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Question

A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter of 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas in the cylinder. Answer: 93.6 kPa, 157 kg

Solution

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The local atmospheric pressure can be determined from the total pressure at the initial state:

P1=Patm+F1A=Patm+m1gD24π\begin{align*} P_{1}&=P_{\text{atm}}+\dfrac{F_{1}}{A}\\ &=P_{\text{atm}}+\dfrac{m_{1}g}{\dfrac{D^{2}}{4}\pi} \end{align*}

Patm=P1m1gD24π=100kPa109.810.1424π103kPa=93.6kPa\begin{align*} P_{\text{atm}}&=P_{1}-\dfrac{m_{1}g}{\dfrac{D^{2}}{4}\pi}\\ &=100\:\text{kPa}-\dfrac{10\cdot9.81}{\dfrac{0.14^{2}}{4}\cdot\pi}\cdot10^{-3}\:\text{kPa}\\ &=\boxed{93.6\:\text{kPa}} \end{align*}

The mass that leads to the doubling of the pressure is:

Δm=m2m1=D2π4g(P2Patm)m1=D2π4g(2P1Patm)m1=0.142π49.81(210093.6)103kg10kg=157kg\begin{align*} \Delta m&=m_{2}-m_{1}\\ &=\dfrac{D^{2}\pi}{4g}(P_{2}-P_{\text{atm}})-m_{1}\\ &=\dfrac{D^{2}\pi}{4g}(2P_{1}-P_{\text{atm}})-m_{1}\\ &=\dfrac{0.14^{2}\pi}{4\cdot9.81}\cdot(2\cdot100-93.6)\cdot10^{3}\:\text{kg}-10\:\text{kg}\\ &=\boxed{157\:\text{kg}} \end{align*}

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