First start with an isothermal expansion at 2.0 atm from 1.0 L to 4.0 L. If this now is adiabatically compressed back to its original volume, what is the change in internal energy here? Also what are some examples of these processes?
To find the change in internal energy during the adiabatic compression, we need to know the properties of the gas and use the First Law of Thermodynamics. The First Law states that the change in internal energy (ΔU) of a system is equal to the heat (Q) transferred to the system minus the work (W) done by the system:
ΔU = Q - W
In an adiabatic process, no heat is transferred to or from the system (Q = 0). Therefore, the change in internal energy simplifies to:
ΔU = -W
Now, to determine the work done during the adiabatic compression, we need to use the ideal gas law and the adiabatic equation. The initial and final states of the gas can be described by the following parameters:
Initial state: P₁ = 2.0 atm, V₁ = 4.0 L
Final state: P₂ = ?, V₂ = 1.0 L (original volume)
The ideal gas law can be written as:
P₁V₁ = nRT₁
Since the process is isothermal, the gas's temperature remains constant throughout, so we can write:
P₁V₁ = nRT
The final pressure (P₂) is unknown, but we know the final volume and temperature of the gas remain the same as the initial volume and temperature. Therefore, the ideal gas law for the final state is:
P₂V₂ = nRT
Since nRT is the same for both initial and final states, we can equate the two equations:
P₁V₁ = P₂V₂
Solving for P₂:
P₂ = (P₁V₁) / V₂
= (2.0 atm * 4.0 L) / 1.0 L
= 8.0 atm
Now, we have the final pressure (P₂ = 8.0 atm), and the initial and final volumes (V₁ = 4.0 L, V₂ = 1.0 L). We can use the adiabatic equation to calculate the work done during the adiabatic compression:
W = (P₂V₂ - P₁V₁) / (γ - 1)
Here, γ is the adiabatic index or the heat capacity ratio for the gas. The value of γ depends on the gas being considered. For monoatomic ideal gases, γ is approximately 5/3 (1.67). For diatomic ideal gases, such as nitrogen or oxygen, γ is approximately 7/5 (1.4).
Let's assume we have a monoatomic ideal gas, so γ = 5/3 (1.67). Plugging in the values:
W = (8.0 atm * 1.0 L - 2.0 atm * 4.0 L) / (5/3 - 1)
= (8.0 - 8.0) / (2/3)
= 0
Therefore, the work done during the adiabatic compression (W) is 0. This means that no work is done on or by the system during this process.
Since ΔU = -W, the change in internal energy (ΔU) is also 0.
Examples of isothermal and adiabatic processes include:
- Isothermal expansion/contraction of a gas in a heat reservoir
- Adiabatic compression/decompression of gas in an engine cylinder
- Expansion/contraction of air in a refrigerator or air conditioner