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.