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Chapter 10: Problem 64

Two small containers, each with a volume of \(100 \mathrm{~cm}^{3}\), contain helium gas at \(0^{-} \mathrm{C}\) and \(1.00\) atm pressure. The two containers are joined by a small open tube of negligible volume, allowing gas to flow from one container to the other. What common pressure will exist in the two containers if the temperature of one container is raised to \(100^{\circ} \mathrm{C}\) while the other container is kept at \(0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
To solve for the common pressure in the two containers when one of them is heated, we apply the Ideal Gas Law to each container, remembering that the number of moles of gas in two combined containers after the temperature change is twice the original. Solving these equations gives the result.

Step by step solution

01

Understand the Given Details

There are two containers each with a volume, V = 100 cm鲁, containing helium gas at a temperature, T鈧� = 0掳C = 273.15 K, and pressure P鈧� = 1.00 atm. Let's assume that n鈧� represents the number of moles of helium gas in each container. When the temperature of one container is raised to T鈧� = 100掳C = 373.15 K while the other is kept at T鈧�, we need to find the new pressure P鈧�.
02

Apply the Ideal Gas Law for Each Container

For container 1 (kept at 0掳C), we apply the Ideal Gas Law: P鈧乂 = n鈧丷T鈧�. Since we're looking for n鈧�, we arrange it to be n鈧� = (P鈧乂) / (RT鈧�). We keep R as the gas constant. For container 2 (heated to 100掳C), the Ideal Gas Law becomes: P鈧俈 = n鈧俁T鈧�. Now note, since the containers are connected, the helium gas in the two containers is the same, hence, n鈧� + n鈧� = n鈧�. Therefore, n鈧� = (2P鈧乂) / (RT鈧�).
03

Calculating the New Pressure

Substitute n鈧� in the equation for container 2 and solve for P鈧� to get: P鈧� = (2P鈧乂) / (RT鈧�). Now, plug in the values: P鈧� = 1.00 atm, V = 100 cm鲁, R = 0.0821 (in atm鈥/mol鈥 terms), T鈧� = 273.15 K, and T鈧� = 373.15 K. Then, calculate P鈧�. This gives the common pressure that will exist in the two containers when one of them is heated.

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Key Concepts

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

Helium Gas
Helium is a fascinating and unique element, often recognized for its lightness and non-reactive nature. It is the second lightest element, and it remains one of the noble gases, which means it does not easily form compounds with other elements. This makes helium an ideal gas for many applications, including scientific experiments and as a coolant in cryogenics.
  • Light and Non-reactive: One of the primary properties of helium is its extremely low density, much lower than air, and its non-flammable nature. This is why helium balloons float.

  • Inertness: Helium's outer shell of electrons is full, which makes it chemically inert. As a result, it doesn't react at normal temperatures and pressures.

  • Availability: Helium is relatively rare on Earth. Most of the helium we use is extracted from natural gas deposits.

In our exercise, helium is used primarily because of its simplicity as an ideal gas, making it perfect for demonstrating concepts of gas laws without additional complications due to reactivity or changes in molecular behavior.
Temperature Change
Temperature plays a crucial role in the behavior of gases. According to the Ideal Gas Law, any change in temperature will impact either the pressure or volume of the gas if all other conditions remain constant. Helium, being an ideal gas, closely follows these principles, allowing us to predict changes precisely when we alter the temperature.
  • Basic Relationship: The relationship between pressure, volume, and temperature is expressed in the Ideal Gas Law formula: \(PV = nRT\), where \(T\) must be in Kelvin for the math to work correctly.

  • Kelvin Conversion: In the exercise, temperatures given in Celsius are converted to Kelvin by adding 273.15. Thus, \(0^{\circ}C\) becomes 273.15 K, and \(100^{\circ}C\) becomes 373.15 K. This conversion is crucial because the Kelvin scale starts at absolute zero, ensuring meaningful and proportional calculations.

  • Effect of Temperature Change: When one container's temperature rises from 0掳C to 100掳C, the increased kinetic energy of the helium atoms causes a change in pressure. This is due to the gas molecules' increased velocity and collision rate with the container walls, which is directly proportional to the absolute temperature.

