To solve the problem step by step, we will follow the principles of thermodynamics, particularly focusing on the adiabatic expansion of an ideal monoatomic gas.
### Step 1: Identify Given Data
- Number of moles (n) = 1 mole
- Initial temperature (T1) = 27°C = 27 + 273 = 300 K
- External pressure (P_ext) = 1 atm
- Initial volume (V1) = 10 dm³
- Final volume (V2) = 20 dm³
### Step 2: Calculate Work Done (W)
For an adiabatic process, the work done by the system when expanding against a constant external pressure can be calculated using the formula:
\[ W = -P_{ext} \Delta V \]
Where:
\[ \Delta V = V_2 - V_1 = 20 \, \text{dm}^3 - 10 \, \text{dm}^3 = 10 \, \text{dm}^3 \]
Convert dm³ to liters (1 dm³ = 1 L):
\[ \Delta V = 10 \, \text{L} \]
Now, substituting the values:
\[ W = -1 \, \text{atm} \times 10 \, \text{L} \]
We need to convert atm·L to Joules:
\[ 1 \, \text{L} \cdot \text{atm} = 101.3 \, \text{J} \]
Thus,
\[ W = -10 \times 101.3 \, \text{J} = -1013 \, \text{J} \]
### Step 3: Calculate Change in Internal Energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = Q + W \]
Since it is an adiabatic process, \( Q = 0 \):
\[ \Delta U = 0 + W = W = -1013 \, \text{J} \]
### Step 4: Calculate Final Temperature (T2)
For a monoatomic ideal gas, the change in internal energy can also be expressed as:
\[ \Delta U = n C_V (T_2 - T_1) \]
Where \( C_V \) for a monoatomic gas is:
\[ C_V = \frac{3}{2} R \]
Substituting the values:
\[ -1013 \, \text{J} = 1 \cdot \frac{3}{2} \cdot 8.314 \, \text{J/K} \cdot (T_2 - 300) \]
Calculating \( C_V \):
\[ C_V = \frac{3}{2} \cdot 8.314 = 12.471 \, \text{J/K} \]
Now substituting \( C_V \):
\[ -1013 = 12.471 (T_2 - 300) \]
Solving for \( T_2 \):
\[ T_2 - 300 = \frac{-1013}{12.471} \]
\[ T_2 - 300 = -81.2 \]
\[ T_2 = 300 - 81.2 = 218.8 \, \text{K} \]
### Step 5: Conclusion
The final temperature \( T_2 \) is approximately 218.8 K, and the work done is -1013 J.
