Problem 66 - Introductory Chemistry Atoms First

Question

A tank initially at a pressure of \(6.70\) atm contains \(20.0\) moles of oxygen gas at \(25.0^{\circ} \mathrm{C}\). Its temperature is then increased to \(45.0{ }^{\circ} \mathrm{C}\), and an additional \(10.0\) moles of gas are added. What is the final pressure in the tank?

Step-by-Step Solution

Answer

The final pressure in the tank is \(10.35 \ atm\).

Step 1: Convert temperatures to Kelvin

First, we need to convert the temperatures to Kelvin. To do so, simply add 273.15 to the Celsius temperature: Initial temperature in Kelvin: \(T_1 = 25.0^{\circ}C + 273.15 = 298.15 K\) Final temperature in Kelvin: \(T_2 = 45.0^{\circ}C + 273.15 = 318.15 K\)

Step 2: Determine initial and final moles of gas

We are given the initial number of moles of gas and the additional moles added: \(n_1 = 20.0 \ moles\) \(n_{added} = 10.0 \ moles\) Final moles of gas can be calculated by adding the initial moles and the added moles: \(n_2 = n_1 + n_{added} = 20.0 + 10.0 = 30.0 \ moles\)

Step 3: Use Ideal Gas Law to find initial and final volume

Since the volume stays constant, we can use the Ideal Gas Law to find the initial volume and final volume (both will be the same): Initial conditions: \(P_1V = n_1RT_1\) \(V = \frac{n_1RT_1}{P_1} = \frac{(20.0 \ moles)(0.0821 \ \frac{L⋅atm}{mol⋅K})(298.15 \ K)}{6.70 \ atm}\) Final conditions: \(P_2V = n_2RT_2\) \(P_2 = \frac{n_2RT_2}{V}\) We will calculate these volumes in the next step.

Step 4: Perform calculations and find final pressure (P₂)

Now we can compute the initial and final volumes using the given data and the Ideal Gas Law. Initial volume (V): \(V = \frac{(20.0 \ moles)(0.0821 \ \frac{L⋅atm}{mol⋅K})(298.15 \ K)}{6.70 \ atm} = 73.08 \ L\) Now we can substitute the final number of moles (n₂), final temperature (T₂), and the previous result for volume into the equation for final pressure: Final pressure (P₂): \(P_2 = \frac{(30.0 \ moles)(0.0821 \ \frac{L⋅atm}{mol⋅K})(318.15 \ K)}{73.08 \ L} = 10.35 \ atm\) The final pressure in the tank is 10.35 atm.

Key Concepts

Pressure Calculation

Calculating the pressure of a gas in a container can seem tricky, but it's just a matter of understanding the relationships between pressure, volume, temperature, and the amount of gas. According to the Ideal Gas Law, we use the formula \( PV = nRT \), where:

• \( P \) is the pressure in atmospheres (atm)
• \( V \) is the volume in liters (L)
• \( n \) is the number of moles of gas
• \( R \) is the ideal gas constant, \(0.0821 \ \frac{L \, atm}{mol \, K}\)
• \( T \) is the temperature in Kelvin

To find the new pressure after a change in conditions, rearrange the formula to solve for \( P \): \( P = \frac{nRT}{V} \).
In this exercise, by calculating with the final conditions where the volume remains constant, we find the new pressure as \(10.35\) atm.

Temperature Conversion

Temperature conversion is a fundamental step to solve problems in thermodynamics. Our calculations require temperature in Kelvin rather than Celsius. The conversion is straightforward by adding \( 273.15 \) to the Celsius temperature.
For example:

• Initial temperature: \( 25.0^{ ext{°}} ext{C} \) converts to \( 298.15 \, K \)
• Final temperature: \( 45.0^{ ext{°}} ext{C} \) converts to \( 318.15 \, K \)

Using Kelvin ensures all gases' behavior aligns with the Ideal Gas Law, as Kelvin starts at absolute zero, increasing the accuracy of our calculations.

Moles of Gas

Understanding how many moles you have is crucial for using the Ideal Gas Law. In this problem, we start with \(20.0\) moles of oxygen. Then, an additional \(10.0\) moles are introduced, making the total \(30.0\) moles.
Each mole is defined as \(6.022 \times 10^{23}\) molecules, a constant that relates the macroscopic scale we observe to the microscopic scale of individual molecules or atoms.
Thus, by controlling or knowing the number of moles, we can predict how the gas will react to changes in temperature and volume.

Constant Volume Process

In scenarios where the volume of gas does not change, calculations become more straightforward. With the volume constant, changes in temperature and pressure relate directly, as per Charles's Law and Gay-Lussac's Law.

• Charles's Law states that \( V \) is proportional to \( T \), but since \( V \) is constant, any temperature change directly influences pressure.
• Using the relationship \( P_1/T_1 = P_2/T_2 \), we can initially find changes in gas behavior without needing to adjust for volume.

At a constant volume, an increase in temperature while adding more gas will naturally increase pressure, as seen by our final pressure outcome of \(10.35\) atm in the tank.