Chapter 1.11, Problem 98RP

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 inside the cylinder Answers: 93.6 kPa, 157 kg

To determine

The local atmospheric pressure of the vertical piston-cylinder device.

The mass of the weights that will double the pressure of the vertical piston-cylinder device.

Answer to Problem 98RP

The local atmospheric pressure of the vertical piston-cylinder device is $\underset{\_}{93.63\text{kPa}}$.

The mass of the weights that will double the pressure of the vertical piston-cylinder device is $\underset{\_}{157\text{kg}}$.

Explanation of Solution

Show the free body diagram of the vertical piston-cylinder device.

Write the expression of vertical force in the piston-cylinder device.

$P_{\text{atm}}=P-\frac{m_{\text{piston}}g}{A}$ (I)

Here, the mass of piston is $m_{\text{piston}}$, the absolute pressure is $P$, the atmospheric pressure is $P_{\text{atm}}$, the cross-sectional area is $A$, and acceleration gravity is $g$.

Write the expression of balance force in the piston-cylinder device.

$P=P_{\text{atm}}+\frac{\left(m_{\text{piston}}+m_{\text{weight}}\right)g}{A}$ (II)

Here, the mass of the weights that will double the pressure is $m_{\text{weight}}$.

Determine the area of the piston-cylinder.

$A=\pi D^2/4$ (III)

Conclusion:

Substitute $14\text{cm}\text{}$ for $D$ in Equation (III).

$\begin{array}{c}A=\pi {\left(14\text{cm}\right)}^2/4\\ =153.9{\text{cm}}^2\times \left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\\ \simeq 0.0154{\text{m}}^2\end{array}$

Substitute $0.0154{\text{m}}^2$ for $A$, $10\text{kg}$ for $m_{\text{piston}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $100\text{kPa}$ for $P$ in Equation (I).

$\begin{array}{c}{P}_{\text{atm}}=100\text{kPa}-\frac{\left(10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{0.0154{\text{m}}^{\text{2}}}\\ =100\text{kPa}-\left(6370.129\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\\ =93.63\text{kPa}\end{array}$

Thus, the local atmospheric pressure of the vertical piston-cylinder device is $\underset{\_}{93.63\text{kPa}}$.

Substitute $0.0154{\text{m}}^2$ for $A$, $10\text{kg}$ for $m_{\text{piston}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $93.63\text{kPa}$ for $P_{\text{atm}}$, and $200\text{kPa}$ for $P$ in Equation (II).

$\begin{array}{l}200\text{kPa}=93.63\text{kPa}-\frac{\left(10\text{kg}+m_{\text{weight}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{0.0154{\text{m}}^{\text{2}}}\\ \left(10\text{kg}+m_{\text{weight}}\right)=\left(93.63\text{kPa}\right)\left(\frac{9.81\text{m}/{\text{s}}^{\text{2}}}{0.0154{\text{m}}^{\text{2}}}\right)\times \left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ m_{\text{weight}}=157\text{kg}\end{array}$

Thus, the mass of the weights that will double the pressure of the vertical piston-cylinder device is $\underset{\_}{157\text{kg}}$.