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Chapter 3: Problem 81

A helium-filled toy balloon has a gauge pressure of 0.200 atm and a volume of 10.0 L. How much greater is the internal energy of the helium in the balloon than it would be at zero gauge pressure?

Short Answer

Expert verified
The internal energy of the helium in the balloon is 1823.85 J greater than it would be at zero gauge pressure.

Step by step solution

01

Identify the given variables and constants

Gauge pressure of the helium-filled toy balloon, P = 0.200 atm Volume of the helium-filled toy balloon, V = 10.0 L We need to convert the gauge pressure to absolute pressure, so we add atmospheric pressure, Patm = 1 atm R, the gas constant for helium = 8.314 J/(mol*K)
02

Convert the pressure and volume of the balloon

Convert the volume of the balloon from liters to cubic meters: V = 10 L 脳 (1 m鲁 / 1000 L) = 0.01 m鲁 Convert the gauge pressure to absolute pressure in pascals: P_abs = (P + Patm) 脳 101325 Pa/atm = (0.200 + 1) 脳 101325 = 121590 Pa
03

Calculate the number of moles of helium in the balloon using the ideal gas law

We will use the equation PV = nRT to calculate the number of moles of helium in the balloon. n = PV / RT We need to find the temperature to calculate the number of moles. Assume the temperature to be T. Since it is not given and not needed for the final result, it will cancel out in the next step.
04

Calculate the change in internal energy

The internal energy of an ideal gas depends only on its temperature and the number of moles. Using the formula for the internal energy of a monatomic ideal gas (like helium), 螖U = (3/2)nR螖T, we can calculate the change in internal energy: Since the pressure is changing, we use the relation 螖T = (T_2 - T_1) = [(P_2 * V_2) / (nR)] - [(P_1 * V_1) / (nR)] Using that and substituting the values into the formula for the change in internal energy: 螖U = (3/2)nR螖T = (3/2)nR[(P_2 * V_2) / (nR)] - [(P_1 * V_1) / (nR)] Since the balloons are at 0 gauge pressure, P1 = 0 and V1 = V2 So, 螖U = (3/2)nR[(P_2 * V) / (nR)] From Step 3, the number of moles n = PV/RT, substituting this equation into 螖U: 螖U = (3/2)[(P_abs * V) / RT] * R[(P_abs * V) / (nR)] The R's and V's cancel out: 螖U = (3/2) * P_abs * V Now substituting the values for P_abs and V: 螖U = (3/2) * 121590 * 0.01 = 1823.85 J The internal energy of the helium in the balloon is 1823.85 J greater than it would be at zero gauge pressure.

