If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

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If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

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Answer 1 · 2016-08-21T06:52:08

From Boyle's Law, $pressure \times volume = constant$ $"pressure"xx"volume"="constant"$

$P_{2} ≅ \frac{2}{7} \times 0.12 \cdot a t m$ $P_2~=2/7xx0.12*atm$

Explanation:

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$

Thus $V_{2}$ $V_2$ $=$ $=$ $\frac{P_{1} V_{1}}{P_{2}}$ $(P_1V_1)/P_2$ $=$ $=$ $\frac{2.31 \cdot L \times 0.12 \cdot a t m}{7.14 \cdot L}$ $(2.31*Lxx0.12*atm)/(7.14*L)$ .

The advantage of using Boyle's Law is that I can use any units I like, $pints$ $"pints"$ , $bars$ $"bars"$ , $mm Hg$ $"mm Hg"$ , $foot pounds$ $"foot pounds"$ , $gallons$ $"gallons"$ so long as I am consistent.

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If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

1 Answer

Explanation:

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