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

Explanation

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

### Step 1: Convert Initial Temperature to Kelvin
The initial temperature is given in Celsius. We need to convert it to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
Given:
\[ T_i = 227 °C \]
\[ T_i = 227 + 273 = 500 K \]

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
For an adiabatic process, \( Q = 0 \). Therefore:
\[ \Delta U = W \]
Since the work is done by the system, it is negative:
\[ \Delta U = -75 \text{ cal} \]

### Step 3: Relate Change in Internal Energy to Temperature Change
The change in internal energy (\( \Delta U \)) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where:
- \( n \) = number of moles
- \( C_v \) = molar heat capacity at constant volume
- \( \Delta T = T_f - T_i \)

### Step 4: Determine \( C_v \) for a Monoatomic Ideal Gas
For a monoatomic ideal gas, the molar heat capacity at constant volume is given by:
\[ C_v = \frac{3}{2} R \]
Given \( R = 2 \text{ cal/(K·mol)} \):
\[ C_v = \frac{3}{2} \times 2 = 3 \text{ cal/(K·mol)} \]

### Step 5: Substitute Values into the Internal Energy Equation
Now we can substitute the known values into the equation for \( \Delta U \):
\[ -75 = n C_v (T_f - T_i) \]
Substituting \( n = 0.10 \) moles and \( C_v = 3 \):
\[ -75 = 0.10 \times 3 \times (T_f - 500) \]

### Step 6: Solve for \( T_f \)
Rearranging the equation gives:
\[ -75 = 0.30 (T_f - 500) \]
Dividing both sides by 0.30:
\[ -250 = T_f - 500 \]
Now, adding 500 to both sides:
\[ T_f = 500 - 250 = 250 K \]

### Final Result
The final temperature of the gas is:
\[ T_f = 250 K \]