A sealed balloon is filled with \(1.00 \mathrm{L}\) helium at \(23^{\circ} \mathrm{C}\) and 1.00 atm. The balloon rises to a point in the atmosphere where the pressure is \(220 .\) torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of \(220 .\) torr?

Short Answer

Expert verified
The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 1.75 L.

Step by step solution

01

Convert given temperatures to Kelvin

We need to convert the given initial and final temperatures from Celsius to Kelvin, as the Ideal Gas Law requires temperatures to be in Kelvin. Initial temperature (Ti): \[T_i = 23^{\circ}C + 273.15 = 296.15 K\] Final temperature (Tf): \[T_f = -31^{\circ}C + 273.15 = 242.15 K\]
02

Find the number of moles of helium in the balloon

Using the Ideal Gas Law, we have: \(PV = nRT\). We are given the initial volume (Vi = 1.00 L), initial pressure (Pi = 1.00 atm), and initial temperature (Ti). First, convert the initial pressure from atm to torr: \[P_i = 1.00 \, atm \times \frac{760 \, torr}{1 \, atm} = 760 \, torr\] Next, we need the value of the gas constant (R) in \(L \cdot torr \cdot mol^{-1} \cdot K^{-1}\): \[R = 62.364 \, L \cdot torr \cdot mol^{-1} \cdot K^{-1}\] Now, rearrange the Ideal Gas Law to solve for n, the number of moles: \[n = \frac{P_iV_i}{RT_i}\]
03

Calculate the number of moles

Plug in the given values and the values we found above into our n equation: \[n = \frac{(760 \, torr)(1.00 \, L)}{(62.364 \, L \cdot torr \cdot mol^{-1} \cdot K^{-1})(296.15 \, K)} = 0.04123 \, mol\]
04

Find the final volume using the Ideal Gas Law

Now we can use the Ideal Gas Law again, this time for the final state (using the final pressure, final temperature, and the number of moles we found). Rearrange the equation to solve for Vf (final volume): \[V_f = \frac{nRT_f}{P_f}\]
05

Calculate the final volume

Plug in the given values and the values we found previously: \[V_f = \frac{(0.04123 \, mol)(62.364 \, L \cdot torr \cdot mol^{-1} \cdot K^{-1})(242.15 \, K)}{220 \, torr} = 2.75 \, L\]
06

Calculate the change in volume

Now, find the change in volume by subtracting the initial volume from the final volume: \[\Delta V = V_f - V_i = 2.75 \, L - 1.00 \, L = 1.75 \, L\] The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 1.75 L.

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Key Concepts

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

Gas Constant (R)
The gas constant, often represented by the symbol 'R', is a crucial value in the Ideal Gas Law equation, which relates the pressure, volume, temperature, and number of moles of a gas. Its value depends on the units used for pressure and volume. For instance, when pressure is measured in atmospheres (atm) and volume in liters (L), R equals approximately 0.0821 L·atm·mol−1·K−1. However, in problems where pressure is given in torr, the value of R is usually taken as 62.364 L·torr·mol−1·K−1.
Understanding the correct value of R is essential for solving problems involving gas laws. Incorrect R can lead to an error in calculating the number of moles, pressure, volume, or temperature of the gas. Therefore, it's always important to double-check the units given for pressure and volume and to correspondingly adjust the value of R used in calculations.
Temperature Conversion
The Ideal Gas Law requires the use of absolute temperatures for accurate calculations. This means temperatures must be converted into Kelvin (K) from degrees Celsius (°C) before they're used in gas law equations. The conversion formula is simple:
Kelvin = Degrees Celsius + 273.15
By using Kelvin, we ensure that our temperature readings reflect the absolute temperature, which is crucial in calculating gas properties correctly. Negative temperatures in Celsius result in lower Kelvin temperatures, which correspond to the decrease of kinetic energy in the gas particles, affecting the gas volume and pressure calculations.
Moles of Gas
The term 'moles of gas' refers to the amount of substance or the quantity of gas particles present in a container. One mole of any substance contains Avogadro's number of particles, which is approximately 6.022×1023 particles. This principle allows us to connect the microscopic behavior of gas molecules with macroscopic physical properties that we can measure.
When we talk about the moles of gas in the context of the Ideal Gas Law, we encounter the variable 'n'. Calculating the moles of gas involves rearranging the Ideal Gas Law (PV=nRT) to solve for 'n', thus providing a direct relationship between the pressure, volume, temperature of the gas, and the amount of gas in moles. Understanding this relationship is vital for predicting how a gas will behave under different conditions.
Change in Gas Volume
The volume of a gas is directly related to the pressure and temperature conditions in which the gas is found. According to the Ideal Gas Law, if the temperature increases and pressure remains constant, the volume of the gas will also increase. Conversely, if the gas is at a higher pressure and the temperature remains constant, the volume will decrease.
In the context of our solved problem, the balloon's volume change is observed as it rises in the atmosphere. Because of the altitude increase, there's a decrease in atmospheric pressure and a drop in temperature. According to the Ideal Gas Law, such changes can result in an increase or decrease in gas volume depending on the extent of temperature and pressure changes. Therefore, understanding how these factors impact gas volume is essential for accurately calculating the change in conditions, just as we computed the balloon's volume adjustment from ground level to altitude.

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Most popular questions from this chapter

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain \(0.586 \mathrm{g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)

A large flask with a volume of \(936 \mathrm{mL}\) is evacuated and found to have a mass of \(134.66 \mathrm{g}\). It is then filled to a pressure of 0.967 atm at \(31^{\circ} \mathrm{C}\) with a gas of unknown molar mass and then reweighed to give a new mass of 135.87 g. What is the molar mass of this gas?

A steel cylinder contains \(5.00\) moles of graphite (pure carbon) and \(5.00\) moles of \(\mathrm{O}_{2}\). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

A gauge on a compressed gas cylinder reads \(2200 \mathrm{psi}\) (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

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