What is the final temperature of `0.10` mole monoatomic ideal gas that performs `75 cal` of work adiabatically.if the initial temperature is `227 ^(@)C` ( use `R=2 cal //K-mol)`

Options

A. 250 K
B. 300 K
C. 350 K
D. 750 K

Explanation

To find the final temperature of a monoatomic ideal gas that performs work adiabatically, we can follow these steps:

### Step 1: Convert the Initial Temperature to Kelvin
The initial temperature is given in Celsius, so we need to convert it to Kelvin.

\[ T_1 = 227 \, ^\circ C + 273 = 500 \, K \]

### Step 2: Identify the Given Values
We have the following values:
- Number of moles, \( n = 0.10 \, \text{mol} \)
- Work done, \( W = -75 \, \text{cal} \) (negative because work is done by the gas)
- Ideal gas constant, \( R = 2 \, \text{cal/K-mol} \)

### Step 3: Determine the Change in Internal Energy
For an adiabatic process, the change in internal energy (\( \Delta U \)) is equal to the work done (\( W \)):
\[ \Delta U = W = -75 \, \text{cal} \]

### Step 4: Calculate the Specific Heat Capacity at Constant Volume (\( C_v \))
For a monoatomic ideal gas, the specific heat capacity at constant volume is given by:
\[ C_v = \frac{3}{2} R = \frac{3}{2} \times 2 = 3 \, \text{cal/K-mol} \]

### Step 5: Use the Formula for Change in Internal Energy
The change in internal energy can also be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \( \Delta T = T_2 - T_1 \).

Substituting the known values:
\[ -75 = 0.10 \times 3 \times (T_2 - 500) \]

### Step 6: Simplify the Equation
\[ -75 = 0.30 \times (T_2 - 500) \]

### Step 7: Solve for \( T_2 \)
First, divide both sides by \( 0.30 \):
\[ -75 / 0.30 = T_2 - 500 \]
\[ -250 = T_2 - 500 \]

Now, add 500 to both sides:
\[ T_2 = 500 - 250 = 250 \, K \]

### Final Answer
The final temperature \( T_2 \) of the monoatomic ideal gas is:
\[ \boxed{250 \, K} \]