One mole of an ideal gas is allowed to expand reversible and adiabatically from a temperatureof`27^(@)C)`if the work done during the process is`3`kJ,the final temperature will be equal to`(C_(v)=20JK^(-1))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have:
- Number of moles (n) = 1 mole
- Initial temperature (T1) = 27°C = 300 K (converted to Kelvin)
- Work done (W) = -3 kJ = -3000 J (since work done during expansion is negative)
- Heat capacity at constant volume (Cv) = 20 J/K

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
In an adiabatic process, \(Q = 0\). Therefore:
\[ \Delta U = W \]
Since the work done is negative (because the gas is expanding), we can write:
\[ \Delta U = -W = 3000 \text{ J} \]

### Step 3: Relate Change in Internal Energy to Temperature Change
The change in internal energy (\(\Delta U\)) for an ideal gas can also be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \(\Delta T = T_2 - T_1\). Substituting the known values:
\[ 3000 = 1 \times 20 \times (T_2 - 300) \]

### Step 4: Solve for the Final Temperature \(T_2\)
Rearranging the equation:
\[ 3000 = 20(T_2 - 300) \]
Dividing both sides by 20:
\[ 150 = T_2 - 300 \]
Now, solving for \(T_2\):
\[ T_2 = 150 + 300 = 450 \text{ K} \]

### Step 5: Convert the Final Temperature to Celsius (if needed)
If required, we can convert the final temperature back to Celsius:
\[ T_2 = 450 - 273 = 177°C \]

### Final Answer
The final temperature \(T_2\) after the adiabatic expansion is **450 K** or **177°C**.

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