Question

Helium gas with a volume of 3.20 L, under a pressure of 0.180 atm and at 41.0\(^\circ\)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 g/mol.

Short answer

Final temperature is 629.15 K and 0.0894 g of helium are present.

Step-by-step solution

Understand the Initial Conditions

Identify the initial conditions given in the problem: the initial volume \( V_1 = 3.20 \text{ L} \), initial pressure \( P_1 = 0.180 \text{ atm} \), and initial temperature \( T_1 = 41.0^{\circ} \text{C}\). Since gas laws require absolute temperature, convert \( T_1 \) from Celsius to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \). Thus, \( T_1 = 41.0 + 273.15 = 314.15 \text{ K} \).

Identify Final Conditions

According to the problem, both the pressure and volume are doubled. So, the final pressure \( P_2 = 2 \times 0.180 \text{ atm} = 0.360 \text{ atm} \), and the final volume \( V_2 = 2 \times 3.20 \text{ L} = 6.40 \text{ L} \).

Use Combined Gas Law

The combined gas law \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) will help us find the final temperature \( T_2 \). Substitute the known values: \( \frac{0.180 \times 3.20}{314.15} = \frac{0.360 \times 6.40}{T_2} \).

Solve for Final Temperature \( T_2 \)

Rearrange the equation from Step 3 to find \( T_2 \): \( T_2 = \frac{0.360 \times 6.40 \times 314.15}{0.180 \times 3.20} \). Simplify: \( T_2 = \frac{1.152 \times 314.15}{0.576} = 629.15 \text{ K} \).

Calculate Moles of Helium

Use the ideal gas law to find the moles of helium: \( PV = nRT \). Solve for \( n \) using the initial conditions: \( n = \frac{P_1V_1}{RT_1} \), where \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). Substitute \( 0.180 \times 3.20 \) and \( 0.0821 \times 314.15 \). \( n = \frac{0.576}{25.792715} \approx 0.02235 \text{ mol} \).

Calculate Mass of Helium

Convert moles of helium to grams using its molar mass: \( \text{mass} = n \times \text{molar mass} = 0.02235 \times 4.00 = 0.0894 \text{ grams} \).

Conclusion

The final temperature of the gas is 629.15 K, and there are 0.0894 grams of helium in the container.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates a gas's pressure ( P ), volume ( V ), temperature ( T ), and amount in moles ( n ). It is expressed as \( PV = nRT \) , where \( R \) is the universal gas constant. This relationship is incredibly useful as it allows us to find any one of the variables if the others are known.
For example, to find the number of moles of helium in a gas sample, we can rearrange the equation to solve for \( n \) : \( n = \frac{PV}{RT} \) . When applying the Ideal Gas Law, ensure units are consistent, typically requiring volume in liters, temperature in Kelvin, pressure in atmospheres, and \( R \) as 0.0821 L atm K⁻¹ mol⁻¹.
This law is based on certain assumptions, such as gas particles occupying no volume and experiencing no intermolecular forces. Although some of these assumptions are not entirely accurate in real-world conditions, the Ideal Gas Law still provides a reliable approximation for many gases under common conditions.

Combined Gas Law

The Combined Gas Law merges several gas laws (Boyle's, Charles's, and Gay-Lussac's) to relate changes in pressure, volume, and temperature for a fixed amount of gas. The equation is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) . It identifies how a gas's state changes under different conditions without changing the quantity of the gas.
In a scenario where both the pressure and volume of a gas sample are doubled, the Combined Gas Law helps us find the new temperature. To do so, we rearrange the formula to solve for \( T_2 \) , the final temperature. This is especially handy when the number of moles remains constant and direct application of individual gas laws is cumbersome.
Remember that temperature must always be in Kelvin for accuracy. The Combined Gas Law provides a versatile tool for analyzing gaseous systems when multiple variables are changing.

Helium

Helium is a colorless, odorless, and tasteless gas that is the second lightest element and the second most abundant in the universe. Its atomic number is 2, and it has a molar mass of 4.00 g/mol. Helium is an inert gas, meaning it doesn't easily react with other elements or compounds, making it very stable.
Helium's low density and inert properties make it useful in various applications, such as filling balloons, serving as a protective gas in welding, and as a coolant in superconducting magnets. In terms of lighter gases, Helium is often the gas of choice due to its unique qualities.
When exploring gas laws, Helium often approximates ideal behavior because of its monoatomic nature and weak intermolecular forces. This makes it an excellent candidate for exercises involving the Ideal and Combined Gas Laws. Understanding Helium's properties can aid in comprehending its behavior under various conditions as described by these gas laws.