At a pressure 42 kPa, the gas in a cylinder has a volume of 18 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 2 liters, what is the new pressure

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24-06-2016
Chemistry

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At a pressure 42 kPa, the gas in a cylinder has a volume of 18 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 2 liters, what is the new pressure

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ratibyuksel ratibyuksel · Answer 1 · 2016-06-24T01:24:25+02:00

The ideal gas law formula is: PV=nRT. As the n(moles of gas), R(Gas constant) and T(temperature) values are constant as it is given in the problem, we can calculate the pressure difference by the following formula: xΔP(difference factor of pressure)=1/xΔV. We can find the difference factor of volume by x=V(0)/v(1),
x=18l/2l=9 which means the pressure is 9 times greater after the volume is down to 2l. To find the final pressure, we can use P(1)=xP(0), when we add the numbers in, it looks like this: P(1)=9*42 kPa=378 kPa