Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
21.04 grams of propane has been used.

Step by step solution

01

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.
02

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, \[ V = \pi \times (0.060)^2 \times 1.00 = 0.0113 \, \text{m}^3 \]
03

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 \( n_f \):\[ n_f \approx 0.170 \, \text{moles} \]
04

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):\[ \Delta m = 0.477 \times 44.1 = 21.04 \, \text{grams} \]
05

Conclusion

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 3.00-L tank contains air at 3.00 atm and 20.0\(^\circ\)C. The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

At an altitude of 11,000 m (a typical cruising altitude for a jet airliner), the air temperature is -56.5\(^\circ\)C and the air density is 0.364 kg/m\(^3\). What is the pressure of the atmosphere at that altitude? (Note: The temperature at this altitude is not the same as at the surface of the earth, so the calculation of Example 18.4 in Section 18.1 doesn't apply.)

We have two equal-size boxes, \(A\) and \(B\). Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at 50\(^\circ\)C while the gas in box \(B\) is at 10\(^\circ\)C. This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? Explain your reasoning. (a) The pressure in \(A\) is higher than in \(B\). (b) There are more molecules in \(A\) than in \(B\). (c) A and B do not contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\). (e) The molecules in \(A\) are moving faster than those in \(B\).

Estimate the number of atoms in the body of a 50-kg physics student. Note that the human body is mostly water, which has molar mass 18.0 g/mol, and that each water molecule contains three atoms.

A person at rest inhales 0.50 L of air with each breath at a pressure of 1.00 atm and a temperature of 20.0\(^\circ\)C. The inhaled air is 21.0% oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of 2000 m but the temperature is still 20.0\(^\circ\)C. Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report 'shortness of breath' at high elevations.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free