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?

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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?

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Answer 1 · 2016-06-16T15:11:31

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

Explanation:

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ at constant temperature.

Thus $V_{2}$ $V_2$ $=$ $=$ $\frac{P_{1} V_{1}}{P_{2}}$ $(P_1V_1)/P_2$

$=$ $=$ $\frac{1.01 \cdot a t m \times V_{1}}{0.562 \cdot a t m}$ $(1.01*atmxxV_1)/(0.562*atm)$

$≅$ $~=$ $2 V_{1}$ $2V_1$ as required.

Note that given Boyle's Law, the proportionality of $P$ $P$ and $V$ $V$ at constant $T$ $T$ , I am quite justified in using non-standard units such as atmospheres, and litres. I could even use $pounds per square inch$ $"pounds per square inch"$ and $pints$ $"pints"$ and $gallons$ $"gallons"$ , 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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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

Explanation:

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