Understanding the effects of temperature changes helps clarify why we expect a different pressure in the connected containers based on the given conditions.
Pressure Calculation
Pressure calculation in gases, especially under varying temperatures, is best understood through the Ideal Gas Law. This law, \(PV = nRT\), ties together pressure (P), volume (V), moles of gas (n), the gas constant (R), and absolute temperature (T). In the context of our exercise, we're focused on finding the equilibrium pressure of helium when temperatures differ in two connected containers.
  • Initial Conditions: Initially, both containers have the same pressure of 1 atm. When connected, the gas can flow between them, seeking equilibrium.

  • Application of Gas Law: For the unchanged container, the number of moles \(n_1\) is calculated using initial conditions, \( n_1 = \frac{P_1V}{RT_1}\).

  • System Equilibrium: The total number of moles remains constant since no gas is added or removed; just the temperatures change. Therefore, \(n_2 = 2n_1\) because the containers are open to each other, allowing mass flow.

  • Common Pressure Calculation: Using the formula \(P_2 = \frac{2P_1V}{RT_2}\), we substitute known values to find the new pressure in the heated condition. This relies on the total moles and the increased temperature in one container, ensuring continuity across the system.

By running through these calculations, one can understand how temperature impacts pressure directly, helping to predict system behavior in real-world applications involving gases like helium.

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Most popular questions from this chapter

A pair of eyeglass frames are made of epoxy plastic (coefficient of linear expansion \(=1.30 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) ). At room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\), the frames have circular lens holes \(2.20 \mathrm{~cm}\) in radius. To what temperature must the frames be heated if lenses \(2.21 \mathrm{~cm}\) in radius are to be inserted into them?

A 1.5-m-long glass tube that is closed at one end is weighted and lowered to the bottom of a freshwater lake. When the tube is recovered, an indicator mark shows that water rose to within \(0.40 \mathrm{~m}\) of the closed end. Determine the depth of the lake. Assume constant temperature.

The density of gasoline is \(7.30 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\) at \(0^{\circ} \mathrm{C}\). Its average coefficient of volume expansion is \(9.60 \times\) \(10^{-4}\left({ }^{\circ} \mathrm{C}\right)^{-1}\), and note that \(1.00 \mathrm{gal}=0.00380 \mathrm{~m}^{3}\). (a) Cal- culate the mass of \(10.0\) gal of gas at \(0^{\circ} \mathrm{C}\). (b) If \(1.000 \mathrm{~m}^{3}\) of gasoline at \(0^{\circ} \mathrm{C}\) is warmed by \(20.0^{\circ} \mathrm{C}_{1}\) calculate its new volume. (c) Using the answer to part (b), calculate the density of gasoline at \(20.0^{\circ} \mathrm{C}\). (d) Calculate the mass of 10,0 gal of gas at \(20.0^{\circ} \mathrm{Ci}\) (e) How many extra kilograms of gasoline would you get if you bought \(10.0\) gal of gasoline at \(0^{\circ} \mathrm{C}\). rather than at \(20.0^{\circ} \mathrm{C}\) from a pump that is not temperature compensated?

Temperature differences on the Rankine scale are identical to differences on the Fahrenheit scale, but absolute zero is given as \(0^{\circ} \mathrm{R}\). (a) Find a relationship converting the temperatures \(T_{f}\) of the Fahrenheit scale to the corresponding temperatures \(T_{n}\) of the Rankine scale. (b) Find a second relationship converting temperatures \(T_{n}\) of the Rankine scale to the temperatures \(T_{K}\) of the Kelvin scale.

Death Valley holds the record for the highest recorded temperature in the United States. On July 10, 1913 , at at place called Fumace Creek Ranch, the temperature rose to \(134^{\circ} \mathrm{F}\). The lowest U.S. temperature ever recorded occurred at Prospect Creek Camp in Alaska on January 23,1971, when the temperature plummeted to \(-79.8^{\circ}\) F. (a) Convert these temperatures to the Celsius scale. (b) Convert the Cielsius temperatures to Kelvin.
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