# What is the new volume for a 50.4*mL volume of gas, at 742*mm*Hg pressure and 293.15*K, for which conditions are changed to "STP"?

Sep 10, 2017

Well, STP specifies a pressure of ${10}^{5} \cdot P a$, and a temperature of $273.15 \cdot K$.....

#### Explanation:

We use the old combined gas law, which says for a GIVEN amount of gas.....

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

And so we solve for ${V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{P}_{2} {T}_{1}}$, which product clearly has units of volume. Why?

And we plug in the numbers knowing that $100 \cdot k P a \equiv 750 \cdot m m \cdot H g$...

${V}_{2} = \frac{742 \cdot m m \cdot H g \times 50.4 \times {10}^{-} 3 L \times 273.15 \cdot K}{750 \cdot m m \cdot H g \times 293.15 \cdot K}$

${V}_{2} = 0.0465 \cdot L \equiv 46.5 \cdot c {m}^{3}$; i.e. the volume is SLIGHTLY compressed......

Note that while this does seem complicated, all I have done is use the appropriate units. That I got an answer in $c {m}^{3}$ suggests that my order of operations was right.