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An automobile tire has a volume of \(0.0185 \mathrm{m}^{3} .\) At a temperature of \(294 \mathrm{K}\) the absolute pressure in the tire is \(212 \mathrm{kPa}\). How many moles of air must be pumped into the tire to increase its pressure to \(252 \mathrm{kPa}\), given that the temperature and volume of the tire remain constant?

Short Answer

Expert verified
Add approximately 0.368 moles of air to increase the pressure.

Step by step solution

01

Identify Variables

To solve this problem, first identify and list the given information: initial volume \(V = 0.0185 \, \mathrm{m}^3\), initial temperature \(T = 294 \, \mathrm{K}\), initial pressure \(P_1 = 212 \, \mathrm{kPa}\), and final pressure \(P_2 = 252 \, \mathrm{kPa}\).
02

Apply Ideal Gas Law

Using the ideal gas law \(PV = nRT\), where \(P\) is pressure in pascals (1 kPa = 1000 Pa), \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant \(8.3145 \, \mathrm{J/(mol \, K)}\), and \(T\) is temperature.
03

Calculate Initial Moles

Convert the initial pressure to pascals \(P_1 = 212,000 \, \mathrm{Pa}\). Use \(n_1 = \frac{P_1 \times V}{R \times T}\) to find the initial moles:\[n_1 = \frac{212,000 \, \mathrm{Pa} \times 0.0185 \, \mathrm{m}^3}{8.3145 \, \mathrm{J/(mol \, K)} \times 294 \, \mathrm{K}}\]
04

Calculate Final Moles

Convert the final pressure to pascals \(P_2 = 252,000 \, \mathrm{Pa}\). Use \(n_2 = \frac{P_2 \times V}{R \times T}\) to calculate the new moles required:\[n_2 = \frac{252,000 \, \mathrm{Pa} \times 0.0185 \, \mathrm{m}^3}{8.3145 \, \mathrm{J/(mol \, K)} \times 294 \, \mathrm{K}}\]
05

Determine Additional Moles

Calculate the additional moles of air necessary to increase the pressure by finding \(n_2 - n_1\).

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Key Concepts

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

Pressure Conversion
Understanding pressure conversion is crucial when working with the ideal gas law. Pressure is often given in kilopascals (kPa), but the formula for the ideal gas law uses pascals (Pa). To convert kilopascals to pascals, multiply by 1000.
For example, if you have a pressure of 212 kPa, converting it to pascals means:
  • Multiply 212 by 1000, resulting in 212,000 Pa.
This conversion ensures that all units are consistent, which is necessary for accurate calculations. This is because the ideal gas constant, \(R\), is typically given in terms of joules per mole per Kelvin (J/(mol K)), and it requires the pressure to be in pascals.
Remembering to adjust your pressure units avoids errors in calculations and helps keep your results precise.
Moles Calculation
Moles calculation is a core aspect of using the ideal gas law to solve problems involving gases. The ideal gas law is expressed as:
\[PV = nRT\]Here, \(n\) represents the number of moles. To find the number of moles using the formula, rearrange it to solve for \(n\):
  • \(n = \frac{PV}{RT}\)
Start by inserting the known values for pressure \(P\), volume \(V\), the ideal gas constant \(R\), and temperature \(T\). In our example, with an initial pressure of 212,000 Pa and a final of 252,000 Pa, respectively, compute the moles as follows for both conditions:
  • Initial moles: \(n_1 = \frac{212,000 \times 0.0185}{8.3145 \times 294}\)
  • Final moles: \(n_2 = \frac{252,000 \times 0.0185}{8.3145 \times 294}\)
This will yield the number of moles before and after pressure change. Remember, solving actual problems may require adjusting for significant digits based on inputs.
Temperature and Volume Constants
In this type of problem, temperature and volume remain constant, which simplifies calculations. Since we don't have to account for changes in these variables, we can focus solely on the relationship between pressure and moles.
The constancy in temperature \((T = 294\, K)\) and volume \((V = 0.0185\, m^3)\) means that they can be factored together into a single constant term:
  • Constant term: \(\frac{V}{RT}\)
When given constant temperature and volume, the formula \(\frac{P}{n} = \text{constant}\) applies. This implies that if you increase moles, pressure increases, as seen in the ideal gas law. This direct relationship simplifies understanding how adding or removing moles will affect the pressure when temperature and volume don't change.
This fixed state of temperature and volume is what allows us to consider and solve the problem using the provided parameters without needing further data.

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Most popular questions from this chapter

A piston held at the temperature \(T\) contains a gas mixture with molecules of three different types; \(A, B,\) and \(C\). The corresponding molecular masses are \(m_{\mathrm{C}}>m_{\mathrm{B}}>m_{\mathrm{A}} .\) Rank these molecular types in order of increasing (a) average kinetic energy and (b) rms speed. Indicate ties where appropriate.

Peter catches a \(4.8-\mathrm{kg}\) striped bass on a fishing line \(0.54 \mathrm{mm}\) in diameter and begins to reel it in. He fishes from a pier well above the water, and his fish hangs vertically from the line out of the water. The fishing line has a Young's modulus of \(5.1 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) (a) What is the fractional increase in length of the fishing line if the fish is at rest? (b) What is the fractional increase in the fishing line's length when the fish is pulled upward with a constant speed of \(1.5 \mathrm{m} / \mathrm{s} ?\) (c) What is the fractional increase in the fishing line's length when the fish is pulled upward with a constant acceleration of \(1.5 \mathrm{m} / \mathrm{s}^{2} ?\)

A reaction vessel contains 8.06 g of \(\mathrm{H}_{2}\) and \(64.0 \mathrm{g}\) of \(\mathrm{O}_{2}\) at a temperature of \(125^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa}\). (a) What is the volume of the vessel? (b) The hydrogen and oxygen are now ignited by a spark, initiating the reaction \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) This reaction consumes all the hydrogen and oxygen in the vessel. What is the pressure of the resulting water vapor when it returns to its initial temperature of \(125^{\circ} \mathrm{C} ?\)

The rms speed of \(\mathrm{O}_{2}\) is \(1550 \mathrm{m} / \mathrm{s}\) at a given temperature. (a) Is the rms speed of \(\mathrm{H}_{2} \mathrm{O}\) at this temperature greater than, less than, or equal to \(1550 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Find the rms speed of \(\mathrm{H}_{2} \mathrm{O}\) at this temperature.

A large punch bowl holds \(3.99 \mathrm{kg}\) of lemonade (which is essentially water) at \(20.5^{\circ} \mathrm{C} .\) A \(0.0550-\mathrm{kg}\) ice cube at \(-10.2^{\circ} \mathrm{C}\) is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

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