Question

A vertical cylindrical tank contains 1.80 mol of an ideal gas under a pressure of 0.300 atm at 20.0\(^\circ\)C. The round part of the tank has a radius of 10.0 cm, and the gas is supporting a piston that can move up and down in the cylinder without friction. There is a vacuum above the piston. (a) What is the mass of this piston? (b) How tall is the column of gas that is supporting the piston?

Short answer

(a) The mass of the piston is approximately 97.3 kg. (b) The height of the gas column is about 4.58 m.

Step-by-step solution

Understand the Problem

We need to find the mass of the piston that the gas can support, and the height of the gas column in the tank. We are given the amount of gas (1.80 mol), the pressure (0.300 atm), the temperature (20°C), and the radius of the tank (10.0 cm). We'll use the ideal gas law and formulas for pressure and force.

Convert Units

Convert the pressure to Pascals and the radius to meters. The pressure is 0.300 atm, which is equivalent to:\[P = 0.300 \times 101325 \, \text{Pa} = 30397.5 \, \text{Pa}.\]The radius is 10.0 cm, which is:\[r = 0.10 \, \text{m}.\]

Apply the Ideal Gas Law

The ideal gas law is given by: \[ PV = nRT. \]We solve for the volume \( V \) of the gas. First, convert the temperature from Celsius to Kelvin:\[T = 20 + 273.15 = 293.15 \, \text{K}.\]Substitute the values:\[(30397.5) V = (1.80) (8.314) (293.15). \]\[ V \approx 0.1439 \, \text{m}^3. \]

Calculate the Area of the Piston

The area \( A \) of the piston is the same as the area of the circle of the tank's radius:\[A = \pi r^2 = \pi (0.10)^2 = 0.0314 \, \text{m}^2.\]

Calculate the Force Exerted by the Gas

The force \( F \) exerted by the gas can be calculated using:\[ F = PA = (30397.5) (0.0314) = 954.5 \, \text{N}.\]

Determine the Mass of the Piston

The mass \( m \) of the piston can be found using the equation for force, where \[F = mg. \]Therefore, \[ m = \frac{F}{g} = \frac{954.5}{9.81} \approx 97.3 \, \text{kg}.\]

Calculate the Height of the Gas Column

Now, to find the height \( h \) of the gas column, use:\[ V = Ah. \]Solve for \( h \):\[ h = \frac{V}{A} = \frac{0.1439}{0.0314} \approx 4.58 \, \text{m}.\]

Key concepts

Pressure conversion

Converting pressure units is crucial, especially when applying formulas that require the pressure in a specific unit like Pascals. In this exercise, the initial pressure is provided in atmospheres (atm), a common unit often used when discussing gases at a macroscopic level. However, for calculations involving the ideal gas law, the pressure needs to be in Pascals (Pa), which is more widely used in scientific formulas due to being a standard SI unit.
To convert from atmospheres to Pascals, use the conversion factor:

• 1 atm = 101,325 Pa.

This gives us the formula:

• \[ P = 0.300 \times 101,325 \, \text{Pa} = 30,397.5 \, \text{Pa}. \]

Understanding how to switch between these two units ensures accuracy across physical equations and helps when learning about physical principles expressed in different systems.

Piston mechanics

Pistons are vital components in many engineering and physics applications because they can change the volume of gas containers, consequently affecting pressure and work.In the given problem, the piston is supported by gas within a cylindrical tank. Key aspects to consider include:

• The area of the piston, which corresponds to the area of the tank's top circle. This is calculated through:\[ A = \pi r^2 = \pi (0.10 \, \text{m})^2 = 0.0314 \, \text{m}^2. \]
• The force exerted by the gas on the piston. This force can be calculated by multiplying the pressure of the gas by the area of the piston:\[ F = PA = 30,397.5 \, \text{Pa} \times 0.0314 \, \text{m}^2 = 954.5 \, \text{N}. \]
• Finally, the mass of the piston can be determined since the force exerted by the gas balances the gravitational force on the piston (assuming equilibrium):\[ m = \frac{F}{g} = \frac{954.5}{9.81} \approx 97.3 \, \text{kg}. \]

Understanding these mechanics helps explain how gases can be used to exert and balance forces in mechanical systems, highlighting the practical applications of physics principles.

Temperature conversion

Temperature conversion is essential in the application of the ideal gas law, as the law requires temperature to be in Kelvin (K), the standard SI unit for temperature.In Celsius, water freezes at 0°C, but in Kelvin, the same point is 273.15 K. To convert Celsius to Kelvin, use the simple conversion formula:

• \[ T(\text{K}) = T(\text{°C}) + 273.15. \]

For this problem, where the temperature is 20°C, the conversion to Kelvin gives:

• \[ T = 20 + 273.15 = 293.15 \, \text{K}. \]

Using Kelvin allows for straightforward and correct use of the ideal gas law:

• \[ PV = nRT, \]where \( R \) is the universal gas constant.

Grasping temperature conversion is pivotal for accurately working with thermal expansion, reactions, and various gas laws across scientific disciplines.