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A large research balloon containing \(2.00 \times 10^{3} \mathrm{~m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. \(\mathbf{P} 19.50\) ). Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

Short Answer

Expert verified
(a) 2222.22 m³; (b) 288.15 K; (c) 0 J (no change in internal energy).

Step by step solution

01

Understand the Ideal Gas Law

The ideal gas law is given by the equation \( PV = nRT \), 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. We'll use this equation to find the unknowns as the balloon rises.
02

Convert Temperature from Celsius to Kelvin

The initial temperature is given in Celsius (\(15.0^{\circ}\mathrm{C}\)). Convert this to Kelvin using the formula: \( T(K) = T(^{\circ}\mathrm{C}) + 273.15 \). \[ T_1 = 15.0 + 273.15 = 288.15 \text{ K} \]
03

Calculate Volume at Higher Altitude

Given that the atmospheric pressure at higher altitude is 0.900 atm, we use the formula:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Since the process is adiabatic, it implies \( T_1 = T_2 \) (since there's no time for heat exchange).Substituting and solving for \( V_2 \):\[ V_2 = V_1 \frac{P_1}{P_2} = 2.00 \times 10^{3} \times \frac{1.00}{0.900} = 2222.22 \text{ m}^3 \]
04

Confirm Temperature Consistency

Since the problem states the ascent is too rapid for heat exchange, the temperature remains constant. Therefore: \[ T_2 = T_1 = 288.15 \text{ K} \]
05

Calculate Change in Internal Energy

The change in internal energy for an ideal gas during an adiabatic process is given by:\[ \Delta U = nC_v\Delta T \] For an adiabatic process where there's no heat exchange, \( \Delta T = 0 \), therefore: \[ \Delta U = 0 \] There is no change in internal energy, as the temperature of the gas does not change.

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Key Concepts

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

Adiabatic Process
An adiabatic process in thermodynamics is fascinating because it happens without any heat exchange between the system and its surroundings. Picture a rising balloon that quickly ascends so that heat can't transfer in or out of the gas inside it. In this situation, energy changes occur only through work done on or by the system.
To understand why the temperature can remain constant despite changes in pressure and volume, think of the adiabatic law for gases. This process often involves significant shifts in pressure and volume, but the rapid pace prevents heat exchange, maintaining a steady temperature.
  • In our balloon example, as the balloon rises, it doesn't have time to gain or lose heat to the surrounding air, categorized as an adiabatic and isothermal process (for simplicity, in this particular case).
  • Mathematically, adiabatic processes for ideal gases can be expressed with equations involving pressure, volume, and adiabatic index (which is not needed here since temperature stays constant).
Adiabatic processes are vital because they describe scenarios where a system changes internally without heat from the outside, as might happen in a fast-moving aircraft or a swiftly rising balloon.
Internal Energy
The internal energy is a key concept in understanding how energy changes within a system like a gas. It encompasses the energy related to the random motions of the particles within a substance. For an ideal gas, internal energy is mostly a function of its temperature. As the molecules of the gas move faster with increasing temperature, the internal energy rises as well.
When no heat is exchanged, as in an adiabatic process, any work done on or by the gas affects its internal energy. However, in our exercise, since the temperature of the gas does not change (as the process is isothermal and rapid), the internal energy remains constant. This is a critical understanding in thermodynamics:
  • For an ideal gas undergoing an adiabatic process with no change in temperature, the change in internal energy ( ΔU ) is zero.
  • No temperature change implies no kinetic energy change at the molecular level, which keeps the internal energy unchanged.
Internal energy's connection to temperature means that when ΔT (change in temperature) is zero, ΔU (change in internal energy) is also zero.
Temperature Conversion
In scientific calculations, especially when dealing with gases, it's usually necessary to convert temperature from Celsius to Kelvin. Kelvin is the absolute temperature scale used in thermodynamics because it starts at absolute zero, where all kinetic motion ceases.
This conversion is simple: add 273.15 to the Celsius temperature to convert it to Kelvin. For example, the initial temperature of the balloon's helium gas was given in Celsius, so to use it in the ideal gas law equations, we convert it like this:
  • T(K) = T(°C) + 273.15
  • The initial temperature of 15.0°C converts to 288.15 K.
Using the Kelvin scale ensures all thermodynamic calculations are consistent, particularly when computing changes in internal energy or applying equations like the ideal gas law. Remember, Kelvin scales directly relate to the energy states of the particles in a system, offering a precise understanding of temperature's role in physical processes.

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Most popular questions from this chapter

A gas in a cylinder is held at a constant pressure of 1.80 \(\times\) 10\(^5\) \(Pa\) and is cooled and compressed from 1.70 m\(^3\) to 1.20 m\(^3\). The internal energy of the gas decreases by 1.40 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [\(Q\)] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0\(^\circ\)C, 392 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a \(pV\)-diagram for the process.

In a test of the effects of low temperatures on the gas mixture, a cylinder filled at 20.0\(^\circ\)C to 2000 psi (gauge pressure) is cooled slowly and the pressure is monitored. What is the expected pressure at -5.00\(^\circ\)C if the gas remains a homogeneous mixture? (a) 500 psi; (b) 1500 psi; (c) 1830 psi; (d) 1920 psi.

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m\(^3\) and the pressure increases from 2.00 \(\times\) 10\(^5\) Pa to 5.00 \(\times\) 10\(^5\) Pa. The second process is a compression to a volume of 0.120 m\(^3\) at a constant pressure of 5.00 \(\times\) 10\(^5\) Pa. (a) In a pV-diagram, show both processes. (b) Find the total work done by the gas during both processes.

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 \(\times\) 10\(^5\) Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(pV\)-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

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