Problem 11 - Physics

Question

An ideal gas at \(15.5^{\circ} \mathrm{C}\) and a pressure of \(1.72 \times 10^{5} \mathrm{~Pa}\) occupies a volume of \(2.81 \mathrm{~m}^{3}\). (a) How many moles of gas are present? (b) If the volume is raised to \(4.16 \mathrm{~m}^{3}\) and the temperature raised to \(28.2{ }^{\circ} \mathrm{C}\), what will be the pressure of the gas?

Step-by-Step Solution

Answer

(a) About 200.60 moles; (b) New pressure is approximately \(1.27 \times 10^5\) Pa.

Step 1: Understand the Problem

We're given a situation involving an ideal gas and need to determine two things: (a) the number of moles present in the initial state and (b) the new pressure if the volume and temperature change.

Step 2: Use Ideal Gas Law for Initial State (Part a)

The Ideal Gas Law is given by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvins. First, convert temperature from Celsius to Kelvin: \( T_1 = 15.5 + 273.15 = 288.65 \text{ K} \). Then solve for \( n \) using: \( n = \frac{PV}{RT} \). Substitute \( P = 1.72 \times 10^5 \text{ Pa} \), \( V = 2.81 \text{ m}^3 \), and \( R = 8.314 \text{ J/mol·K} \) into the equation to find \( n \).

Step 3: Calculate Initial Moles (Part a)

Substitute and calculate: \( n = \frac{(1.72 \times 10^5) \times 2.81}{8.314 \times 288.65} \approx 200.60 \text{ moles} \). Thus, there are approximately 200.60 moles of gas present.

Step 4: Determine New Conditions (Part b)

For the new conditions, use the initial equation \( PV = nRT \). Keeping \( n \) constant, we apply the ideal gas law to the new conditions: new volume \( V_2 = 4.16 \text{ m}^3 \), and new temperature \( T_2 = 28.2 + 273.15 = 301.35 \text{ K} \). Use the transformed Ideal Gas Law to find new pressure \( P_2 \).

Step 5: Use Ideal Gas Law for New State (Part b)

Since the number of moles \( n \) remains constant, equate the relationships for initial and new conditions: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). Solve for \( P_2 \) to find the new pressure.

Step 6: Calculate New Pressure (Part b)

Rearrange and solve: \( P_2 = \frac{P_1V_1T_2}{T_1V_2} \). Substitute values: \( P_2 = \frac{(1.72 \times 10^5) \times 2.81 \times 301.35}{288.65 \times 4.16} \approx 1.27 \times 10^5 \text{ Pa} \). Thus, the new pressure is approximately \(1.27 \times 10^5 \text{ Pa}\).

Key Concepts

Moles of Gas

To determine the number of moles of gas in a given volume, you can use the Ideal Gas Law, which is expressed as \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. For accurate calculations, it's crucial to first convert the temperature from Celsius to Kelvin by adding 273.15. Given the initial conditions of the problem, we have:

• Pressure \( P = 1.72 \times 10^5 \text{ Pa} \)
• Volume \( V = 2.81 \text{ m}^3 \)
• Temperature \( T = 15.5 + 273.15 = 288.65 \text{ K} \)

Using these values and knowing the ideal gas constant \( R = 8.314 \text{ J/mol·K} \), we substitute into the formula and solve for \( n \) (moles):\[ n = \frac{PV}{RT} = \frac{(1.72 \times 10^5) \times 2.81}{8.314 \times 288.65} \approx 200.60 \text{ moles}\] This calculation tells us there are approximately 200.60 moles of gas present.

Pressure Calculation

In scenarios where the volume and temperature of a gas change, calculating the new pressure can be required. Here, the number of moles remains constant, and again, the Ideal Gas Law is the key to finding the solution. The equation remains the same: \( PV = nRT \), but adapted to the new state, and using the fact that \( n \) (the amount of gas) stays constant, we derive the relation: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]To find the new pressure \( P_2 \), we simply rearrange the equation:\[ P_2 = \frac{P_1V_1T_2}{T_1V_2}\]Let's submit known values to calculate:

• Old Pressure \( P_1 = 1.72 \times 10^5 \text{ Pa} \)
• Old Volume \( V_1 = 2.81 \text{ m}^3 \)
• Old Temperature \( T_1 = 288.65 \text{ K} \)
• New Volume \( V_2 = 4.16 \text{ m}^3 \)
• New Temperature \( T_2 = 28.2 + 273.15 = 301.35 \text{ K} \)

Substitute these values:\[ P_2 = \frac{(1.72 \times 10^5) \times 2.81 \times 301.35}{288.65 \times 4.16} \approx 1.27 \times 10^5 \text{ Pa}\]This means the new pressure of the gas after changing volume and temperature is about \(1.27 \times 10^5 \text{ Pa}\).

Volume and Temperature Change

When you increase or decrease the volume of a gas while the temperature changes, the pressure will also change if the number of moles remains constant. According to the Ideal Gas Law, the relation between volume, temperature, and pressure can be expressed directly using a combined form.
Volume expands when temperature increases if the gas is allowed to expand. In such cases:

• If volume goes up, pressure falls if temperature conditions don't change drastically, since pressure and volume are inversely proportional when temperature is constant (according to Boyle's Law).
• If temperature increases, pressure can increase or volume can increase if it's allowed to expand (in line with Charles' Law).

When tackling problems with both volume and temperature changing, it’s important to understand which parameters are held constant, allowing you to predict how other related properties will vary.
In this exercise, the volume was increased from \(2.81 \text{ m}^3\) to \(4.16 \text{ m}^3\), and the temperature was also raised from \(15.5^{\circ} \text{C} \) to \(28.2^{\circ} \text{C} \). Using the derived relationship from the Ideal Gas Law, you can solve for new pressure parameters, which clearly demonstrates the interconnectedness between pressure, volume, and temperature.