Problem 11 - Physics

Question

\(\mathbf{1}\) 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) Approximately 202.42 moles; (b) New pressure is about \(1.40 \times 10^5 \text{ Pa}\).

Step 1: Convert Temperatures to Kelvin

The initial and final temperatures are given in degrees Celsius. We need to convert these to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \]- Initial Temperature: \[ T_1 = 15.5 + 273.15 = 288.65 \text{ K} \]- Final Temperature: \[ T_2 = 28.2 + 273.15 = 301.35 \text{ K} \]

Step 2: Use Ideal Gas Law to Find Moles

Apply the Ideal Gas Law to solve for the number of moles \(n\): \[ PV = nRT \]Given:- Pressure \( P = 1.72 \times 10^5 \text{ Pa} \)- Volume \( V = 2.81 \text{ m}^3 \)- Gas constant \( R = 8.314 \text{ J/(mol K)} \)- Temperature \( T_1 = 288.65 \text{ K} \)Substitute these values into the equation and solve for \(n\):\[ n = \frac{PV}{RT} = \frac{1.72 \times 10^5 \times 2.81}{8.314 \times 288.65} \approx 202.42 \text{ moles} \]

Step 3: Use Combined Gas Law for New Conditions

The Combined Gas Law equation is:\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]Given:- \( P_1 = 1.72 \times 10^5 \text{ Pa} \)- \( V_1 = 2.81 \text{ m}^3 \)- \( T_1 = 288.65 \text{ K} \)- \( V_2 = 4.16 \text{ m}^3 \)- \( T_2 = 301.35 \text{ K} \)Rearrange the equation to solve for \( P_2 \):\[ P_2 = \frac{P_1V_1T_2}{V_2T_1} = \frac{1.72 \times 10^5 \times 2.81 \times 301.35}{4.16 \times 288.65} \approx 1.40 \times 10^5 \text{ Pa} \]

Step 4: Conclude the Solution

You have calculated that the number of moles \( n \) of the gas is approximately 202.42 moles and the new pressure \( P_2 \) under the changed conditions is approximately \( 1.40 \times 10^5 \text{ Pa} \).

Key Concepts

Moles

In chemistry, a mole is a unit that helps us count particles that are incredibly tiny, like atoms or molecules. It simplifies dealing with large numbers of these particles. A mole is defined as containing exactly 6.022 x 10\(^23\) particles, known as Avogadro's number.

When we talk about gases, the amount of substance, or moles, is crucial to determining how it behaves under various conditions. The Ideal Gas Law equation, which we'll look at next, is a handy way to find out how many moles of a gas you have when you know the pressure, volume, and temperature.

The Ideal Gas Law is:

• \( PV = nRT \)

Here, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

Using this formula, you can rearrange to solve for the number of moles as:

• \( n = \frac{PV}{RT} \)

By using the given data from the problem such as the pressure, volume, and converting Celsius to Kelvin, we can calculate the moles of gas present.

Kelvin temperature

Understanding temperature is key in gas law calculations, and Kelvin is the standard unit used. The Kelvin scale is an absolute scale, meaning it starts from zero, which is the coldest possible temperature. It helps us avoid negative numbers in thermodynamic formulas.

To convert from Celsius, which we often use in everyday life, to Kelvin, you add 273.15 to the Celsius temperature. For example, if you start with a temperature of 15.5°C, you convert it to Kelvin like this:

• \( T(K) = T(^{\circ}C) + 273.15 \)
• For 15.5°C, \( T(K) = 15.5 + 273.15 = 288.65 \ \text{K} \)

Getting familiar with this conversion is very important because Kelvin makes calculations more straightforward in physics and chemistry.

No temperature can drop below 0 Kelvin, making it a reliable scale for scientific measurements.

Pressure calculation

Pressure in gases is an important concept and it often changes with volume and temperature. In our problem, after finding out the initial conditions, we needed to calculate the new pressure using the conditions that changed.

The pressure of a gas is measured using various units like Pascals (Pa), which is a common unit in the SI system. To find the new pressure when the volume and temperature change, we don’t use the Ideal Gas Law directly, but rather an adjusted form called the Combined Gas Law.The Combined Gas Law formula helps us relate changes in pressure, volume, and temperature by:

• \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

Where:

• \( P_1 \) and \( P_2 \) are the initial and final pressures,
• \( V_1 \) and \( V_2 \) are the initial and final volumes, and
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

To solve for a new pressure \( P_2 \), rearrange the formula:

• \( P_2 = \frac{P_1V_1T_2}{V_2T_1} \)

Substituting the known values into this equation provides the final pressure calculation.

Combined Gas Law

The Combined Gas Law is a powerful tool in chemistry that relates three fundamental properties of gases: pressure, volume, and temperature. It is called "combined" because it brings together several individual gas laws—Boyle's Law, Charles's Law, and Avogadro's Law—into a comprehensive equation.

This law can predict how a gas will behave when changes in conditions occur, such as heating, cooling, compression, or expansion. Here's the formula you’ll typically see:

• \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

The formula assumes that the number of moles remains constant, meaning no gas is added or removed from the system.

To use the Combined Gas Law effectively:

• First, make sure that all temperatures are in Kelvin.
• Second, convert pressures using a consistent pressure unit like Pascals or atmospheres.
• Finally, apply the formula by inputting your known initial and final conditions to find the unknown variable.

This approach allows you to find out how one property of the gas changes as another property is varied, making it highly practical for real-world applications.