Problem 2 - University Physics with Modern Physics

Question

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 \(\mathrm{g} / \mathrm{mol} .\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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: \(\frac{0.180 \times 2.60}{314.15} = \frac{0.360 \times 5.20}{T_2}\).

Step 4: Solve for Final Temperature

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

Step 5: 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 \(n\):\[n = \frac{P_1 V_1}{R T_1} = \frac{0.180 \times 2.60}{0.0821 \times 314.15}.\]

Step 6: Calculate Moles of Helium

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

Step 7: Convert Moles to Grams

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

Key Concepts

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.