What volume would a sample of gas that has a volume of $1.63 \cdot L$ $1.63L$ at $793 \cdot mm Hg$ $793"mm Hg"$ occupy at $755 \cdot mm Hg$ $755*"mm Hg"$ ?

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What volume would a sample of gas that has a volume of $1.63 \cdot L$ $1.63L$ at $793 \cdot mm Hg$ $793"mm Hg"$ occupy at $755 \cdot mm Hg$ $755*"mm Hg"$ ?

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Answer 1 · 2018-02-10T19:57:43

Again it is highly non-standard (and dangerous) to quote a pressure in $m m \cdot H g$ $mm*Hg$ for pressures OVER 1 atmosphere.

Explanation:

We know that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$ $1*atm-=760*mm*Hg$ ...or that one atmosphere will support a column of mercury that is $760 \cdot m m$ $760*mm$ high...pressures higher than one atmosphere ARE NEVER quoted in $m m \cdot H g$ $mm*Hg$ (except in these scenarios where some punter who has NEVER used a mercury manometer mistakenly proposes an illegitimate measurement).

And so $793 \cdot Torr \equiv \frac{793 \cdot Torr}{760 \cdot Torr \cdot a t m^{- 1}} = 1.04 \cdot a t m$ $793*"Torr"-=(793*"Torr")/(760*"Torr"*atm^-1)=1.04*atm$

And given $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ ...

$V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{\frac{755 \cdot Torr}{760 \cdot Torr \cdot a t m^{- 1}}}{1.04 \cdot a t m} \times 1.63 \cdot L = 1.56 \cdot L$ $V_2=(P_1V_1)/P_2=((755*"Torr")/(760*"Torr"*atm^-1))/(1.04*atm)xx1.63*L=1.56*L$ ..

Reasonably, the volume has reduced slightly. But your chemistry teacher has no business asking questions with these whack units.

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What volume would a sample of gas that has a volume of $1.63 \cdot L$ $1.63L$ at $793 \cdot mm Hg$ $793"mm Hg"$ occupy at $755 \cdot mm Hg$ $755*"mm Hg"$ ?

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Explanation:

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What volume would a sample of gas that has a volume of 1.63⋅L1.63*L at 793⋅mm Hg793*"mm Hg" occupy at 755⋅mm Hg755*"mm Hg"?

1 Answer

Explanation:

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What volume would a sample of gas that has a volume of $1.63 \cdot L$ $1.63L$ at $793 \cdot mm Hg$ $793"mm Hg"$ occupy at $755 \cdot mm Hg$ $755*"mm Hg"$ ?