Question #42a61

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Question #42a61

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Answer 1 · 2017-08-08T19:00:25

We assume temperature is constant.........and get $V_{2} ≅ 12 \cdot L$ $V_2~=12*L$

Explanation:

And $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ according to old $Boyle's Law$ $"Boyle's Law"$

And thus $V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{573 \cdot m m \cdot H g \times 5.30 \cdot L}{255 \cdot m m \cdot H g}$ $V_2=(P_1V_1)/P_2=(573*mm*Hgxx5.30*L)/(255*mm*Hg)$

$= ? ? \cdot L$ $=??*L$

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Question #42a61

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