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?

1 Answer
Nov 22, 2014

The new volume would be "394 mL".

Explanation:

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

(P_1V_1)/T_1=(P_2V_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 "273.15 K" and "100 kPa".

Given/Known:

V_1= "435mL"

T_1 = "25"^o"C" + "273.15" = "298K"

P_1 = "98.7 kPa"

T_2 = "273.15K"

P_2 = "100 kPa"

Unknown:

"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)