A gas occupies $1.46 L$ $"1.46 L"$ at a pressure of $1.00 bar$ $"1.00 bar"$ . What is the volume when the gas is dropped into the ocean to a depth such that the pressure is increased to $60.0 bar$ $"60.0 bar"$ ?

Question

A gas occupies $1.46 L$ $"1.46 L"$ at a pressure of $1.00 bar$ $"1.00 bar"$ . What is the volume when the gas is dropped into the ocean to a depth such that the pressure is increased to $60.0 bar$ $"60.0 bar"$ ?

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Answer 1 · 2017-07-20T05:16:41

The final volume $(V_{2})$ $(V_2)$ is $0.0243 L$ $"0.0243 L"$ .

Refer to the explanation for the process.

Explanation:

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

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

Organize the data.

$P_{1} = 1.00 bar$ $P_1="1.00 bar"$

$V_{1} = 1.46 L$ $V_1="1.46 L"$

$P_{2} = 60.0 bar$ $P_2="60.0 bar"$

$V_{2} = ?$ $V_2="?"$

Solution

Rearrange the equation to isolate $V_{2}$ $V_2$ . Substitute the data into the equation and solve.

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

$V_2=(1.00color(red)cancel(color(black)("bar"))xx1.46"L")/(60.0color(red)cancel(color(black)("bar")))="0.0243 L"$ (rounded to three significant figures)

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A gas occupies $1.46 L$ $"1.46 L"$ at a pressure of $1.00 bar$ $"1.00 bar"$ . What is the volume when the gas is dropped into the ocean to a depth such that the pressure is increased to $60.0 bar$ $"60.0 bar"$ ?

1 Answer

Explanation:

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1 Answer

Explanation:

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A gas occupies $1.46 L$ $"1.46 L"$ at a pressure of $1.00 bar$ $"1.00 bar"$ . What is the volume when the gas is dropped into the ocean to a depth such that the pressure is increased to $60.0 bar$ $"60.0 bar"$ ?