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?

Question

Naomi Aronson · Accepted Answer

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

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. The combined gas law equation can be used to provide an answer to this question: #(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. Assumed/Known: #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"# Not sure: #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) Take 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.

Kai Bowman · Answer

undefined The volume of the gas at STP (Standard Temperature and Pressure) would be 770.0 mL.