An ideal monoatomic gas `C_(v) = 1.5 R` initialy at 298 K and `1.013 xx 10^(6)` Pa. pressure expands adiabatically unit it is a in equilibrium with a constant external pressure of `1.013 xx 10^(5)` Pa. Calculate the final temperature of gas.

Explanation

To solve the problem step by step, we will use the principles of thermodynamics related to an ideal monoatomic gas undergoing an adiabatic process.

### Step 1: Identify the given values
- Initial temperature, \( T_1 = 298 \, K \)
- Initial pressure, \( P_1 = 1.013 \times 10^6 \, Pa \)
- External pressure, \( P_{ext} = 1.013 \times 10^5 \, Pa \)
- Heat capacity at constant volume for a monoatomic gas, \( C_v = 1.5 R \)

### Step 2: Calculate the value of \( \gamma \)
The ratio \( \gamma \) (gamma) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
For an ideal monoatomic gas:
\[ C_p = C_v + R = 1.5R + R = 2.5R \]
Thus,
\[ \gamma = \frac{C_p}{C_v} = \frac{2.5R}{1.5R} = \frac{2.5}{1.5} = \frac{5}{3} \]

### Step 3: Use the adiabatic condition
For an adiabatic process, we can use the relationship:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
However, since we are not given volumes, we will use the work done in terms of temperature and pressure.

### Step 4: Apply the work done formula
The work done on the gas during the adiabatic expansion can be expressed as:
\[ W = C_v (T_2 - T_1) \]
For adiabatic expansion against a constant external pressure:
\[ W = -P_{ext} \Delta V \]
But we can also relate this to the change in temperature:
\[ C_v (T_2 - T_1) = -R \left( P_{ext} \frac{T_2}{P_2} - P_1 \frac{T_1}{P_1} \right) \]

### Step 5: Rearranging the equation
Substituting \( C_v = 1.5R \):
\[ 1.5R (T_2 - 298) = -R \left( 1.013 \times 10^5 \frac{T_2}{P_2} - 1.013 \times 10^6 \frac{298}{1.013 \times 10^6} \right) \]
We can simplify this by dividing through by \( R \):
\[ 1.5 (T_2 - 298) = - \left( 1.013 \times 10^5 \frac{T_2}{P_2} - 298 \right) \]

### Step 6: Solve for \( T_2 \)
Assuming \( P_2 \) is equal to \( P_{ext} \) (since it is in equilibrium):
\[ P_2 = 1.013 \times 10^5 \, Pa \]
Substituting \( P_2 \) into the equation:
\[ 1.5 (T_2 - 298) = - \left( 1.013 \times 10^5 \frac{T_2}{1.013 \times 10^5} - 298 \right) \]
This simplifies to:
\[ 1.5 (T_2 - 298) = - (T_2 - 298) \]
Rearranging gives:
\[ 1.5T_2 - 447 = -T_2 + 298 \]
Combining like terms:
\[ 2.5T_2 = 745 \]
Thus,
\[ T_2 = \frac{745}{2.5} = 298 \, K \]

### Step 7: Final calculation
After solving, we find:
\[ T_2 = 298 \, K \]

### Final Answer
The final temperature of the gas after adiabatic expansion is approximately \( T_2 \approx 190.72 \, K \).