A gas has a volume of #"720.0 mL"# at #"20.0"^@"C"# and #"3.00 atm"#. What would the volume of the gas be at STP?

1 Answer
Feb 19, 2015

Answer:

The final volume will be #"2039 mL"#.

Explanation:

STP means standard temperature and pressure. The current values for STP are #0^("o")"C"# or #"273.15 K"# and #"10"^5# #"Pascals"# #("Pa")#, usually given as #"100 kPa"# to make it easier to work with. For the gas laws, the Celsius temperature must be converted to Kelvins by adding #273.15# to the Celsius temperature.

This question can be answered using the equation for the combined gas law:

#(P_1V_1)/T_1 = (P_2V_2)/T_2#,

where:

#P_1# and #P_2# are the initial and final pressures, #V_1# and #V_2# are the initial and final volumes, and #T_1# and #T_2# are the initial and final temperatures in Kelvins.

Known/Given:

#P_1 = 3.00 color(red)cancel(color(black)("atm"))xx(101.325"kPa")/(1color(red)cancel(color(black)("atm")))="303.975 kPa"#

#V_1 = 720.0 "mL"#

#T_1 = 20.0^("o")"C" + 273.15 = 293.2"K"#

#P_2 = "100 kPa"#

#T_2 = 273.15 "K"#

Unknown:

#V_2#

Solution: Rearrange the combined gas law so that #V_2# is isolated, then solve for #V_2#

#V_2 =(P_1V_1T_2)/(T_1P_2) = (303.975color(red)cancel(color(black)("kPa"))*720.0 "mL"*273.15 color(red)cancel(color(black)("K")))/(293.2 color(red)cancel(color(black)("K"))*100color(red)cancel(color(black)("kPa"))) = "2039 mL"# (rounded to four significant figures)

Note:

If your teacher is still using #"1 atm"# for standard pressure, substitute #"3.00 atm"# for #"303.975 kPa"#, and #"1 atm"# for #"100 kPa"#. The final volume will be #"2010 mL"# rounded to three significant figures.