# A 150 ml sample of gas is at 620 mm Hg. The volume is decreased to 100 ml at a constant temperature. What is the new pressure of gas ? How would you set this up to solve it , and what is the answer ?

May 22, 2018

Well, $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high....

#### Explanation:

And so we gots a $150 \cdot m L$ volume of gas at $\frac{620 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 0.816 \cdot a t m$..

And old Boyle's law holds that ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ for a given mass of gas....

And so ${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{0.816 \cdot a t m \times 150 \cdot m L}{100 \cdot m L} = 1.22 \cdot a t m$...

I changed the pressure to atmospheres because I would not accept a pressure OVER $1 \cdot a t m$ that quoted a measurement in $m m \cdot H g$. Units of $m m \cdot H g$ are used for ONE atmosphere or BELOW, and it is inappropriate to use them in scenarios where $P \text{>>} 1 \cdot a t m$....