Density varies with temperature because volume varies with temperature. As temperature increases, volume increases, and vice-versa (Charles' law). Also volume varies with pressure. The greater the pressure the lower the volume and vice-versa (Boyle's law).
#"density"="mass"/"volume"#
The current density is #(4.2color(white)(.)"g")/(13color(white)(.)
"L")="0.32 g/L"#
Current STP is #0^@"C"# or #"273.15 K"#, and #"100 kPa"#. For gas problems, the Kelvin temperature is used. The Celsius temperature and pressure in torrs will have to be converted to Kelvins and kilopascals.
To convert Celsius temperature to Kelvins, add #273.15#.
#"1 kPa=7.50061683 torr"#
In order to answer this question, the combined gas law is used:
#(V_1P_1)/T_1=(V_2P_2)/T_2#, where #V, P, and T# are volume, pressure, and temperature. The first set of variables will be the values given in the question. The second set will be values at STP.
Given/Known
#V_1="13 L"#
#P_1=672color(red)cancel(color(black)("torr"))xx(1 "kPa")/(7.50061683 color(red)cancel(color(black)("torr")))="89.593 kPa"#
#T_1="70"^@"C"+273.15="343 K"#
#P_2="100 kPa"#
#T_2="273.15 K"#
Unknown: #V_2# (volume at STP)
Solve equation
Rearrange the equation to isolate #V_2#. Substitute the known values into the equation and solve.
#V_2=(V_1P_1T_2)/(T_1P_2)#
#V_2=((13"L")xx(89.593color(red)cancel(color(black)"kPa"))xx(273.15color(red)cancel(color(black)"K")))/((343color(red)cancel(color(black)"K")xx(100 color(red)cancel(color(black)"kPa"))##=##"9.3 L"# (rounded to two significant figures)
Back to Density
Mass is not affected by temperature or pressure, so the mass is still #"4.2 g"#
#"density"=(4.2color(white)(.)"g")/(9.3color(white)(.)"L")="0.45 g/L"#
The decrease in temperature from #"343 K"# to #"273.15"# caused a decrease in volume and therefore an increase in density. Also, the increase in pressure from #"89.593 kPa"# to #"100 kPa"# contributed to a decrease in volume.
