/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A large cylindrical tank contain... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A large cylindrical tank contains 0.750 \(\mathrm{m}^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(7.50 \times 10^{3}\) Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 \(\mathrm{m}^{3}\) and the temperature is increased to \(157^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The final pressure is \( 9.01 \times 10^3 \) Pa.

Step by step solution

01

Identify Initial and Final States

First, recognize the initial conditions and final conditions of the gas. Initially, the volume \( V_1 = 0.750 \, \text{m}^3 \), pressure \( P_1 = 7.50 \times 10^3 \, \text{Pa} \), and temperature \( T_1 = 27^\circ \text{C} \). Convert \( T_1 \) to Kelvin: \( T_1 = 27 + 273 = 300 \, \text{K} \). For the final state, the volume \( V_2 = 0.480 \, \text{m}^3 \) and temperature \( T_2 = 157^\circ \text{C} \). Convert \( T_2 \) to Kelvin: \( T_2 = 157 + 273 = 430 \, \text{K} \).
02

Utilize the Combined Gas Law

Apply the combined gas law, which is \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Here, we need to find the final pressure \( P_2 \). Rearrange the equation to solve for \( P_2 \): \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \]
03

Plug in the Values and Calculate

Substitute the known values into the equation: \[ P_2 = \frac{(7.50 \times 10^3 \, \text{Pa}) \times (0.750 \, \text{m}^3) \times (430 \, \text{K})}{(300 \, \text{K}) \times (0.480 \, \text{m}^3)} \] Calculate \( P_2 \) to find:\[ P_2 = \frac{(3225 \times 430)}{144} \approx 9.01 \times 10^3 \, \text{Pa} \]
04

Confirm the Result

Verify the units and the calculation for accuracy. The units of pressure (Pa) are consistent throughout. Double-check the multiplication and division steps to ensure all calculations were performed correctly.

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 91Ó°ÊÓ!

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 principle in chemistry and physics, represented by the equation \( PV = nRT \). This law describes how gases behave under various conditions of pressure (\( P \)), volume (\( V \)), and temperature (\( T \)). The ideal gas constant \( R \) and the number of moles \( n \) are also factored in. However, in exercises involving changing conditions, we often use the Combined Gas Law, which is adapted from the Ideal Gas Law. This law relates initial and final states of a gas, expressing that the ratio of the product of pressure, volume, and temperature remains constant. By knowing these properties, we can predict how one variable will change given changes in the others, which is key in solving problems like the one in the exercise. The relationship simplifies to \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) when the number of moles is constant.
Pressure Calculation
Calculating pressure is an essential part of understanding gas behaviors. In the given exercise, you start with an initial pressure of \(7.50 \times 10^3 \) Pascal (Pa) and need to find the new pressure when conditions change. The formula used is derived from the Combined Gas Law as:
  • \( P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \)
This formula helps calculate the final pressure (\( P_2 \)) by considering how volume and temperature are altered. Substituting the known values into the equation will allow you to solve for \( P_2 \), showing how pressure increases under certain conditions. The units of pressure, typically in Pascals (Pa), must remain consistent throughout your calculations to ensure accuracy.
Temperature Conversion
Temperature plays a crucial role in gas law calculations. It must be represented in Kelvin for mathematical accuracy because Kelvin is the absolute temperature scale. To convert from Celsius to Kelvin, add 273, as seen in the exercise where:
  • Initial temperature \( T_1 = 27^\circ \text{C} \rightarrow 300 \text{ K} \)
  • Final temperature \( T_2 = 157^\circ \text{C} \rightarrow 430 \text{ K} \)
This conversion is essential to ensure the equations and calculations are correct because using Celsius directly would yield incorrect results. Kelvin ensures that the scale starts at absolute zero, allowing the gas laws to apply universally.
Volume Change
Understanding how changes in volume affect pressure and temperature is crucial. In this exercise, the volume of nitrogen gas is decreased from \(0.750 \, \text{m}^3\) to \(0.480 \, \text{m}^3\). This reduction in volume, when paired with an increase in temperature, predicts an increase in pressure according to the Combined Gas Law.
  • Decrease in volume since pressure tends to increase when a gas is compressed.
  • Relationship between volume and pressure is inversely proportional, meaning if volume decreases, pressure tends to increase if the temperature is held constant or increases.
Thus, volume change, alongside temperature change, allows accurate calculation of new pressure conditions, illustrating the interplay between the variables of gases.

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

At an altitude of \(11,000 \mathrm{m}\) (a typical cruising altitude for a jet airliner), the air temperature is \(-56.5^{\circ} \mathrm{C}\) and the air density is 0.364 \(\mathrm{kg} / \mathrm{m}^{3} .\) 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.)

If a certain amount of ideal gas occupies a volume \(V\) at STP on earth, what would be its volume (in terms of \(V )\) on Venus, where the temperature is \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

A metal tank with volume 3.10 \(\mathrm{L}\) will burst if the absolute pressure of the gas it contains exceeds 100 atm. (a) If 11.0 mol of an ideal gas is put into the tank at a temperature of \(23.0^{\circ} \mathrm{C},\) to what temperature can the gas be warmed before the tank ruptures? You can ignore the thermal expansion of the tank. (b) Based on your answer to part (a), is it reasonable to ignore the thermal expansion of the tank? Explain.

Hydrogen on the Sun. The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27}\) kg. (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2},\) where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant \((\) see Example 13.5 of Section 13.3\() .\) Use the data in Appendix \(F\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

Planetary Atmospheres. (a) Calculate the density of the atmosphere at the surface of Mars (where the pressure is 650 \(\mathrm{Pa}\) and the temperature is typically \(253 \mathrm{K},\) with a \(\mathrm{CO}_{2}\) atmosphere), Venus (with an average temperature of 730 \(\mathrm{K}\) and pressure of 92 atm, with a \(\mathrm{CO}_{2}\) atmosphere), and Saturn's moon Titan (where the pressure is 1.5 atm and the temperature is \(-178^{\circ} \mathrm{C},\) with a \(\mathrm{N}_{2}\) atmosphere). (b) Compare each of these densities with that of the earth's atmosphere, which is 1.20 \(\mathrm{kg} / \mathrm{m}^{3} .\) Consult the periodic chart in Appendix D to determine molar masses.

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.