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Valentin [98]
4 years ago
12

At a pressure 42 kPa, the gas in a cylinder has a volume of 11 liters. Assuming temperature remains the same, if the volume of t

he gas is decreased to 9 liters, what is the new pressure?
Physics
1 answer:
nlexa [21]4 years ago
5 0
Boyle's law<span> is a gas </span>law<span>, stating that the pressure and volume of a gas have an inverse relationship , when temperature is held constant. That is PV = constant. Therefore, (PV)initial = (PV)final. 42x11 = 9x P(final). P(final) = 42x11/9 = 51.34kPa. </span>
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Answer:

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NOTE: We use the relation 1rpm*\frac{2\pi}{60s}=\frac{rad}{s} to convert 280 rev/min(rpm) to 29.32 rad/s

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