## Piston Basics

Two cylinders containing a gas are connected as shown in the figure below. Piston A has a 5 cm diameter, while Piston B has a 3 cm diameter. The atmospheric pressure is 101 kPa.

- If the mass of Piston A is 18 kg, what is the Piston B’s mass?
- What is the pressure in the system?

##
__
__**Hint**

**Hint**

$$$Pressure=\frac{Force}{Area}=\frac{mass\times acceleration}{area}$$$

##
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__**Hint 2**

**Hint 2**

$$$P_0+\frac{m_Ag}{A_A}=P_0+\frac{m_Bg}{A_B}$$$

where
$$P_0$$
is the atmospheric pressure,
$$m$$
is the mass,
$$g$$
is the acceleration due to gravity, and
$$A$$
is the area.

As a system, the pressures in Piston A and Piston B are equivalent. So,

$$$P_A=P_B=\frac{Force}{Area}=\frac{mass\times acceleration}{area}$$$

$$$P_0+\frac{m_Ag}{A_A}=P_0+\frac{m_Bg}{A_B}$$$

where
$$P_0$$
is the atmospheric pressure,
$$m$$
is the mass,
$$g$$
is acceleration due to gravity, and
$$A$$
is the area. Solving for Piston B’s mass:

$$$m_B=\frac{A_B}{A_A}m_A=\frac{3^2}{5^2}(18)=6.48\:kg$$$

Solving for system pressure,

$$$P_A=P_B=P_0+\frac{m_Bg}{A_B}=101+\frac{(6.48)(9.81)(10)^{-3}}{\pi (0.03)^2/4}=190\:kPa$$$

- 6.48 kg
- 190 kPa