A helium balloon has a volume of #"2600 cm"^3# when the temperature is #21^@"C"#. What is the volume of the balloon after it’s placed in a freezer with a temperature of #-15^@"C"#?

Question

Mateo Aguilar · Accepted Answer

The volume of the balloon when it is placed in a freezer at #-15^@"C"# is #"2300 cm"^3"#.

This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. This means that if the volume increases, so does the temperature, and vice versa. The equation is: #V_1/T_1=V_2/T_2#, where: #V_1# and #T_1# are the initial volume and temperature, and #V_2# and #T_2# are the final volume and temperature. The temperature must be in Kelvins, so the Celsius temperatures will be converted to Kelvins. Known #V_1="2600 cm"^3"# #T_1="21"^@"C+273.15"="294 K"# #T_2="-15"^@"C+273.15"="258 K"# Unknown #V_2# Solution Rearrange the equation to isolate #"V_2#. Plug in the known values and solve. #V_2=(V_1T_2)/T_1# #V_2=(("2600 cm"^3xx258color(red)cancel(color(black)("K"))))/(294color(red)cancel(color(black)("K")))="2300 cm"^3"# (rounded to two significant figures)