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Chapter 18: Problem 11

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of \(0.600 \mathrm{~L}\) at \(19.0^{\circ} \mathrm{C}\). What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{~K}) ?\)

Short Answer

Expert verified
When calculated, we find that the volume V2 equals approximately 0.158 L.

Step by step solution

01

Conversion of Temperature to Kelvin

Firstly convert the initial temperature from Celsius to Kelvin. To do this, add 273.15 to the Celsius temperature. The initial temperature is \(19.0^{\circ} \mathrm{C}\), so the corresponding temperature in Kelvin would be \(19.0 + 273.15 = 292.15 \mathrm{K}\).
02

Apply Charles's Law

Charles's law states that the volume of a gas is directly proportional to its temperature as long as the pressure remains constant. The formula for Charles's Law is \( V1 / T1 = V2 / T2 \), where V represents the volume and T represents the absolute temperature. Given that \(V1 = 0.600 \mathrm{~L}\), \(T1 = 292.15 \mathrm{K}\), and \(T2 = 77.3 \mathrm{~K}\), we can substitute these values into the equation to find V2.
03

Calculation of Final Volume

Let's solve \( V1 / T1 = V2 / T2 \) for \( V2 \). We get \( V2 = (V1 * T2) / T1 \). Substituting our known values, we get \( V2 = (0.600 \mathrm{~L} * 77.3 \mathrm{~K}) / 292.15 \mathrm{~K} \).

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental principle which combines several gas laws, including Charles's Law, to relate the pressure, volume, temperature, and number of moles of an ideal gas. The law is commonly written as PV = nRT, where P stands for pressure, V is the volume, T is the absolute temperature in Kelvin, n is the number of moles, and R is the gas constant.

In simpler terms, the ideal gas law provides a mathematical relationship that enables us to predict the behavior of a gas sample when subjected to changes in pressure, volume, or temperature, assuming the gas behaves ideally. This means the gas particles are point masses with no volume and there are no intermolecular forces between them. Although no gas is truly 'ideal', many gases, such as helium in the given exercise, behave nearly ideally under a wide range of conditions.
Temperature Conversion
One of the keys to working with gas laws鈥攕uch as Charles's Law鈥攊s correct temperature conversion. Since gas volumes are directly related to temperature, and this relationship relies on an absolute temperature scale, we need to convert temperatures to the Kelvin scale. Absolute zero on this scale corresponds to a complete lack of thermal energy.

To convert Celsius to Kelvin, we add 273.15, acknowledging that 0 K is equivalent to -273.15掳C. This step ensures that temperature ratios are accurately calculated, which is vital when we're comparing how the same sample of gas behaves at two different temperatures. For instance, in our exercise, the conversion from 19.0掳C to Kelvin was crucial for the correct application of Charles's Law.
Gas Volume and Temperature Relationship
Underpinning our balloon's behavior is Charles's Law, which formalizes the direct proportionality between gas volume and temperature. Charles's Law states that for a given amount of gas at a constant pressure, the volume is directly proportional to its absolute temperature. The law can be expressed as \( V_1 / T_1 = V_2 / T_2 \) where \( V \) represents the volume and \( T \) the absolute temperature in Kelvin.

If we increase the temperature, the volume increases accordingly and vice versa. For instance, cooling our balloon filled with helium decreases its volume as observed when it is brought to the boiling point of liquid nitrogen. This relationship is essential to understanding not just academic problems but also real-world applications like air conditioning, refrigeration, and even meteorology where understanding how gas volumes will change with temperature is critical.

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Most popular questions from this chapter

A balloon of volume \(750 \mathrm{~m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of \(1.90 \mathrm{~m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{~Pa}\), how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at \(15.0^{\circ} \mathrm{C} ?\) The molar mass of hydrogen \(\left(\mathrm{H}_{2}\right)\) is \(2.02 \mathrm{~g} / \mathrm{mol} .\) The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is \(1.23 \mathrm{~kg} / \mathrm{m}^{3} .\) See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass \(4.00 \mathrm{~g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

Calculate the mean free path of air molecules at \(3.50 \times 10^{-13} \mathrm{~atm}\) and \(300 \mathrm{~K}\). (This pressure is readily attainable in the laboratory; see Exercise \(18.21 .\) ) As in Example \(18.8,\) model the air molecules as spheres of radius \(2.0 \times 10^{-10} \mathrm{~m}\).

An empty cylindrical canister \(1.50 \mathrm{~m}\) long and \(90.0 \mathrm{~cm}\) in diameter is to be filled with pure oxygen at \(22.0^{\circ} \mathrm{C}\) to store in a space station. To hold as much gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is \(32.0 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilograms does this gas increase the mass to be lifted?

A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height \(h\) has the value \(3.50 \mathrm{~m}\), the absolute pressure \(p\) of the compressed air is \(4.20 \times 10^{5} \mathrm{~Pa}\). Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5} \mathrm{~Pa}\). (a) What is the speed with which water flows out of the hose when \(h=3.50 \mathrm{~m} ?\) (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h=3.00 \mathrm{~m}\) and for \(h=2.00 \mathrm{~m} .\) (c) At what value of \(h\) does the flow stop?

The atmosphere of Mars is mostly \(\mathrm{CO}_{2}\) (molar mass \(44.0 \mathrm{~g} / \mathrm{mol}\) ) under a pressure of \(650 \mathrm{~Pa}\), which we shall assume remains constant. In many places the temperature varies from \(0.0^{\circ} \mathrm{C}\) in summer to \(-100^{\circ} \mathrm{C}\) in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the \(\mathrm{CO}_{2}\), molecules and (b) the density (in \(\mathrm{mol} / \mathrm{m}^{3}\) ) of the atmosphere?
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