One mole of an ideal monatomic gas at 300 K and 1 atm is compressed
adiabatically by application of a constant external pressure of 25 atm until mechanical equiliibrium
is achieved. Calculate the final temperature of the gas, as well as q, w, deltaU and deltaH
for this process.
To solve this problem, we can use the equations for an adiabatic process, as well as the first law of thermodynamics.
First, let's find the final temperature of the gas using the equation for an adiabatic process:
T₁V₁^(γ-1) = T₂V₂^(γ-1)
Where:
T₁ is the initial temperature of the gas (300 K)
T₂ is the final temperature of the gas (to be determined)
V₁ is the initial volume of the gas (which we can assume is 1 mole at 1 atm)
V₂ is the final volume of the gas (which we can assume is the same as V₁)
γ is the heat capacity ratio for monatomic ideal gases, which is equal to 5/3
Plugging in the given values into the equation, we have:
(300 K)(1)^(5/3-1) = T₂(1)^(5/3-1)
Simplifying the equation, we get:
300^(2/3) = T₂
Calculating this value, we find that T₂ ≈ 238.26 K.
Next, let's calculate q (heat), w (work), ΔU (change in internal energy), and ΔH (change in enthalpy) for this adiabatic compression process.
Since the process is adiabatic, q = 0 (no heat exchange with the surroundings).
To calculate the work done, we can use the equation:
w = P(V₂ - V₁)
However, since the process is done at constant external pressure, we need to use the equation:
w = -Pext(V₂ - V₁)
Plugging in the values, we have:
w = - (25 atm)(1 - 1) = 0
Therefore, no work is done in this process (w = 0).
To calculate the change in internal energy (ΔU), we can use the first law of thermodynamics:
ΔU = q + w
Since q = 0 and w = 0, we can conclude that ΔU = 0.
Finally, to calculate the change in enthalpy (ΔH), we can use the relation:
ΔH = ΔU + PΔV
In an adiabatic process, we assume there is no heat exchange and thus ΔU = 0. Additionally, since the process is done at constant external pressure, we can rewrite the equation as:
ΔH = 0 + Pext(V₂ - V₁)
Plugging in the values, we have:
ΔH = (25 atm)(1 - 1) = 0
Therefore, ΔH = 0.
In summary:
- The final temperature of the gas is approximately 238.26 K.
- The heat exchange (q) is 0.
- The work done (w) is 0.
- The change in internal energy (ΔU) is 0.
- The change in enthalpy (ΔH) is 0.
To calculate the final temperature (Tf) of the gas, as well as the other quantities, we need to use various thermodynamic equations and principles.
First, we need to understand that the process described is an adiabatic compression. In an adiabatic process, there is no heat exchange with the surroundings (q = 0), but work can be done on or by the system (w ≠ 0).
To calculate the final temperature, we can use the adiabatic process equation:
Tf/Ti = (Pf/Pi)^((γ-1)/γ)
Where:
- Tf and Ti are the final and initial temperatures, respectively.
- Pf and Pi are the final and initial pressures, respectively.
- γ is the ratio of specific heat capacities (Cp/Cv) for the gas.
For an ideal monatomic gas, γ is equal to 5/3.
Given:
- Initial temperature (Ti) = 300 K
- Initial pressure (Pi) = 1 atm
- External pressure (Pf) = 25 atm
- γ = 5/3
Using the adiabatic process equation, we can substitute the given values:
Tf/300 = (25/1)^((5/3-1)/(5/3))
Simplifying the equation:
Tf/300 = (25/1)^(2/5)
Tf/300 = 5^2
Tf/300 = 25
Finally, solving for Tf:
Tf = (25)(300)/1
Tf = 7500 K
Therefore, the final temperature (Tf) of the gas after the adiabatic compression is 7500 K.
Now, let's calculate the other quantities:
1. Heat Exchange (q): Since the process is adiabatic, q = 0.
2. Work Done (w): To calculate work, we need to use the equation for an adiabatic process:
w = (PfVf - PiVi)/(γ - 1)
Since the process is adiabatic and the external pressure is constant, we know that PfVf = PiVi.
Therefore, the work done (w) is:
w = 0
3. Change in Internal Energy (ΔU): To calculate ΔU, we can use the first law of thermodynamics:
ΔU = q - w
Since q = 0 and w = 0, we conclude that ΔU = 0.
4. Change in Enthalpy (ΔH): For an ideal gas, ΔU = ΔH because there is no phase change or chemical reaction occurring.
Therefore, ΔH = 0.
In summary:
- Final temperature (Tf) = 7500 K
- Heat exchange (q) = 0
- Work done (w) = 0
- Change in internal energy (ΔU) = 0
- Change in enthalpy (ΔH) = 0
T2 = 3180 K
q = 0 J
w = 37034.29 J
dU = 37034.29 J
dH = 59860.8 J