Problem 57 - University Physics with Modern Physics

Question

A cylinder 1.00 m tall with inside diameter 0.120 \(\mathrm{m}\) is used to hold propane gas (molar mass 44.1 \(\mathrm{g} / \mathrm{mol}\) ) for use in a barbecue. It is initially filled with gas until the gauge pressure is \(1.30 \times 10^{6} \mathrm{Pa}\) and the temperature is \(22.0^{\circ} \mathrm{C} .\) The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is \(2.50 \times 10^{5}\) Pa. Calculate the mass of propane that has been used.

Step-by-Step Solution

Answer

The mass of propane used is approximately 214.2 grams.

Step 1: Understand the given values

First, identify the given values: the initial gauge pressure \( P_1 = 1.30 \times 10^6 \) Pa, the final gauge pressure \( P_2 = 2.50 \times 10^5 \) Pa, the volume of the cylinder \( V \), the temperature \( T = 22.0^{\circ} \mathrm{C} = 295.15 \) K (converted to Kelvin), and the molar mass of propane \( M = 44.1 \) g/mol. The diameter of the cylinder \( d = 0.120 \) m, and its height \( h = 1.00 \) m.

Step 2: Calculate the volume of the cylinder

Use the formula for the volume of a cylinder: \( V = \pi r^2 h \). First, calculate the radius \( r = d/2 = 0.060 \) m. Then, compute the volume: \( V = \pi (0.060)^2 (1.00) \approx 0.0113 \) m³.

Step 3: Convert gauge pressure to absolute pressure

Gauge pressure is relative to atmospheric pressure. Add atmospheric pressure (\( P_{atm} = 1.013 \times 10^5 \) Pa) to both the initial and final gauge pressures to get absolute pressures: \( P_{1,abs} = 1.30 \times 10^6 + 1.013 \times 10^5 = 1.4013 \times 10^6 \) Pa; \( P_{2,abs} = 2.50 \times 10^5 + 1.013 \times 10^5 = 3.513 \times 10^5 \) Pa.

Step 4: Use the Ideal Gas Law

Apply the ideal gas law \( PV = nRT \) for both initial and final states. \( R = 8.314 \) J/(mol·K) is the gas constant. Calculate initial moles \( n_1 = \frac{P_{1,abs}V}{RT} \) and final moles \( n_2 = \frac{P_{2,abs}V}{RT} \) using the absolute pressures calculated. Substitute \( V = 0.0113 \) m³ and \( T = 295.15 \) K into these equations.

Step 5: Calculate the initial and final moles

Substitute the values in: \( n_1 = \frac{1.4013 \times 10^6 \times 0.0113}{8.314 \times 295.15} \approx 6.48 \) mol and \( n_2 = \frac{3.513 \times 10^5 \times 0.0113}{8.314 \times 295.15} \approx 1.62 \) mol.

Step 6: Calculate the mass of propane used

Find the change in moles: \( \Delta n = n_1 - n_2 = 6.48 - 1.62 = 4.86 \) mol. Convert moles to mass using the molar mass \( m = \Delta n \times 44.1 \approx 214.2 \) g. Therefore, the mass of propane used is approximately 214.2 grams.

Key Concepts

Gas Pressure

Gas pressure is a measure of the force that the gas molecules exert when they collide with the surfaces of a container. In this exercise, you'll encounter both gauge pressure and absolute pressure. Knowing the difference is crucial.
Gauge pressure is the pressure of the gas relative to atmospheric pressure. To convert gauge pressure to absolute pressure, add atmospheric pressure to the gauge pressure value.- **Gauge Pressure**: The exercise gives us the initial gauge pressure as \(1.30 \times 10^6\) Pa and the final gauge pressure as \(2.50 \times 10^5\) Pa.- **Atmospheric Pressure**: It's typically around \(1.013 \times 10^5\) Pa at sea level. Adding this to the gauge pressures converts them to absolute pressures.- **Importance**: Using absolute pressure allows us to apply it within the Ideal Gas Law to find an unknown, like the number of moles.

Cylinder Volume

Understanding the volume of a cylinder is essential to solving gas-related problems with the Ideal Gas Law. The formula for the volume \( V \) of a cylinder is \( V = \pi r^2 h \).
- **Components of the Formula**: - \( r \) is the radius of the cylinder. - \( h \) is the height of the cylinder.- **Calculation Example**: For a cylinder with a diameter of 0.120 m, the radius \( r = 0.060 \) m, and the height \( h = 1.00 \) m. - Substitute into the formula: \[ V = \pi \times (0.060)^2 \times 1.00 \] - This results in approximately \(0.0113\, \text{m}^3\).Understanding how to calculate volume is crucial whenever applying pressure values in the Ideal Gas Law, because the law includes \(V\), the volume.

Molar Mass

Molar mass is the mass of one mole of a substance. For propane, the molar mass \( M \) is 44.1 g/mol, as given in the problem. In gas calculations, molar mass is often used to convert moles of gas to grams, providing a practical understanding of amounts.- **Purpose in Calculations**: - Convert between moles \( n \) and mass \( m \) using: \[ m = n \times M \]- **Example Calculation**: - If the change in moles \( \Delta n \) is calculated to be \(4.86\) moles, the equation becomes: \[ m = 4.86 \times 44.1 \] - This results in approximately 214.2 grams, representing the mass of propane used during the specified use.Understanding molar mass is key to translating theoretical model results, like moles, into real-world units of mass.

Gauge Pressure

Gauge pressure indicates the pressure of a gas relative to atmospheric pressure. The gauge pressure doesn't include the atmospheric pressure in its measurement, which distinguishes it from absolute pressure. This exercise helps you see how the gauge pressure changes before and after a partial release of gas from the cylinder.
- **Initial and Final Pressures**: - Initial gauge pressure given is \(1.30 \times 10^6\) Pa. - Final gauge pressure after some gas is used is \(2.50 \times 10^5\) Pa.- **Addition of Atmospheric Pressure**: - To get the absolute pressure, add atmospheric pressure to both gauge measurements: - This transforms them into absolute terms required for the Ideal Gas Law calculations.The distinction is critical when solving problems that involve multiple systems of pressure, especially when calculations require transformations between these states using consistent units of Pa (Pascals).