Problem 9 - University Physics with Modern Physics

Question

A large cylindrical tank contains 0.750 \(\mathrm{m}^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(7.50 \times 10^{3}\) Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 \(\mathrm{m}^{3}\) and the temperature is increased to \(157^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The final pressure is \( 9.01 \times 10^3 \) Pa.

Step 1: Identify Initial and Final States

First, recognize the initial conditions and final conditions of the gas. Initially, the volume \( V_1 = 0.750 \, \text{m}^3 \), pressure \( P_1 = 7.50 \times 10^3 \, \text{Pa} \), and temperature \( T_1 = 27^\circ \text{C} \). Convert \( T_1 \) to Kelvin: \( T_1 = 27 + 273 = 300 \, \text{K} \). For the final state, the volume \( V_2 = 0.480 \, \text{m}^3 \) and temperature \( T_2 = 157^\circ \text{C} \). Convert \( T_2 \) to Kelvin: \( T_2 = 157 + 273 = 430 \, \text{K} \).

Step 2: Utilize the Combined Gas Law

Apply the combined gas law, which is \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Here, we need to find the final pressure \( P_2 \). Rearrange the equation to solve for \( P_2 \): \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \]

Step 3: Plug in the Values and Calculate

Substitute the known values into the equation: \[ P_2 = \frac{(7.50 \times 10^3 \, \text{Pa}) \times (0.750 \, \text{m}^3) \times (430 \, \text{K})}{(300 \, \text{K}) \times (0.480 \, \text{m}^3)} \] Calculate \( P_2 \) to find:\[ P_2 = \frac{(3225 \times 430)}{144} \approx 9.01 \times 10^3 \, \text{Pa} \]

Step 4: Confirm the Result

Verify the units and the calculation for accuracy. The units of pressure (Pa) are consistent throughout. Double-check the multiplication and division steps to ensure all calculations were performed correctly.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics, represented by the equation \( PV = nRT \). This law describes how gases behave under various conditions of pressure (\( P \)), volume (\( V \)), and temperature (\( T \)). The ideal gas constant \( R \) and the number of moles \( n \) are also factored in. However, in exercises involving changing conditions, we often use the Combined Gas Law, which is adapted from the Ideal Gas Law. This law relates initial and final states of a gas, expressing that the ratio of the product of pressure, volume, and temperature remains constant. By knowing these properties, we can predict how one variable will change given changes in the others, which is key in solving problems like the one in the exercise. The relationship simplifies to \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) when the number of moles is constant.

Pressure Calculation

Calculating pressure is an essential part of understanding gas behaviors. In the given exercise, you start with an initial pressure of \(7.50 \times 10^3 \) Pascal (Pa) and need to find the new pressure when conditions change. The formula used is derived from the Combined Gas Law as:

• \( P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \)

This formula helps calculate the final pressure (\( P_2 \)) by considering how volume and temperature are altered. Substituting the known values into the equation will allow you to solve for \( P_2 \), showing how pressure increases under certain conditions. The units of pressure, typically in Pascals (Pa), must remain consistent throughout your calculations to ensure accuracy.

Temperature Conversion

Temperature plays a crucial role in gas law calculations. It must be represented in Kelvin for mathematical accuracy because Kelvin is the absolute temperature scale. To convert from Celsius to Kelvin, add 273, as seen in the exercise where:

• Initial temperature \( T_1 = 27^\circ \text{C} \rightarrow 300 \text{ K} \)
• Final temperature \( T_2 = 157^\circ \text{C} \rightarrow 430 \text{ K} \)

This conversion is essential to ensure the equations and calculations are correct because using Celsius directly would yield incorrect results. Kelvin ensures that the scale starts at absolute zero, allowing the gas laws to apply universally.

Volume Change

Understanding how changes in volume affect pressure and temperature is crucial. In this exercise, the volume of nitrogen gas is decreased from \(0.750 \, \text{m}^3\) to \(0.480 \, \text{m}^3\). This reduction in volume, when paired with an increase in temperature, predicts an increase in pressure according to the Combined Gas Law.

• Decrease in volume since pressure tends to increase when a gas is compressed.
• Relationship between volume and pressure is inversely proportional, meaning if volume decreases, pressure tends to increase if the temperature is held constant or increases.

Thus, volume change, alongside temperature change, allows accurate calculation of new pressure conditions, illustrating the interplay between the variables of gases.