Question #1231e

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Question #1231e

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Answer 1 · 2017-04-10T17:16:43

671.4mL to 1 decimal place.

Explanation:

As temperature and number of moles of gas remained unchanged, use the formula $P_{1} \cdot V_{1} = P_{2} \cdot V_{2}$ $P_1*V_1 = P_2*V_2$ , where $P_{1}$ $P_1$ and $P_{2}$ $P_2$ are the old and new pressures and $V_{1}$ $V_1$ and $V_{2}$ $V_2$ are the old and new volumes respectively, to calculate the volume of gas.

You must convert the pressures into a single unit. The best option is to convert the new pressure, given in mmHg, into atm. As we know, $1 a t m = 760 m m H g$ $1 atm = 760 mmHg$
Therefore, $738 m m H g = \frac{738}{760} a t m$ $738mmHg = 738/760 atm$ . For the sake of accuracy, we are going to used this fraction rather than the irrational decimal value.
$P_{1} \cdot V_{1} = P_{2} \cdot V_{2}$ $P_1*V_1 = P_2*V_2$
$P_{1} = 1 a t m$ $P_1 = 1 atm$
$V_{1} = 652 m L$ $V_1 = 652mL$
$P_{2} = \frac{738}{760} a t m$ $P_2 = 738/760 atm$
$V_{2} = ?$ $V_2 = ?$

Putting in all the known values, we have:
$1 a t m \cdot 652 m L = \frac{738}{760} a t m \cdot V_{2} m L$ $1atm*652mL = 738/760 atm * V_2mL$
$V_{2} = \frac{1 a t m \cdot 652 m L}{\frac{738}{760} a t m}$ $V_2 = (1atm*652mL)/(738/760atm)$
$V_{2} = 671.436 m L$ $V_2 = 671.436mL$

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Question #1231e

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