/*! 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 54 (II) An ideal monatomic gas, con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) An ideal monatomic gas, consisting of \(2.8 \mathrm{~mol}\) of volume \(0.086 \mathrm{~m}^{3},\) expands adiabatically. The initial and final temperatures are \(25^{\circ} \mathrm{C}\) and \(-68^{\circ} \mathrm{C}\). What is the final volume of the gas?

Short Answer

Expert verified
The final volume of the gas is approximately \(0.151 \mathrm{~m}^3\).

Step by step solution

01

Convert Temperatures to Kelvin

To use the temperatures in thermodynamic equations, we must convert them from Celsius to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \).\( T_i = 25 + 273.15 = 298.15 \ \text{K} \)\( T_f = -68 + 273.15 = 205.15 \ \text{K} \)
02

Use the Adiabatic Process Formula

For an adiabatic process, the formula \( TV^{\gamma-1} = \text{constant} \) holds, where \( \gamma \) is the heat capacity ratio.For a monatomic ideal gas, \( \gamma = \frac{5}{3} \).Thus, \( T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1} \).
03

Rearrange for Final Volume

To find the final volume \( V_f \), rearrange the adiabatic process formula:\[ V_f = V_i \left( \frac{T_i}{T_f} \right)^{\frac{1}{\gamma-1}} \]
04

Calculate Final Volume

Substitute the known values into the equation:\( V_f = 0.086 \left( \frac{298.15}{205.15} \right)^{\frac{1}{\frac{5}{3}-1}} \)\( V_f = 0.086 \left( \frac{298.15}{205.15} \right)^{1.5} \)Calculate the expression:\( V_f \approx 0.086 \times 1.454^{1.5} \approx 0.086 \times 1.755 \approx 0.151 \ \text{m}^3 \)

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.

Monatomic Ideal Gas
A monatomic ideal gas is a theoretical gas composed of individual atoms, with no molecular complexity. For example, noble gases like helium or neon often serve as real-world approximations of monatomic gases.Monatomic gases are assumed to follow the ideal gas law perfectly, without interactions beyond elastic collisions.
When dealing with monatomic ideal gases, it is important to recognize the significance of their atomic simplicity, which minimizes degrees of freedom to just three translational motions. This simplicity allows for straightforward calculations of physical properties such as pressure, volume, and temperature using the ideal gas law formula:
  • \( PV = nRT \)
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. These core properties make monatomic ideal gases an essential and practical tool for exploring basic thermodynamic processes.
Heat Capacity Ratio
The heat capacity ratio, often denoted by \( \gamma \), is a key concept in understanding gas behavior during different thermodynamic processes.For a monatomic ideal gas, \( \gamma \) is the ratio of specific heats at constant pressure \( C_p \) and constant volume \( C_v \).
The formula for the heat capacity ratio is given by:
  • \( \gamma = \frac{C_p}{C_v} \)
For monatomic gases, this value is \( \frac{5}{3} \).
In an adiabatic process, which has no heat exchange with the surroundings, the heat capacity ratio helps to determine how changes in temperature affect the volume and pressure.This relationship is crucial in formulating the adiabatic equation \( TV^{\gamma-1} = \text{constant} \), which describes how the temperature and volume are related when no heat is lost or gained.Understanding these ratios enables predictions of the gas behavior in naturally occurring and engineered processes.
Thermodynamic Equations
Thermodynamic equations are the mathematical expressions that describe the relationships between various physical properties of thermodynamic systems.These equations help us understand how systems transfer heat and work within different processes, such as isothermal, isobaric, and adiabatic processes.
Key equations in thermodynamics describe how parameters like temperature, pressure, volume, and energy interact:
  • Ideal Gas Law: \( PV = nRT \)
  • First Law of Thermodynamics: \( \Delta U = Q - W \)
  • Adiabatic Process Equation: \( PV^\gamma = \text{constant} \)

Each thermodynamic equation serves a unique role;for example, in adiabatic processes, the adiabatic equation is applied because it takes into account no heat transfer \( Q = 0 \).Applying the right equation accurately predicts the changes in a system's state and guides us in understanding real-world phenomena through simplified models.

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

(II) At a crime scene, the forensic investigator notes that the \(7.2-\mathrm{g}\) lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature \(\left(20^{\circ} \mathrm{C}\right),\) what does the investigator calculate as the minimum muzzle velocity of the gun?

(II) \(A\) hot iron horseshoe \((\) mass \(=0.40 \mathrm{~kg}),\) just forged (Fig. \(19-28\) ), is dropped into \(1.05 \mathrm{~L}\) of water in a \(0.30-\mathrm{kg}\) iron pot initially at \(20.0^{\circ} \mathrm{C}\). If the final equilibrium temperature is \(25.0^{\circ} \mathrm{C}\), estimate the initial temperature of the hot horseshoe.

(II) One and one-half moles of an ideal monatomic gas expand adiabatically, performing \(7500 \mathrm{~J}\) of work in the process. What is the change in temperature of the gas during this expansion?

A microwave oven is used to heat \(250 \mathrm{~g}\) of water. On its maximum setting, the oven can raise the temperature of the liquid water from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) in \(1 \mathrm{~min} 45 \mathrm{~s}(=105 \mathrm{~s}) \cdot(a) \mathrm{At}\) what rate does the oven input energy to the liquid water? (b) If the power input from the oven to the water remains constant, determine how many grams of water will boil away if the oven is operated for 2 min (rather than just 1 min 45 s).

(I) A 1.00 -mol sample of an ideal diatomic gas, originally at \(1.00 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\), expands adiabatically to 1.75 times its initial volume. What are the final pressure and temperature for the gas? (Assume no molecular vibration.)

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.