To solve the problem of finding the final temperature of an ideal monoatomic gas after it expands adiabatically, we can follow these steps:
### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. Therefore, the heat transfer \( q = 0 \). According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is equal to the work done \( W \) on or by the system:
\[
\Delta U = W
\]
### Step 2: Relate Change in Internal Energy to Temperature
For an ideal monoatomic gas, the change in internal energy can be expressed as:
\[
\Delta U = n C_v \Delta T
\]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T = T_2 - T_1 \).
For a monoatomic gas, \( C_v = \frac{3}{2} R \).
### Step 3: Calculate the Work Done
The work done during the expansion against a constant external pressure is given by:
\[
W = -P \Delta V = -P (V_2 - V_1)
\]
where \( P \) is the external pressure, \( V_2 \) is the final volume, and \( V_1 \) is the initial volume. Here, \( P = 1 \, \text{atm} \), \( V_2 = 2 \, \text{L} \), and \( V_1 = 1 \, \text{L} \).
Calculating \( \Delta V \):
\[
\Delta V = V_2 - V_1 = 2 \, \text{L} - 1 \, \text{L} = 1 \, \text{L}
\]
Thus, the work done is:
\[
W = -1 \, \text{atm} \times 1 \, \text{L} = -1 \, \text{L atm}
\]
To convert this to Joules, we use the conversion factor \( 1 \, \text{L atm} = 101.325 \, \text{J} \):
\[
W = -101.325 \, \text{J}
\]
### Step 4: Set Up the Equation
Now we can set up the equation using the first law of thermodynamics:
\[
n C_v (T_2 - T_1) = W
\]
Substituting the values:
\[
1 \cdot \frac{3}{2} R (T_2 - T) = -101.325
\]
Substituting \( R = 8.314 \, \text{J/(mol K)} \):
\[
\frac{3}{2} \cdot 8.314 (T_2 - T) = -101.325
\]
### Step 5: Solve for Final Temperature
Now, we can solve for \( T_2 \):
\[
12.471 (T_2 - T) = -101.325
\]
\[
T_2 - T = \frac{-101.325}{12.471}
\]
Calculating the right side:
\[
T_2 - T \approx -8.12
\]
Thus,
\[
T_2 \approx T - 8.12
\]
### Final Result
The final temperature \( T_2 \) of the gas after the adiabatic expansion will be:
\[
T_2 = T - 8.12 \, \text{K}
\]