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Key Concepts

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

Internal Energy Calculation
When assessing the internal energy of a gas, particularly in the case of ideal gases such as helium, it's crucial to understand how temperature and moles factor into this concept. For ideal gases, the internal energy is not influenced by pressure or volume directly, but rather by the temperature and the quantity of gas, expressed in moles.
In the case of monatomic gases like helium, the formula to calculate the change in internal energy, 螖U, is depicted as:
  • 螖U = \( \frac{3}{2} nR螖T \)
Where \( n \) represents the number of moles, \( R \) is the universal gas constant, and \( 螖T \) denotes the change in temperature. This means any shift in internal energy is inherently connected to changes in temperature. In this specific scenario, since volume changes but temperature is assumed to remain constant, 螖T can be deduced through changes in pressure and volume, manifested in our calculations.
Gauge Pressure Conversion
Gauge pressure measures the pressure of gas above atmospheric pressure. To effectively solve gas law problems, it's necessary to convert gauge pressure to absolute pressure. Absolute pressure is the sum of gauge pressure and atmospheric pressure.
  • In formula terms: \( P_{abs} = (P_{gauge} + P_{atm}) \)
Atmospheric pressure at sea level is approximately 1 atm, equivalent to 101325 Pa. Converting gauge pressure into absolute pressure involves adding 1 atm in the desired units.
In the exercise, gauge pressure is provided as 0.200 atm. To find the absolute pressure in pascals, it was calculated as:
\[ P_{abs} = (0.200 \, \text{atm} + 1 \, \text{atm}) \times 101325 \, \text{Pa/atm} = 121590 \, \text{Pa} \]
Converting pressure is a vital step to precisely apply it in further calculations, especially when dealing with ideal gas laws.
Helium Gas Properties
Helium is a widely known inert gas with exceptional properties, often used in applications that require low reactivity. As a monatomic element, its simple particle nature makes it ideal for demonstrating ideal gas laws. Helium's key properties:
  • It is highly stable and non-reactive.
  • Being the second-lightest element, it boasts a low density.
  • Its boiling point is extremely low, near -269掳C at standard pressure, meaning it remains a gas at everyday temperatures.
In the context of thermodynamics and ideal gas law exercises, helium鈥檚 role as a monoatomic gas allows for the use of simplified equations when determining internal energy changes. Its real-life applications extend from medical imaging to being utilized in balloons, as its non-flammable nature and low density make it ideal for such purposes.
Monatomic Ideal Gases
Monatomic ideal gases, like helium, consist of individual atoms, not bonded to others. For these gases, the theoretical simplifications made in physics become effective practical tools for solving real-world problems.
In this setting:
  • The internal energy relies solely on temperature and is given by \( U = \frac{3}{2}nRT \).
  • These computations assume the gas molecules do not interact except during elastic collisions.
One primary assumption is that such gases obey the ideal gas law, where pressure, volume, and temperature interplay within the context:
  • \( PV = nRT \)
This allows straightforward projection of changes in properties like volume due to temperature or pressure variations, rendering them useful in both theoretical and applied physics.
Monatomic gases offer simplicity, making calculations both feasible and intuitive鈥攆or students and engineers alike, allowing accurate predictions under a range of conditions.

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Most popular questions from this chapter

The insulated cylinder shown below is closed at both ends and contains an insulating piston that is free to move on frictionless bearings. The piston divides the chamber into two compartments containing gases A and B. Originally, each compartment has a volume of \(5.0 \times 10^{-2} \mathrm{m}^{3}\) and contains a monatomic ideal gas at a temperature of \(0^{\circ} \mathrm{C}\) and a pressure of 1.0 atm. (a) How many moles of gas are in each compartment? (b) Heat Q is slowly added to A so that it expands and \(\mathrm{B}\) is compressed until the pressure of both gases is 3.0 atm. Use the fact that the compression of \(\mathrm{B}\) is adiabatic to determine the final volume of both gases. (c) What are their final temperatures? (d) What is the value of Q?

An amount of n moles of a monatomic ideal gas in a conducting container with a movable piston is placed in a large thermal heat bath at temperature \(T_{1}\) and the gas is allowed to come to equilibrium. After the equilibrium is reached, the pressure on the piston is lowered so that the gas expands at constant temperature. The process is continued quasi-statically until the final pressure is \(4 / 3\) of the initial pressure \(p_{1}\). (a) Find the change in the internal energy of the gas. (b) Find the work done by the gas. (c) Find the heat exchanged by the gas, and indicate, whether the gas takes in or gives up heat.

Compare the charge in internal energy of an ideal gas for a quasi-static adiabatic expansion with that for a quasi-static isothermal expansion. What happens to the temperature of an ideal gas in an adiabatic expansion?

Ideal gases A and B are stored in the left and right chambers of an insulated container, as shown below. The partition is removed and the gases mix. Is any work done in this process? If the temperatures of \(A\) and \(B\) are initially equal, what happens to their common temperature after they are mixed?

One mole of an ideal gas is initially in a chamber of volume \(1.0 \times 10^{-2} \mathrm{m}^{3}\) and at a temperature of \(27^{\circ} \mathrm{C}\) (a) How much heat is absorbed by the gas when it slowly expands isothermally to twice its initial volume? (b) Suppose the gas is slowly transformed to the same final state by first decreasing the pressure at constant volume and then expanding it isobarically. What is the heat transferred for this case? (c) Calculate the heat transferred when the gas is transformed quasi-statically to the same final state by expanding it isobarically, then decreasing its pressure at constant volume.
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