If a volume of $1.7 \cdot a t m$ $1.7atm$ at $26.0 \cdot L$ $26.0L$ is compressed to $10.7 \cdot L$ $10.7*L$ , what is the new pressure?

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If a volume of $1.7 \cdot a t m$ $1.7atm$ at $26.0 \cdot L$ $26.0L$ is compressed to $10.7 \cdot L$ $10.7*L$ , what is the new pressure?

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Answer 1 · 2017-06-18T17:48:29

Well, we use [old Boyle's Law.........](https://en.wikipedia.org/wiki/Boyle%27s_law)

Explanation:

Which holds that for a given mass of gas at constant temperature, then $P V = k$ $PV=k$ , where $k$ $k$ is some constant...........and so under equivalent conditions of temperature......

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

And here we solve for $P_{2} = \frac{P_{1} V_{1}}{V_{2}}$ $P_2=(P_1V_1)/V_2$

$= \frac{1.7 \cdot a t m \times 26.0 \cdot L}{10.7 \cdot L}$ $=(1.7*atmxx26.0*L)/(10.7*L)$

$=4.1*atm.........$

Why is it reasonable that pressure has increased?

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If a volume of $1.7 \cdot a t m$ $1.7atm$ at $26.0 \cdot L$ $26.0L$ is compressed to $10.7 \cdot L$ $10.7*L$ , what is the new pressure?

1 Answer

Explanation:

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If a volume of 1.7*atm1.7⋅atm1.7*atm at 26.0*L26.0⋅L26.0*L is compressed to 10.7*L10.7⋅L10.7*L, what is the new pressure?

1 Answer

Explanation:

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If a volume of $1.7 \cdot a t m$ $1.7atm$ at $26.0 \cdot L$ $26.0L$ is compressed to $10.7 \cdot L$ $10.7*L$ , what is the new pressure?