Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.

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Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.

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Answer 1 · 2014-12-15T18:47:54

The answer is $1.8 L$ $1.8L$ .

Standard Temperature and Pressure stipulates a pressure of $100.0 k P a$ $100.0kPa$ . In order to solve this problem, we'll use Boyle's law, $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1 = P_2V_2$ , which states that the pressure and volume of a gas are proportional to one another.

Let's calculate the volume by converting $m m H g$ $mmHg$ to $k P a$ $kPa$ first.

$18 m m H g \cdot \frac{1.0 a t m}{760 m m H g} \cdot \frac{101.325 k P a}{1.0 a t m} = 2.4 k P a$ $18 mmHg * (1.0 atm)/(760 mmHg) * (101.325 kPa)/(1.0 atm) = 2.4kPa$

So, $V_{2} = \frac{P_{1}}{P_{2}} \cdot V_{1} = \frac{2.4 k P a}{100.0 k P a} \cdot 77 L = 1.8 L$ $V_2 = P_1/P_2 * V_1 = (2.4 kPa)/(100.0 kPa) * 77L = 1.8L$

Now calculate the volume by converting $k P a$ $kPa$ to $m m H g$ $mmHg$ .

$100.0 k P a \cdot \frac{1 a t m}{101.325 k P a} \cdot \frac{760 m m H g}{1 a t m} = 750 m m H g$ $100.0 kPa * (1 atm)/(101.325 kPa) * (760 mmHg)/(1 atm) = 750 mmHg$

$V_{2} = \frac{P_{1}}{P_{2}} \cdot V_{1} = \frac{18 m m H g}{750 m m H g} \cdot 77 L = 1.8 L$ $V_2 = P_1/P_2 * V_1 = (18 mmHg)/(750 mmHg) * 77L = 1.8L$

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Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.

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