Problem 58 - University Physics with Modern Physics

Question

High-Altitude Research. 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. Pl9.58. 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?

Step-by-Step Solution

Answer

(a) The volume at the higher altitude is \(2.22 \times 10^{3} \text{ m}^3\). (b) The temperature remains \( 15^{\circ} \text{C} \). (c) The change in internal energy is zero.

Step 1: Understand the Initial Conditions

At the initial condition, we have helium gas with a volume of \( V_i = 2.00 \times 10^{3} \ \text{m}^3 \) at a pressure of \( P_i = 1.00 \ \text{atm} \) and a temperature of \( T_i = 15.0^{\circ} \text{C} \). Since the ideal gas law requires temperature in Kelvin, we need to convert it: \[ T_i = 15 + 273.15 = 288.15 \ \text{K} \].

Step 2: Apply the Ideal Gas Law for Volume

To find the volume of helium at the new altitude, apply the ideal gas law: \( \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} \), where \( P_f = 0.900 \ \text{atm} \). Assuming the altitude change is too rapid for significant heat exchange, the process is adiabatic, so \( T_f = T_i \) initially.Thus, we get:\[ V_f = V_i \cdot \frac{P_i}{P_f} \cdot \frac{T_f}{T_i} = 2.00 \times 10^{3} \cdot \frac{1.00}{0.900} \cdot \frac{288.15}{288.15} = 2.22 \times 10^{3} \ \text{m}^3 \].

Step 3: Calculate the Temperature at Higher Altitude

As the balloon rises, the process can be considered adiabatic due to the rapid ascent. Typically, for an adiabatic process, \( \gamma = \frac{C_p}{C_v} \) needs to be considered. Still, since the question already keeps the temperature constant by instruction, \( T_f = T_i = 288.15 \ \text{K} \). Thus, the temperature does not change and remains:\( T_f = 288.15 \ \text{K} \) or equivalently \( 15^{\circ} \text{C} \).

Step 4: Calculate the Change in Internal Energy

The change in internal energy for an ideal gas in an adiabatic process is given by the formula:\( \Delta U = nC_v\Delta T \).Since \( \Delta T = 0 \) for this rapid ascent, \( \Delta U = 0 \). Therefore, the internal energy of the helium does not change as it rises to the higher altitude.

Key Concepts

Adiabatic Process

An adiabatic process is a thermodynamic scenario where no heat is exchanged with the surroundings. This means that all the energy changes in the system are internal. In the context of the helium-filled balloon, the rapid ascent means there isn’t enough time for heat to be lost or gained from the outside air.
This also fits with the original exercise that assumes the balloon’s ascent is too quick to allow major heat transfer. So, during the ascent, any changes in the gas state, like pressure or volume, result from internal energy changes rather than external heat exchange.

• No Heat Exchange: The system is isolated from the external thermal environment.
• Internal Energy Driven: Changes are due to work done on or by the system.

Even though pressure and volume change, the temperature can remain remarkably stable due to the lack of heat exchange.

Internal Energy

Internal energy in an ideal gas is tied directly to its temperature. The formula \( \Delta U = nC_v\Delta T \) captures this relationship, where \( \Delta U \) is the change in internal energy, \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( \Delta T \) is the change in temperature.
During an adiabatic process, if there is no change in temperature as we see with the helium gas here, the internal energy does not change.

• Temperature Link: Directly related to the kinetic energy of gas molecules.
• No Temperature Change: Means no internal energy change.

Thus, for this helium-filled balloon rising quickly, because the temperature remains the same, so does the internal energy.

Helium Gas

Helium is a noble gas and one of the lightest elements, making it ideal for lifting when used in balloons. It behaves closely like an ideal gas, especially under conditions of temperature and pressure changes such as those found in rapid altitude change situations.
Being monoatomic, helium has a simpler behavior compared to multi-atomic gases. This simplicity allows physicists to predict and model its behavior accurately using the ideal gas law.

• Monoatomic Nature: Leads to straightforward thermodynamic calculations.
• Low Density: Ideal for applications needing upthrust, like balloons.

Its unique properties make helium a must-use in applications that require reliable gas expansion with consistent behavior across temperature and pressure changes.

Volume Change

Volume change in gases, especially under the ideal gas laws, is a direct response to changes in pressure and temperature. The law is mathematically represented as \( PV = nRT \), where pressure \( P \), volume \( V \), and temperature \( T \) are interconnected for a given amount of gas \( n \) and ideal gas constant \( R \).
In the case of helium in the balloon, the pressure decreases as it rises to a higher altitude, so the volume must increase to keep the equation balanced, assuming temperature stays constant due to adiabatic conditions.

• Pressure-Volume Relation: Inverse proportionality in a stable temperature setting.
• Adiabatic Consistency: Temperature remains unchanged although volume increases.

As it finds itself in a new environment with different pressure levels, the balloon will expand, increasing its volume in response.