Question #66e5c

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Question #66e5c

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Answer 1 · 2017-09-04T02:49:57

Well....the volume should be COMPRESSED down to approx. $100 \cdot m L$ $100*mL$ .

Explanation:

According to old Boyle's law, $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ at constant $T$ $T$ .

We know that $1 \cdot a t m$ $1*atm$ of pressure will support a column of mercury that is $760 \cdot m m$ $760*mm$ high....and we can use a mercury column as a measurement of pressure on this basis......

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

$= \frac{\frac{8.45 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}} \times 4.60 \cdot L}{\frac{368 \cdot m m \cdot H g}{760 \cdot m m \cdot H g}} = ? ? L$ $=((8.45*mm*Hg)/(760*mm*Hg*atm^-1)xx4.60*L)/((368*mm*Hg)/(760*mm*Hg))=??L$ .

Answer 2 · 2017-09-04T03:13:22

$0.1056 L$ $0.1056L$

Explanation:

Using the Ideal Gas Laws,
$P_{1} \frac{V_{1}}{T_{1}} = P_{2} \frac{V_{2}}{T_{2}}$ $P_1V_1/T_1 = P_2V_2/T_2$ , so at constant temperature we have

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1 = P_2V_2$ ; $\frac{P_{1} V_{1}}{P_{2}} = V_{2}$ $(P_1V_1)/P_2 = V_2$

$V_{2} = \frac{8.45 \times 4.60}{368}$ $V_2 = (8.45 xx 4.60)/368$ ; $V_{2} = 0.1056 L$ $V_2 = 0.1056L$

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Question #66e5c

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