Problem 67 - University Physics with Modern Physics

Question

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

Step-by-Step Solution

Answer

(a) Final pressure: 2.00 atm, Final temperature: 273.15 K. (b) Final volume: 0.75 L.

Step 1: Initial Setup and Known Values

We know that the initial volume of the gas, \(V_1\), is 3.00 L. The initial pressure, \(P_1\), is 1.00 atm and the temperature, \(T_1\), is 0.00°C, which is 273.15 K when converted to Kelvin. We are given that the volume is compressed to half, making the final volume, \(V_2\), equal to 1.50 L.

Step 2: Applying the Ideal Gas Law

The ideal gas law, \(PV = nRT\), can be used in its combined form for initial and final states, expressed as \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). We need to find both the final temperature, \(T_2\), and the final pressure, \(P_2\).

Step 3: Calculating the Final Pressure (\(P_2\))

Using Boyle's Law (since temperature changes haven't been accounted yet), where \(P_1V_1 = P_2V_2\), solve for \(P_2\): \[P_2 = \frac{P_1V_1}{V_2} = \frac{1.00\, \text{atm} \times 3.00\, \text{L}}{1.50\, \text{L}} = 2.00\, \text{atm}\]. Thus, the final pressure \(P_2 = 2.00\, \text{atm}\).

Step 4: Calculating the Final Temperature (\(T_2\))

Using the combined gas law \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), solve for \(T_2\): \[T_2 = \frac{P_2V_2T_1}{P_1V_1} = \frac{2.00\, \text{atm} \times 1.50\, \text{L} \times 273.15\, \text{K}}{1.00\, \text{atm} \times 3.00\, \text{L}} = 273.15\, \text{K}\]. Therefore, the final temperature is 273.15 K.

Step 5: Determining Final Volume After Cooling to 0°C

Cooling the gas at constant pressure back to \(0.00^{\circ}\text{C}\) (273.15 K), apply Charles's Law, where \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), to find the new volume, \(V\): \[V = \frac{T_2 \times V_2}{T_1} = \frac{273.15\, \text{K} \times 1.50\, \text{L}}{546.30\, \text{K}} = 0.75\, \text{L}\]. Thus, the gas volume is reduced to 0.75 L.

Key Concepts

Boyle's Law and its Relation to Gas Pressure and Volume

Boyle's Law is an important concept in understanding the behavior of gases. It describes how the pressure of a gas tends to increase as the volume decreases, provided the temperature remains constant. The law is mathematically expressed as \( P_1 V_1 = P_2 V_2 \). Here, \( P \) represents the pressure and \( V \) the volume of the gas. This direct relationship tells us that if the volume of a gas is reduced, its pressure rises, assuming no change in temperature. For example, when compressing the nitrogen gas in a sealed cylinder, its volume decreases while its pressure increases. By applying Boyle’s Law, we determined that when the volume was halved from 3.00 L to 1.50 L, the initial pressure of 1.00 atm rose to 2.00 atm.

Charles's Law and Volume Change with Temperature

Charles's Law provides insight into how gases expand and contract with temperature change. It shows that the volume of a gas is directly proportional to its temperature measured in Kelvin, as long as the pressure remains constant. The formula \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) describes this relationship, where \( V \) is the volume and \( T \) is the temperature in Kelvin. When the compressed nitrogen gas was cooled back to 0°C, Charles's Law was applied to find the new volume. Even though the pressure was kept constant, cooling the gas caused it to contract from 1.50 L to 0.75 L, hence highlighting how temperature changes can impact gas volume considerably.

Understanding Nitrogen Gas in Ideal Gas Law Applications

Nitrogen gas often serves as a common example in demonstrations of the Ideal Gas Law because it typically behaves like an ideal gas under many conditions. The Ideal Gas Law equation \( PV = nRT \) relates pressure (\( P \)), volume (\( V \)), the number of moles (\( n \)), the ideal gas constant (\( R \)), and temperature (\( T \)) to describe the state of a gas. In our scenario, nitrogen gas in a piston-cylinder setup allowed us to explore various thermodynamic changes. It illustrates how gases, similarly, follow predictable patterns under varying conditions of pressure, volume, and temperature. Even though real gases may not perfectly adhere to ideal behavior, such approximations enable practical applications in evaluating, predicting, or calculating gas behavior in diverse fields.