Problem 2 - University Physics with Modern Physics

Question

Helium gas with a volume of 2.60 \(\mathrm{L}\) , under a pressure of 1.30 atm and at a temperanure 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

Final temperature is approximately 628.3 K. Mass of helium is approximately 0.524 g.

Step 1: Convert Initial Temperature to Kelvin

The initial temperature is given in Celsius, which needs to be converted to Kelvin. The conversion formula is: \( T(K) = T(°C) + 273.15 \). Thus, \( T(1) = 41.0 + 273.15 = 314.15 \ \mathrm{K}.\)

Step 2: Apply the Combined Gas Law

The combined gas law is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). With \( P_2 = 2 \cdot 1.30\, \mathrm{atm},\) and \( V_2 = 2 \cdot 2.60\, \mathrm{L} \), we need to find \( T_2 \). Rearrange the formula to solve for \( T_2 \): \( T_2 = \frac{P_2V_2T_1}{P_1V_1} \).

Step 3: Substitute and Calculate Final Temperature

Substitute known values into the rearranged formula: \( T_2 = \frac{(2 \times 1.30) \times (2 \times 2.60) \times 314.15}{1.30 \times 2.60} \). Work through the calculation: \( T_2 = \frac{6.76 \times 314.15}{3.38} \approx 628.3\, \mathrm{K}.\)

Step 4: Calculate Moles of Helium

Use the ideal gas law \( PV = nRT \) to find the number of moles. First rearrange to solve for \( n \): \( n = \frac{PV}{RT} \). Use \( P = 1.30 \ \mathrm{atm}, \ V = 2.60 \ \mathrm{L}, \) and \( R = 0.0821 \ \mathrm{L} \cdot \mathrm{atm}/\mathrm{mol} \cdot \mathrm{K},\) with initial temperature \( T_1 = 314.15\, \mathrm{K}.\)

Step 5: Substitute and Calculate Moles

Substitute the values into the ideal gas law formula: \( n = \frac{1.30 \times 2.60}{0.0821 \times 314.15} \). Calculate \( n = \frac{3.38}{25.79} \approx 0.131 \ \mathrm{mol}.\)

Step 6: Calculate Mass of Helium

Use the molar mass of helium to find the mass: \( \text{mass} = n \times \text{molar mass} \). Thus, \( \text{mass} = 0.131 \times 4.00 \approx 0.524 \ \mathrm{g}.\)

Key Concepts

Combined Gas Law

The Combined Gas Law is an essential concept in chemistry that combines three different gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It allows us to understand how pressure, volume, and temperature interact with each other for a given amount of gas, provided no gas particles are added or removed.
Where the ideal gas law formula is represented as:

• \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).
• P represents pressure, V is volume, and T is temperature in Kelvin.
• Subscripts 1 and 2 refer to initial and final states, respectively.

To properly utilize the Combined Gas Law, temperature must always be expressed in Kelvin. This preserves the absolute scale of temperature. In our exercise, Helium gas originally at 314.15K saw its pressure and volume both double, requiring us to solve for the final temperature \( T_2 \). The final temperature determined via substitution is approximately 628.3K, demonstrating how precise adjustments in conditions alter the thermodynamic state.

Helium Gas Calculations

Calculating the properties of helium, like any other gas, can be effectively done using the ideal gas equation: \( PV = nRT \), where \( n \) represents the number of moles, \( R \) is the universal gas constant, \( 0.0821 \ \mathrm{L \, atm/mol \, K} \), and \( P \), \( V \), and \( T \) denote pressure, volume, and temperature respectively.
The initial scenario outlines the use of conditions where Helium occupies 2.6 L, is under 1.30 atm of pressure at 314.15K. By substituting these into the ideal gas law, we solve for moles \( n \) as:

• \( n = \frac{PV}{RT} = \frac{1.30 \times 2.60}{0.0821 \times 314.15} \approx 0.131 \ \mathrm{mol} \).

Through this computation, we understand that the moles of helium involved are 0.131, which is crucial for determining further properties like mass.

Molar Mass Calculations

Conducting molar mass calculations enables chemists to convert between moles of a substance and its mass, and this is vital for any quantitative analysis involving gases. The molar mass of helium is known to be 4.00 g/mol.
Once the amount in moles is known, the mass of helium can be calculated using the simple equation:

• \( \text{mass} = n \times \text{molar mass} \).

For our example, using the determined amount of Helium, which is 0.131 mol, we calculate the mass by:

• \( \text{mass} = 0.131 \times 4.00 \approx 0.524 \ \mathrm{g} \).

This calculation illustrates the conversion of a molecular quantity into a tangible mass, showcasing how closely the notion of moles interconnects with chemical mass.