/*! 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 in \(^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(1.50 \times 10^{5} \mathrm{Pa}\) (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 approximately \(3.36 \times 10^5\) Pa.

Step by step solution

01

Define the Ideal Gas Law

The ideal gas law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
02

Convert Temperatures to Kelvin

The initial temperature \( T_1 \) in Celsius is 27°C, which is equal to \( T_1 = 27 + 273.15 = 300.15 \) K. The final temperature \( T_2 \) is 157°C, which is equal to \( T_2 = 157 + 273.15 = 430.15 \) K.
03

Apply the Combined Gas Law

Since \( n \) and \( R \) are constant, we use \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) to find \( P_2 \). Here, \( P_1 = 1.50 \times 10^5 \) Pa, \( V_1 = 0.750 \) m³, and \( T_1 = 300.15 \) K. \( V_2 = 0.480 \) m³ and \( T_2 = 430.15 \) K.
04

Solve for Final Pressure \( P_2 \)

Rearrange the equation to solve for \( P_2 \): \[ P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \]. Substitute the values: \( P_2 = 1.50 \times 10^5 \times \frac{0.750}{0.480} \times \frac{430.15}{300.15} \).
05

Calculate the Value of \( P_2 \)

Calculate the expression: \( P_2 = 1.50 \times 10^5 \times 1.5625 \times 1.433 \). Compute to get \( P_2 \approx 3.36 \times 10^5 \) Pa.

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.

Gas Laws
The key concept behind the given exercise is the Ideal Gas Law, which is a cornerstone of gas behavior studies. This law combines several other gas laws and provides a generalized equation to describe the state of an ideal gas. The law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is the volume occupied by the gas, \( n \) represents the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Explaining each variable helps to understand their role:
  • **Pressure (P)**: This is the force exerted by the gas particles against the walls of the container per unit area.
  • **Volume (V)**: Represents the space the gas occupies.
  • **Moles (n)**: A measure of the quantity of gas particles present.
  • **Temperature (T)**: Impacts the speed and energy of the gas particles, which in turn affects pressure and volume.
Understanding these variables and their interrelations is crucial for solving problems related to changing states of gases. For cases where the amount of gas and the constant \( R \) remain the same, the Combined Gas Law \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) is often used to solve problems as in this exercise, enabling the calculation of a new pressure after changes in volume or temperature.
Thermodynamics
Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. It incorporates how gas laws integrate into broader thermodynamic concepts, explaining how energy is transferred in the form of heat and work, which affects pressure and volume changes in gases.

In this exercise, the temperature of the gas initially at \( 27^{\circ} \mathrm{C} \) increases to \( 157^{\circ} \mathrm{C} \). To facilitate calculations using the Ideal Gas Law, temperature is converted into Kelvin, which sets the absolute scale avoiding negative values that are non-physical in this context. The Kelvin scale is directly related to energy; higher Kelvin numbers indicate more kinetic energy, impacting gas behavior.

Thermodynamic processes influencing gas laws can be described through terms like **isothermal**, where temperature remains constant, or **adiabatic**, where no heat exchange occurs. In this case, it's neither, as both temperature and volume change, adhering to **non-isothermal** conditions. Understanding these concepts helps in predicting outcomes when modifying temperature and volume in gas systems.
Pressure Calculations
Pressure calculation is a focal theme in gas laws. In this problem, the **final pressure** of the gas is determined by changes in both volume and temperature. The formula used, derived from the Combined Gas Law, simplifies to \[ P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \].

Each component of this equation represents a factor contributing to the new pressure:
  • **Initial Pressure \( (P_1) \)**: Acts as a starting point for calculations.
  • **Volume Ratio \( (\frac{V_1}{V_2}) \)**: A smaller final volume \( V_2 \) compared to the initial volume \( V_1 \) will increase pressure, assuming constant temperature initially under Boyle’s Law expectations.
  • **Temperature Ratio \( (\frac{T_2}{T_1}) \)**: A higher final temperature \( T_2 \), compared to \( T_1 \), increases pressure as gases expand according to Charles’s Law.
Thus, converting and combining these ratios leads to the calculated final pressure \( P_2 \), showing how these physical changes impact gas behavior in a quantifiable manner. Understanding these calculations is essential for predicting and controlling gas systems in practical applications, like in the case of pistons within engines or measuring atmospheric pressure changes.

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

For carbon dioxide gas \(\left(\mathrm{CO}_{2}\right),\) the constants in the van der Waals equation are \(a=0.364 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.27 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (a) If 1.00 \(\mathrm{mol}\) of \(\mathrm{CO}_{2} \mathrm{gas}\) at 350 \(\mathrm{K}\) is confined to a volume of 400 \(\mathrm{cm}^{3}\) , find the pressure of the gas using the ideal-gas equation and the van der Waals equation. (b) Which equation gives a lower pressure? Why? What is the percentage difference of the van der Waals equation result from the ideat-gas equation result? (c) The gas is kept at the same temperature as it expands to a volume of 4000 \(\mathrm{cm}^{3} .\) Repeat the calculations of parts (a) and (b). (d) Explain how your calculations show that the van der Waals equation is equivalent to the ideat-gas equation if \(n / V\) is small.

The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C}\) and a pressure of 1.00 \(\mathrm{atm}\) . (a) How many liters does 1.00 \(\mathrm{mol}\) of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth- atmospheres. Of course she would use the Venusian-atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

A cylinder 1.00 \(\mathrm{m}\) 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.

A physicist places a piece of ice at \(0.00^{\circ} \mathrm{C}\) and a beaker of water at \(0.00^{\circ} \mathrm{C}\) inside a glass box and closes the lid of the box. All the air is then removed from the box. If the ice, water, and beaker are all maintained at a temperature of \(0.00^{\circ} \mathrm{C},\) describe the final equilibrium state inside the box.

A cubical cage 1.25 \(\mathrm{m}\) on each side contains 2500 angry bees, each flying randomly at 1.10 \(\mathrm{m} / \mathrm{s}\) . We can model these insects as spheres 1.50 \(\mathrm{cm}\) in diameter. On the average, (a) how far does a typical bee travel between collisions, (b) what is the average time between collisions, and (c) how many collisions per second does a bee make?

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.