If the pressure exerted on a gas in a weather balloon decreases from 1.01 atm to 0.562 atm as it rises, by what factor will the volume of the gas in the balloon increase as it rises?

1 Answer
anor277
Jun 16, 2016

This is a simple manifestation of Boyle's Law The volume of the balloon should almost DOUBLE.

Explanation:

#P_1V_1=P_2V_2# at constant temperature.

Thus #V_2# #=# #(P_1V_1)/P_2#

#=# #(1.01*atmxxV_1)/(0.562*atm)#

#~=# #2V_1# as required.

Note that given Boyle's Law, the proportionality of #P# and #V# at constant #T#, I am quite justified in using non-standard units such as atmospheres, and litres. I could even use #"pounds per square inch"# and #"pints"# and #"gallons"#, and I presume old Boyle did use these units. When these problems are applied to other situations, the utility of using standard SI units becomes more obvious.

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