Question

A cylinder 1.00 m tall with inside diameter 0.120 m is used to hold propane gas (molar mass 44.1 g/mol) for use in a barbecue. It is initially filled with gas until the gauge pressure is 1.30 \(\times\) 10\(^6\) Pa at 22.0\(^\circ\)C. The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is 3.40 \(\times\) 10\(^5\) Pa. Calculate the mass of propane that has been used.

Short answer

21.04 grams of propane has been used.

Step-by-step solution

Understand the Problem

We have a cylinder filled with propane gas, and we need to calculate the change in mass as some gas is used. The given parameters include initial and final gauge pressures, cylinder dimensions, temperature, and molar mass of propane. The task requires using these parameters to find out how much propane has been consumed.

Calculate the Volume of the Cylinder

The volume of the cylinder can be calculated using the formula for the volume of a cylinder: \[ V = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height. Given the diameter is 0.120 m, the radius \( r \) is 0.060 m. Therefore, \[ V = \pi \times (0.060)^2 \times 1.00 = 0.0113 \, \text{m}^3 \]

Calculate Initial and Final Moles of Gas

Using the Ideal Gas Law \( PV = nRT \), we calculate the number of moles \( n \). The absolute pressure \( P \) is the gauge pressure plus atmospheric pressure (\( 1.01 \times 10^5 \) Pa). Using \( R = 8.314 \, \text{J/mol K} \) and \( T = 273.15 + 22.0 \, \text{K} = 295.15 \, \text{K} \):Initial moles (\( n_i \)): \[\left(P_i + 1.01 \times 10^5 \right) \times V = n_i \times R \times T \]\[ (1.30 \times 10^6 + 1.01 \times 10^5) \times 0.0113 = n_i \times 8.314 \times 295.15 \]Solve for \( n_i \):\[ n_i \approx 0.647 \, \text{moles} \]Final moles (\( n_f \)):\[\left(P_f + 1.01 \times 10^5 \right) \times V = n_f \times R \times T \]\[ (3.40 \times 10^5 + 1.01 \times 10^5) \times 0.0113 = n_f \times 8.314 \times 295.15 \]Solve for \( n_f \):\[ n_f \approx 0.170 \, \text{moles} \]

Calculate Mass of Propane Used

The change in moles of propane \( \Delta n = n_i - n_f = 0.647 - 0.170 = 0.477 \, \text{moles} \).Convert to mass using the molar mass of propane (44.1 g/mol):\[ \Delta m = 0.477 \times 44.1 = 21.04 \, \text{grams} \]

Conclusion

The mass of propane that has been used is approximately 21.04 grams.

Key concepts

Cylinder Volume Calculation

When dealing with cylinders, knowing how to calculate their volume is essential, especially when studying gas laws. The volume of a cylinder can be found using the formula \( V = \pi r^2 h \). Here, \( r \) represents the radius of the cylinder, and \( h \) represents its height.

For this exercise, the cylinder has an internal diameter of 0.120 meters. To find the radius, you would divide the diameter by 2, giving us a radius of 0.060 meters. The height of this cylinder is 1.00 meter.

Plugging these values into the formula, the volume is calculated as follows:

• Radius: 0.060 meters
• Height: 1.00 meter
• Volume: \( V = \pi \times (0.060)^2 \times 1.00 = 0.0113 \; \text{m}^3 \)

Understanding these calculations is critical as it sets the stage for further calculations involving gases contained within the cylinder.

Molar Mass of Propane

The molar mass of a substance is the mass of one mole of its molecules. For propane, this molar mass is 44.1 g/mol. Propane is a common gas used in cooking and heating; it's composed of three carbon atoms and eight hydrogen atoms, giving it the molecular formula \( C_3H_8 \).

To find out how much gas has been used, you need to convert moles into grams using the molar mass. In this exercise, once we know the number of moles consumed, multiplying it by the molar mass of propane gives us the mass of the gas used.

• Molar mass of Propane: 44.1 g/mol
• Moles of Propane used: \( \Delta n = 0.647 - 0.170 = 0.477 \; \text{moles} \)
• Mass of Propane used: \( \Delta m = 0.477 \times 44.1 = 21.04 \; \text{grams} \)

This knowledge connects the number of particles present to the macroscopic measurements of mass, important for practical applications like fueling a barbecue.

Pressure and Temperature Relations

The Ideal Gas Law, expressed as \( PV = nRT \), is a fundamental principle when examining the behavior of gases. It showcases the relationship between pressure \( P \), volume \( V \), the number of moles \( n \), the ideal gas constant \( R \), and temperature \( T \).

In this problem, we focus on understanding how pressure changes relate to the change in the number of moles within a constant volume and temperature. Here, the temperature remains constant at 22 degrees Celsius, or 295.15 Kelvin.

• Initial gauge pressure: \( 1.30 \times 10^6 \; \text{Pa} \)
• Final gauge pressure: \( 3.40 \times 10^5 \; \text{Pa} \)
• Absolute pressure: Sum of gauge pressure plus atmospheric pressure
• Atmospheric pressure: \( 1.01 \times 10^5 \; \text{Pa} \)

By calculating the initial and final moles of gas, we see how adjusting pressure impacts the quantity of gas present. This understanding links directly to real-life scenarios such as burning fuel under consistent atmospheric conditions, highlighting the relevance of gas laws in practical settings.