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Chapter 24: Problem 23

Air is pumped into an automobile tyre's tube up to a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(20^{\circ} \mathrm{C}\). During the day the temperature rises to \(40^{\circ} \mathrm{C}\) and the tube expands by \(2 \%\). Calculate the pressure of the air in the tube at this temperature.

Short Answer

Expert verified
The pressure is approximately 209.73 kPa.

Step by step solution

01

Understand the problem

We need to find the new pressure in the tire once the temperature has increased and the volume has expanded slightly. The initial pressure is given as 200 kPa and the temperature as 20掳C. The final temperature is 40掳C with a volume increase of 2%.
02

Convert temperatures to Kelvin

To use the Ideal Gas Law accurately, we must first convert the temperatures from Celsius to Kelvin. The conversion is \( K = 掳C + 273.15 \). So the initial temperature is \( T_1 = 20 + 273.15 = 293.15\, K \) and the final temperature is \( T_2 = 40 + 273.15 = 313.15\, K \).
03

Understand volume change

A volume expansion of 2% means that the final volume \( V_2 = 1.02 \times V_1 \), where \( V_1 \) is the initial volume. This will affect our gas law calculations.
04

Apply Combined Gas Law

The Combined Gas Law relates pressure, volume, and temperature:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]We are given that \( P_1 = 200 \) kPa, \( V_2 = 1.02 V_1 \), \( T_1 = 293.15 \) K, \( T_2 = 313.15 \) K, and need to find \( P_2 \).
05

Solve for the new pressure \(P_2\)

Rearrange the Combined Gas Law to solve for \( P_2 \):\[P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}\]Substitute the known values:\[P_2 = 200 \times \frac{1}{1.02} \times \frac{313.15}{293.15}\]Simplify the expression:\[P_2 \approx 200 \times 0.9804 \times 1.0682\]\[P_2 \approx 209.73 \text{ kPa}\]
06

Conclude the calculation

The pressure in the tire at \( 40^{\circ} \mathrm{C} \) with a 2% volume increase is approximately 209.73 kPa.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics, connecting pressure, volume, temperature, and the number of moles of a gas. It is expressed by the equation \( PV = nRT \), where:
  • \(P\) is the pressure of the gas
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the universal gas constant
  • \(T\) is the temperature in Kelvin
This law is a simplification for ideal gases, assuming no interactions between molecules and that they occupy negligible space.
To apply the Ideal Gas Law, temperature must be measured in Kelvin, given the direct proportionality in the equation. However, real-world conditions often require adjustments using more advanced equations like the Combined Gas Law, especially when variables such as temperature and volume fluctuate.
Pressure and Temperature Relationship
The relationship between pressure and temperature in gases is explained by Gay-Lussac's Law, a facet of the broader Ideal Gas Law. This law states that for a fixed volume of gas, the pressure is directly proportional to its temperature, provided the amount of gas remains constant.
Mathematically, it is expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Here, \(P_1\) and \(T_1\) are the initial pressure and temperature, while \(P_2\) and \(T_2\) are conditions after a change.
In the exercise, when the temperature inside the tire increases from 20掳C to 40掳C, we expect the pressure to increase if the volume was held constant. However, volume expansion introduces a variable changing the dynamics slightly.
Keeping account of temperature changes in Kelvin is critical, as it ensures direct proportionality aligning with ideal gas assumptions. Understanding this foundational relationship aids in predicting gas behavior in varying conditions.
Volume Expansion in Gases
Volume expansion in gases is a crucial concept when analysing changes in gas behavior under varying conditions. When a gas experiences a change in temperature, it may expand if not contained within a rigid volume. This expansion can impact the pressure within a fixed volume larger or more compressible than the initial one.
In practical scenarios, like in the given exercise, a 2% volume increase indicates a situation where the gas has slightly more space due to either an elastic or flexible boundary such as a car tire. This is modeled in gas calculations by adjusting the initial volume \(V_1\) to find \(V_2 = 1.02V_1\) due to the 2% increase.
By applying the Combined Gas Law, \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), we adjust for this volume change and temperature increase, allowing us to find the new pressure where both temperature and volume have altered. Recognizing volume expansion helps accurately calculate resultant pressures in non-ideal conditions.

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Most popular questions from this chapter

The average translational kinetic energy of air molecules is \(0.040 \mathrm{eV} \quad\left(1 \mathrm{eV}=1 \cdot 6 \times 10^{-19} \mathrm{~J}\right) . \quad\) Calculate the temperature of the air. Boltzmann constant \(k=1 \cdot 38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\)

A bucket full of water is placed in a room at \(15^{\circ} \mathrm{C}\) with initial relative humidity \(40 \%\). The volume of the room is \(50 \mathrm{~m}^{3}\). (a) How much water will evaporate ? (b) If the room temperature is increased by \(5^{\circ} \mathrm{C}\), how much more water will evaporate ? The saturation vapour pressure of water at \(15^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\) are \(1.6 \mathrm{kPa}\) and \(2 \cdot 4 \mathrm{kPa}\) respectively.

Figure (24-E1) shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of \(1: 3\). Find the ratio of the pressures in the two parts of the vessel.

One mole of an ideal gas undergoes a process $$ p=\frac{p_{0}}{1+\left(V / V_{0}\right)^{2}} $$ where \(p_{0}\) and \(V_{0}\) are constants. Find the temperature of the gas when \(V=V_{0}\).

An electric bulb of volume \(250 \mathrm{cc}\) was sealed during manufacturing at a pressure of \(10^{-3} \mathrm{~mm}\) of mercury at \(27^{\circ} \mathrm{C}\). Compute the number of air molecules contained in the bulb. Avogadro constant \(=6 \times 10^{23} \mathrm{~mol}^{-1}\), density of mercury \(=13600 \mathrm{~kg} \mathrm{~m}^{-3}\) and \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).
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