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Helium gas with a volume of \(2.60 \mathrm{L},\) under a pressure of 0.180 atm and at a temperature of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\)

Short Answer

Expert verified
The final temperature is 628.3 K, and the amount of helium is 0.0728 grams.

Step by step solution

01

Understand the Initial Conditions

The problem gives us that the initial volume \(V_1\) is 2.60 L, the initial pressure \(P_1\) is 0.180 atm, and the initial temperature \(T_1\) is 41.0°C. We need to convert the temperature to Kelvin: \(T_1 = 41.0 + 273.15 = 314.15\) K.
02

Determine Final Conditions

Both pressure and volume are doubled, so \(P_2 = 2 \times 0.180 = 0.360\) atm, and \(V_2 = 2 \times 2.60 = 5.20\) L. We need to determine the new temperature \(T_2\) in Kelvin.
03

Apply the Ideal Gas Law

Using the combined gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Substituting known values: \(\frac{0.180 \times 2.60}{314.15} = \frac{0.360 \times 5.20}{T_2}\).
04

Solve for Final Temperature

Rearrange the equation to solve for \(T_2\):\[T_2 = \frac{0.360 \times 5.20 \times 314.15}{0.180 \times 2.60} = 628.3 \mathrm{K}.\]
05

Use Ideal Gas Law to Find Moles of Helium

From the ideal gas law, \(PV = nRT\), where \(R = 0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}\). Use initial conditions to find \(n\):\[n = \frac{P_1 V_1}{R T_1} = \frac{0.180 \times 2.60}{0.0821 \times 314.15}.\]
06

Calculate Moles of Helium

Perform the calculation to find \(n\):\[n = \frac{0.468}{25.779} \approx 0.0182\] moles.
07

Convert Moles to Grams

Using the molar mass of helium, convert moles to grams:\[\text{mass} = n \times \text{molar mass} = 0.0182 \times 4.00 = 0.0728 \text{ grams}.\]

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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 colorless, odorless, and tasteless gas commonly used in balloons and as a protective gas in many industrial applications. It is a noble gas, which means it is very unreactive and stable. In the context of gas laws, helium behaves nearly ideally because its interactions are minimal. This makes it perfect for studying gas behavior under changing conditions like temperature, volume, and pressure. Helium's simplicity and ideal behavior in gas law calculations stem from its monatomic nature—each particle is a single atom. This feature contributes to it having minimal intermolecular forces, allowing calculations like those involving the Ideal Gas Law to be more straightforward, as we focus solely on pressure, volume, and temperature.
Temperature Conversion
Temperature in gas law problems is typically measured in Kelvin since it is an absolute scale. Kelvin does not involve negative numbers and is essential for calculations in thermodynamics for this reason. To convert Celsius to Kelvin, which is often necessary, you simply add 273.15 to the Celsius temperature. For example, to convert 41.0°C to Kelvin, you calculate 41.0 + 273.15, resulting in 314.15 K. This conversion is crucial, as using temperatures in Kelvin in the Ideal Gas Law ensures we are using a scale that is proportional to energy, consistent with the physical principles judged by the law.
Molar Mass
Molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance (mol). It is an important factor when converting between the mass of a gas and the amount of moles. For helium, the molar mass is 4.00 g/mol. This value signifies that one mole of helium atoms weighs exactly 4.00 grams. In problems involving gas laws, knowing the molar mass is useful for converting moles obtained from the Ideal Gas Law into grams for practical measurements or implications. The relation is straightforward: if you have the number of moles of a gas, simply multiply by the molar mass to get the grams, as performed in the exercise.
Combined Gas Law
The Combined Gas Law merges Charles's Law, Boyle's Law, and Gay-Lussac's Law. It says that the relationship between pressure, volume, and temperature for a given amount of gas can be described by the expression \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), where \(P\) is pressure, \(V\) is volume, and \(T\) is temperature in Kelvin.This law is especially helpful when comparing two states of a gas where the number of moles doesn’t change. When both the pressure and volume of helium are doubled, the resulting equation aids in calculating the new temperature while the amount of gas remains constant.In the exercise, we used the Combined Gas Law to find the final temperature of helium gas after changes in pressure and volume, demonstrating its utility in addressing real-world scenarios efficiently.

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Most popular questions from this chapter

You have two identical containers, one containing gas \(A\) and the other gas \(B .\) The masses of these molecules are \(m_{A}=3.34 \times 10^{-27} \mathrm{kg}\) and \(m_{B}=5.34 \times 10^{-26} \mathrm{kg} .\) Both gases are under the same pressure and are at \(10.0^{\circ} \mathrm{C}\) . (a) Which molecules \((A\) or \(B)\) have greater translational kinetic energy per molecule and rms speeds? (b) Now you want to raise the temperature of only one of these containers so that both gases will have the same only one of these containers so that both gases will have the same rms speed. For which gas should you raise the temperature? (c) At what temperature will you accomplish your goal? (d) Once you have accomplished your goal, which molecules \((A\) or \(B)\) now have greater average translational kinetic energy per molecule?

The size of an oxygen molecule is about 2.0 \(\times 10^{-10} \mathrm{m}\) Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures \((T=300 \mathrm{K})\) .

A container with volume 1.48 \(\mathrm{L}\) is initially evacuated. Then it is filled with 0.226 \(\mathrm{g}\) of \(\mathrm{N}_{2} .\) Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to high degree of accuracy. If the root-mean-square speed of the gas molecules is \(182 \mathrm{m} / \mathrm{s},\) what is the pressure of the gas?

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) Compute the increase in gravitational potential energy for a nitrogen molecule (molar mass 28.0 \(\mathrm{g} / \mathrm{mol} )\) for an increase in elevation of 400 \(\mathrm{m}\) near the earth's surface. (b) At what temperature is this equal to the average kinetic energy of a nitrogen molecule? (c) Is it possible that a nitrogen molecule near sea level where \(T=15.0^{\circ} \mathrm{C}\) could rise to an altitude of 400 \(\mathrm{m} ?\) Is it likely that it could do so without hitting any other molecules along the way? Explain.

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