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
Answer 1

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

#V_1/T_1=V_2/T_2#,

#V_1# and #T_1# are the initial volume and temperature, and #V_2# and #T_2# are the final volume and temperature.

#V_1="2600 cm"^3"#

#T_1="21"^@"C+273.15"="294 K"#

#T_2="-15"^@"C+273.15"="258 K"#

#V_2#

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)

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