If $2.4 \cdot 10^{5}$ $2.410^5$ $L$ $L$ of gas is at $180$ $180$ $m m H g$ $mmHg$ , what is the pressure when the gas is compressed to $1.8 \cdot 10^{3}$ $1.810^3$ $L$ $L$ at constant temperature?

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If $2.4 \cdot 10^{5}$ $2.410^5$ $L$ $L$ of gas is at $180$ $180$ $m m H g$ $mmHg$ , what is the pressure when the gas is compressed to $1.8 \cdot 10^{3}$ $1.810^3$ $L$ $L$ at constant temperature?

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Answer 1 · 2016-06-08T00:28:16

The new pressure is $2.4 \times 10^{4}$ $2.4xx10^(4)$ $m m H g$ $mmHg$

Explanation:

Let's start off with identifying our known and unknown variables.
The first volume we have is $2.4 \times 10^{5}$ $2.4xx10^(5)$ L, the first pressure is 180 mmHg, and the second volume is $1.8 \times 10^{3}$ $1.8xx10^(3)$ . Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.

The equation we use is $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$
where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure.

We do this by dividing both sides by $V_{2}$ $V_2$ in order to get $P_{2}$ $P_2$ by itself like so:
$P_{2} = \frac{P_{1} \times V_{1}}{V_{2}}$ $P_2=(P_1xxV_1)/V_2$

Now all we do is plug and chug!
$P_2=(180\mmHg xx 2.4xx10^(5)\ cancel"L")/(1.8xx10^(3)\cancel"L")$ = $2.4xx10^(4)$ $mmHg$

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If $2.4 \cdot 10^{5}$ $2.410^5$ $L$ $L$ of gas is at $180$ $180$ $m m H g$ $mmHg$ , what is the pressure when the gas is compressed to $1.8 \cdot 10^{3}$ $1.810^3$ $L$ $L$ at constant temperature?

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Explanation:

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If $2.4 \cdot 10^{5}$ $2.410^5$ $L$ $L$ of gas is at $180$ $180$ $m m H g$ $mmHg$ , what is the pressure when the gas is compressed to $1.8 \cdot 10^{3}$ $1.810^3$ $L$ $L$ at constant temperature?