# A gas, collected by water displacement, occupies 781 cm^3 at 14°C and 99.1 kPa. What volume will the dry gas occupy at STP?

Jun 22, 2018

The question is INCOMPLETE...we need ${P}_{\text{saturated vapour pressure}}$ at $14$ ""^@C...we also need to specify $\text{STP}$...

#### Explanation:

${P}_{\text{SVP}} = 12.0 \cdot m m \cdot H g$ at $14$ ""^@C

Now the quoted pressure is …….

$\frac{99.1 \cdot k P a}{101.3 \cdot k P a \cdot a t {m}^{-} 1} \times 760 \cdot m m \cdot H g \cdot a t {m}^{-} 1 = 743.5 \cdot m m \cdot H g$

And ${P}_{\text{Total"=P_"SVP"+P_"gas}}$

${P}_{\text{gas"=P_"Total"-P_"SVP}} = \left(743.5 - 12.0\right) \cdot m m \cdot H g = 731.5 \cdot m m \cdot H g$

And $\text{STP}$ specifies a pressure of $0.987 \cdot a t m \equiv 750.2 \cdot m m \cdot H g$, and $T \equiv 273.15 \cdot K$

That out of the way...$\frac{{P}_{1} {V}_{1}}{T} _ 1 \equiv \frac{{P}_{2} {V}_{2}}{T} _ 2$

We solve for ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$

$\frac{743.5 \cdot m m \cdot H g \times 781 \cdot c {m}^{3}}{287 \cdot K} \times \frac{273.15 \cdot K}{750.2 \cdot m m \cdot H g}$

$= 736.6 \cdot c {m}^{3}$...