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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 L at 19.0\(^\circ\)C. What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen (77.3 K)?

Short Answer

Expert verified
The volume of the balloon at 77.3 K is approximately 0.159 L.

Step by step solution

01

Understand the Problem

We need to find the new volume of the helium-filled balloon when it is cooled from the initial temperature of 19.0\(^\circ\)C to the boiling point of liquid nitrogen, which is 77.3 K. We'll assume the pressure remains constant, similar to atmospheric pressure.
02

Convert Temperatures to Kelvin

The initial temperature is given as 19.0\(^\circ\)C. To convert Celsius to Kelvin, use the formula: \[ T(K) = T(\degree C) + 273.15 \]So, the initial temperature in Kelvin is: \[ 19.0 + 273.15 = 292.15\, K \]
03

Utilize Charles's Law

Charles's Law states that, for a given amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin. The law can be expressed as: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Where \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.
04

Apply the Known Values

Using Charles's Law, substitute the known values:- Initial volume \(V_1 = 0.600\, L\)- Initial temperature \(T_1 = 292.15\, K\)- Final temperature \(T_2 = 77.3\, K\)The equation becomes:\[ \frac{0.600}{292.15} = \frac{V_2}{77.3} \]
05

Solve for Final Volume \(V_2\)

Rearrange the formula to solve for \(V_2\):\[ V_2 = 0.600 \times \frac{77.3}{292.15} \]Perform the calculation:\[ V_2 = 0.600 \times 0.2646 \approx 0.1588\, L \]
06

Conclusion

The volume of the balloon when cooled to the boiling point of liquid nitrogen is approximately 0.159 L.

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Key Concepts

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

Ideal Gas
Gases like helium often behave like ideal gases under standard conditions. An ideal gas is a theoretical concept used to simplify the relationships between pressure, volume, and temperature in a gas. In an ideal gas, molecules are assumed to have negligible volume and do not interact except through elastic collisions. This simplification allows us to use mathematical models, such as Charles's Law, to understand how gases behave under changing conditions.
Helium is known as a noble gas and is close to an ideal gas, meaning it obeys the gas laws very closely. This makes it easier to predict its behavior as we change the temperature, volume, or pressure in scenarios like the balloon example here. The ideal gas assumption helps us ignore complexities such as intermolecular forces for calculations.
Temperature Conversion
In gas law calculations, it is crucial to convert temperatures from Celsius to Kelvin. Why? Because the Kelvin scale starts at absolute zero where all molecular motion theoretically stops and does not have negative numbers, making it proper for direct proportional relationships.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, an initial temperature of 19.0°C is converted as follows:
  • Add 273.15: \( T(K) = 19.0 + 273.15 \)
  • The result: \( T = 292.15 \, K \)
Now, you have the temperature in the correct unit for further calculations using gas laws such as Charles's Law.
Constant Pressure
In scenarios where a balloon is subject to atmospheric conditions, we often assume the pressure is constant. This is because the pressure inside and outside the balloon tends to equalize.
The constant pressure assumption is crucial in applying Charles's Law, which deals with volume and temperature changes when pressure is held constant. If you change balloon conditions such as temperature without affecting air pressure significantly, the gas volume changes appropriately in response to new temperatures. This means we can focus solely on the volume and temperature ratios in our calculations without worrying about pressure shifts.
Balloon Volume Calculation
Balloon volume calculations under changing temperature can be done using Charles's Law. Charles's Law states that the volume of an ideal gas at constant pressure is directly proportional to its temperature in Kelvin.
This means, when the temperature decreases, so does the volume of the gas inside the balloon. The relationship is mathematically expressed as:
  • \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
  • Where \( V_1 \) and \( V_2 \) are initial and final volumes, and \( T_1 \) and \( T_2 \) are initial and final temperatures.
In the balloon example, cooling the helium to the boiling point of liquid nitrogen requires solving for \( V_2 \):
  • Given \( V_1 = 0.600 \, L \), \( T_1 = 292.15 \, K \), \( T_2 = 77.3 \, K \)
  • Calculate \( V_2 \) using \( V_2 = 0.600 \times \frac{77.3}{292.15} \)
  • The final volume is approximately \( V_2 = 0.159 \, L \)
This clearly illustrates how temperature directly affects the volume of gases under constant pressure conditions.

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Most popular questions from this chapter

(a) Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol. (b) The actual specific heat of water vapor at low pressures is about 2000 J/kg \(\cdot\) K. Compare this with your calculation and comment on the actual role of vibrational motion.

(a) Calculate the density of the atmosphere at the surface of Mars (where the pressure is 650 Pa and the temperature is typically 253 \(K\), with a \(CO_2\) atmosphere), Venus (with an average temperature of 730 \(K\) and pressure of 92 atm, with a \(CO_2\) atmosphere), and Saturn's moon Titan (where the pressure is 1.5 atm and the temperature is -178\(^\circ\)C, with a \(N_2\) atmosphere). (b) Compare each of these densities with that of the earth's atmosphere, which is 1.20 kg/m\(^3\). Consult Appendix D to determine molar masses.

During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine \(Squalus\) sank at a point where the depth of water was 73.0 m. The temperature was 27.0\(^\circ\)C at the surface and 7.0\(^\circ\)C at the bottom. The density of seawater is 1030 kg/m\(^3\). (a) A diving bell was used to rescue 33 trapped crewmen from the \(Squalus\). The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

An ideal gas has a density of 1.33 \(\times\) 10\({^-}{^6}\) g/cm\(^3\) at 1.00 \(\times\) 10\({^-}{^3}\) atm and 20.0\(^\circ\)C. Identify the gas.

At the surface of Venus the average temperature is a balmy 460\(^\circ\)C due to the greenhouse effect (global warming!), the pressure is 92 earth-atmospheres, and the acceleration due to gravity is 0.894\(g{_e}{_a}{_r}{_t}{_h}\). The atmosphere is nearly all CO\(_2\) (molar mass 44.0 g/mol), and the temperature remains remarkably constant. Assume that the temperature does not change with altitude. (a) What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Venus-atmospheres and earth-atmospheres. (b) What is the root-mean-square speed of the CO\(_2\) molecules at the surface of Venus and at an altitude of 1.00 km?

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