To tackle this problem, we will break it down step by step, addressing each part methodically. We have a gas with a specific heat ratio (γ) of 1.32, an initial volume of 1 liter, an initial temperature of 273 K, and an initial pressure of 1.00 atm. Let's begin with part (a).
Adiabatic Compression of the Gas
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between the pressure and volume for an ideal gas undergoing an adiabatic process can be expressed as:
- P₁V₁^γ = P₂V₂^γ
- T₁V₁^(γ-1) = T₂V₂^(γ-1)
Given that the initial volume (V₁) is 1 L, the final volume (V₂) after compression to half its original volume will be:
V₂ = V₁ / 2 = 1 L / 2 = 0.5 L
Now, we can use the first equation to find the final pressure (P₂). The initial pressure (P₁) is 1.00 atm.
Using the formula:
P₂ = P₁ * (V₁ / V₂)^γ
Substituting the values:
P₂ = 1.00 atm * (1 L / 0.5 L)^1.32
P₂ = 1.00 atm * (2)^1.32 ≈ 1.00 atm * 2.5 = 2.5 atm
Finding the Final Temperature
To find the final temperature (T₂), we can use the second equation:
T₂ = T₁ * (V₁ / V₂)^(γ-1)
Substituting the known values:
T₂ = 273 K * (1 L / 0.5 L)^(1.32 - 1)
T₂ = 273 K * (2)^0.32 ≈ 273 K * 1.25 = 341.25 K
Cooling the Gas at Constant Pressure
Now, moving on to part (b), where the gas is cooled back to 273 K at constant pressure. When cooling occurs at constant pressure, we can use the ideal gas law:
PV = nRT
Since the pressure remains constant, we can rearrange the equation to find the final volume (V₃):
V₃ = nRT / P
First, we need to calculate the number of moles (n) of the gas. Using the ideal gas law at the initial conditions:
n = P₁V₁ / RT₁
Substituting the values with R = 0.0821 L·atm/(K·mol):
n = (1.00 atm) * (1 L) / (0.0821 L·atm/(K·mol) * 273 K)
n ≈ 0.0446 moles
Now we can find the final volume at 273 K and constant pressure (1.00 atm):
V₃ = (0.0446 moles) * (0.0821 L·atm/(K·mol)) * (273 K) / (1.00 atm)
V₃ ≈ 1.00 L
Calculating Total Work Done on the Gas
For part (c), the work done on the gas during an adiabatic process can be calculated using the formula:
W = (P₂V₂ - P₁V₁) / (γ - 1)
Substituting the values:
W = (2.5 atm * 0.5 L - 1.00 atm * 1 L) / (1.32 - 1)
Converting atm·L to Joules (1 atm·L = 101.325 J):
W = [(2.5 * 0.5 * 101.325) - (1.00 * 1 * 101.325)] / (0.32)
W = [(126.65625 - 101.325) / 0.32]
W = 79.2302 J / 0.32 ≈ 247.28 J
To summarize:
- The final pressure after adiabatic compression is approximately 2.5 atm.
- The final temperature after adiabatic compression is approximately 341.25 K.
- After cooling at constant pressure, the final volume returns to approximately 1.00 L.
- The total work done on the gas during the adiabatic process is approximately 247.28 J.
By breaking down each part of the problem, we can see how the relationships between pressure, volume, and temperature interact during these processes. This provides a clearer understanding of the behavior of gases under different conditions.
