To solve the problem step-by-step, we will use the principles of thermodynamics and the properties of an ideal monoatomic gas.
### Step 1: Identify the given data
- Initial temperature, \( T_i = 298 \, K \)
- Initial pressure, \( P_i = 1.013 \times 10^6 \, Pa \)
- Final pressure, \( P_f = 1.013 \times 10^5 \, Pa \)
- Heat capacity at constant volume for a monoatomic gas, \( C_v = 1.5R \)
### Step 2: Understand the adiabatic process
In an adiabatic process, the change in internal energy (\( \Delta U \)) is equal to the work done on the gas. The relationship can be expressed as:
\[
\Delta U = -W
\]
Where \( W \) is the work done by the gas.
### Step 3: Relate internal energy change to temperature
For an ideal gas, the change in internal energy can also be expressed as:
\[
\Delta U = nC_v(T_f - T_i)
\]
Where \( n \) is the number of moles of the gas.
### Step 4: Calculate the work done
The work done by the gas during expansion against a constant external pressure can be expressed as:
\[
W = P_{external} \Delta V = P_f (V_f - V_i)
\]
### Step 5: Express volumes in terms of temperature and pressure
Using the ideal gas law, we can express the volumes as:
\[
V_i = \frac{nRT_i}{P_i} \quad \text{and} \quad V_f = \frac{nRT_f}{P_f}
\]
Thus, the change in volume \( \Delta V \) can be expressed as:
\[
\Delta V = V_f - V_i = \frac{nRT_f}{P_f} - \frac{nRT_i}{P_i}
\]
### Step 6: Substitute into the work equation
Substituting the expression for \( \Delta V \) into the work equation gives:
\[
W = P_f \left( \frac{nRT_f}{P_f} - \frac{nRT_i}{P_i} \right)
\]
This simplifies to:
\[
W = nR(T_f - \frac{P_f}{P_i}T_i)
\]
### Step 7: Set up the equation for internal energy change
Equating the two expressions for internal energy change:
\[
nC_v(T_f - T_i) = -W
\]
Substituting for \( W \):
\[
nC_v(T_f - T_i) = -nR(T_f - \frac{P_f}{P_i}T_i)
\]
### Step 8: Cancel \( n \) and rearrange
Cancel \( n \) from both sides (assuming \( n \neq 0 \)):
\[
C_v(T_f - T_i) = -R(T_f - \frac{P_f}{P_i}T_i)
\]
Substituting \( C_v = 1.5R \):
\[
1.5R(T_f - T_i) = -R(T_f - \frac{P_f}{P_i}T_i)
\]
Dividing through by \( R \):
\[
1.5(T_f - T_i) = -(T_f - \frac{P_f}{P_i}T_i)
\]
### Step 9: Solve for \( T_f \)
Rearranging gives:
\[
1.5T_f - 1.5T_i = -T_f + \frac{P_f}{P_i}T_i
\]
Combining like terms:
\[
2.5T_f = 1.5T_i + \frac{P_f}{P_i}T_i
\]
Factoring out \( T_i \):
\[
T_f = \frac{T_i(1.5 + \frac{P_f}{P_i})}{2.5}
\]
### Step 10: Substitute known values
Substituting \( T_i = 298 \, K \), \( P_f = 1.013 \times 10^5 \, Pa \), and \( P_i = 1.013 \times 10^6 \, Pa \):
\[
T_f = \frac{298 \left( 1.5 + \frac{1.013 \times 10^5}{1.013 \times 10^6} \right)}{2.5}
\]
Calculating \( \frac{P_f}{P_i} = \frac{1.013 \times 10^5}{1.013 \times 10^6} = 0.1 \):
\[
T_f = \frac{298 \left( 1.5 + 0.1 \right)}{2.5} = \frac{298 \times 1.6}{2.5}
\]
Calculating:
\[
T_f = \frac{476.8}{2.5} = 190.72 \, K
\]
### Final Answer
The final temperature of the gas is approximately \( T_f \approx 190.7 \, K \).
