# A gas has a volume of 435 mL at 25^@"C" and "98.7 kPa". What would the volume of the gas be at STP?

Nov 22, 2014

The new volume would be $\text{394 mL}$.

#### Explanation:

This problem can be solved using the combined gas law with the equation:

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

The temperature scale used in the gas laws is the Kelvin scale, so the Celsius temperatures will need to be converted to Kelvins.

STP for the gas laws are $\text{273.15 K}$ and $\text{100 kPa}$.

Given/Known:

${V}_{1} = \text{435mL}$

${T}_{1} = \text{25"^o"C" + "273.15" = "298K}$

${P}_{1} = \text{98.7 kPa}$

${T}_{2} = \text{273.15K}$

${P}_{2} = \text{100 kPa}$

Unknown:

${\text{V}}_{2}$

Solution:

Rearrange the formula to isolate ${V}_{2}$. Plug in the known values and solve.

"V"_2 = (V"_1P_1T_2)/(P_2T_1)

V_2 = ((435"mL")xx(98.7"kPa")xx(273.15"K"))/((100"kPa")xx(298"K")) = "394mL" (rounded to three significant figures)