$3.761 \times 10^{1} . L$ $3.761xx10^1color(white)(.)"L"$ of an ideal gas is in a balloon at $1.000 \times 10^{0} . atm$ $1.000xx10^0 color(white)(.)"atm"$ . The weather changes, and the volume of the balloon changes to $5.00 \times 10^{1} L$ $5.00xx10^1 "L"$ . What is the new pressure, assuming no change in temperature?

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$3.761 \times 10^{1} . L$ $3.761xx10^1color(white)(.)"L"$ of an ideal gas is in a balloon at $1.000 \times 10^{0} . atm$ $1.000xx10^0 color(white)(.)"atm"$ . The weather changes, and the volume of the balloon changes to $5.00 \times 10^{1} L$ $5.00xx10^1 "L"$ . What is the new pressure, assuming no change in temperature?

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Answer 1 · 2017-04-26T01:01:03

The new pressure is $7.52 \times 10^{- 1} . atm$ $7.52xx10^(-1)color(white)(.)"atm"$ .

Explanation:

This is an example of Boyle's law , which states that the volume of a gas held at constant amount and temperature, varies inversely with the pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:

$P_{1} V_{1} = P_{2} V_{2}$ $"P_1V_1=P_2V_2$

where $P$ $P$ is pressure and $V$ $V$ is volume.

Organize the information:

Known
$P_{1} = 1.000 \times 10^{0} atm$ $P_1=1.000xx10^0"atm"$
$V_{1} = 3.761 \times 10^{1} L$ $V_1=3.761xx10^1 "L"$
$V_{2} = 5.00 \times 10^{1} L$ $V_2=5.00xx10^1"L"$

Unknown: $P_{2}$ $P_2$

Solution
Rearrange the equation to isolate $P_{2}$ $P_2$ , substitute the known values into the equation and solve.

$P_{2} = \frac{P_{1} V_{1}}{V_{2}}$ $P_2=(P_1V_1)/V_2$

$P_2=(1.000xx10^0"atm"xx3.761xx10^1color(red)cancel(color(black)"L"))/(5.00xx10^1color(red)cancel(color(black)"L"))=7.52xx10^(-1)"atm"$ rounded to three significant figures

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