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3 February, 15:10

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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  1. 3 February, 18:36
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    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
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