Question #4a240

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Question #4a240

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Answer 1 · 2016-12-18T09:56:44

$V_{2} ≅ 40 \cdot L$ $V_2~=40*L$

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$

All temperature is reported on the $absolute scale$ $"Absolute scale"$ ; for $pressure$ $"pressure"$ we use the relationship $760$ $760$ $m m$ $mm$ $H g$ $Hg$ $\equiv$ $-=$ $1 \cdot a t m .$ $1*atm.$

$V_{2} = \frac{T_{2}}{P_{2}} \times \frac{P_{1} V_{1}}{T_{1}} = \frac{256.8 \cdot K}{\frac{385 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}}} \times \frac{1}{303.5 \cdot K} \times (\frac{731 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}}) \times 26.5 \cdot L ≅ 40 \cdot L$ $V_2=T_2/(P_2)xx(P_1V_1)/T_1=(256.8*K)/((385*mm*Hg)/(760*mm*Hg*atm^-1))xx1/(303.5*K)xx((731*mm*Hg)/(760*mm*Hg*atm^-1))xx26.5*L~=40*L$

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Question #4a240

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