvin}$?