Askiitians Tutor Team
Last Activity: 4 Months ago
To tackle this problem, we need to analyze the two processes involved: the adiabatic expansion of nitrogen gas and the subsequent isochoric (constant volume) compression back to its initial state. We'll also calculate the work done on the gas during these processes and illustrate the p-v diagram for clarity.
Understanding the Processes
First, let's break down the two stages:
- Adiabatic Expansion: In this phase, the nitrogen gas expands without exchanging heat with its surroundings. This means that all the work done during the expansion comes from the internal energy of the gas.
- Isochoric Compression: After the gas has expanded, it is compressed back to its original state at constant volume. During this process, the gas does not do any work since the volume remains unchanged.
Adiabatic Expansion Calculation
For an adiabatic process, we can use the following formula to relate the initial and final states of the gas:
P1 * V1^γ = P2 * V2^γ
Where:
- P1 = initial pressure = 6 atm
- V1 = initial volume
- P2 = final pressure after expansion
- V2 = final volume = 2 * V1
- γ (gamma) = specific heat ratio = 1.4
Since we don't have the initial volume, we can express it in terms of the number of moles and the ideal gas law:
PV = nRT
For nitrogen gas, we can calculate the number of moles (n) using its molar mass (approximately 28 g/mol):
n = mass / molar mass = 3 g / 28 g/mol ≈ 0.107 moles
Now, we can express the initial volume (V1) using the ideal gas law:
V1 = nRT / P1
Substituting values (R = 0.0821 L·atm/(K·mol) and T = 160 K):
V1 = (0.107 moles * 0.0821 L·atm/(K·mol) * 160 K) / 6 atm ≈ 0.43 L
Now, we can find V2:
V2 = 2 * V1 = 2 * 0.43 L = 0.86 L
Next, we can find the final pressure (P2) using the adiabatic relation:
6 atm * (0.43 L)^1.4 = P2 * (0.86 L)^1.4
Calculating this gives:
P2 ≈ 2.67 atm
Work Done During Adiabatic Expansion
The work done (W) during an adiabatic process can be calculated using the formula:
W = (P2 * V2 - P1 * V1) / (γ - 1)
Substituting the values we have:
W = [(2.67 atm * 0.86 L) - (6 atm * 0.43 L)] / (1.4 - 1)
Calculating this gives:
W ≈ -3.43 L·atm
Since work done on the gas is considered positive, we take the absolute value:
W ≈ 3.43 L·atm
Isochoric Compression Work
During the isochoric compression, the work done is zero because the volume does not change:
W_isochoric = 0
Final Work Done on the Gas
The total work done on the gas during the entire process is simply the work done during the adiabatic expansion, as the work during the isochoric process is zero:
Total Work Done = 3.43 L·atm
Visualizing the Process: p-v Diagram
To illustrate the processes, we can sketch a p-v diagram:
- Start at point A (P1, V1).
- Draw a curve downwards to point B (P2, V2) representing the adiabatic expansion.
- From point B, draw a vertical line back to point A representing the isochoric compression.
This diagram visually represents how the pressure decreases during the adiabatic expansion while the volume increases, followed by a return to the original state at constant volume.
In summary, the work done on the nitrogen gas during the adiabatic expansion is approximately 3.43 L·atm, and the isochoric process does not contribute any additional work. The p-v diagram effectively illustrates the changes in pressure and volume throughout the process.
