One mole of an ideal gas at an initial temperature true of `TK` does `6R` joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is `5//3`, the final temperature of the gas will be

Options

A. `(T+4)K`
B. `(T-4)K`
C. `(T+2.4)K`
D. `(T-2.4)`

Explanation

To solve the problem step by step, we will use the principles of thermodynamics, specifically focusing on the adiabatic process for an ideal gas.

### Step 1: Understand the given data
- We have 1 mole of an ideal gas.
- Initial temperature \( T_1 = T_K \) (in Kelvin).
- Work done \( W = 6R \) joules.
- The ratio of specific heats \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \).

### Step 2: Use the formula for work done in an adiabatic process
For an adiabatic process, the work done can be expressed as:
\[ W = \frac{R}{1 - \gamma} (T_2 - T_1) \]
Where:
- \( T_2 \) is the final temperature.
- \( T_1 \) is the initial temperature.
- \( R \) is the ideal gas constant.

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ 6R = \frac{R}{1 - \frac{5}{3}} (T_2 - T_K) \]

### Step 4: Simplify the equation
First, calculate \( 1 - \gamma \):
\[ 1 - \frac{5}{3} = \frac{3}{3} - \frac{5}{3} = -\frac{2}{3} \]
Now substitute this back into the equation:
\[ 6R = \frac{R}{-\frac{2}{3}} (T_2 - T_K) \]

### Step 5: Cancel \( R \) from both sides
Since \( R \) is a common factor, we can cancel it out:
\[ 6 = \frac{1}{-\frac{2}{3}} (T_2 - T_K) \]
This simplifies to:
\[ 6 = -\frac{3}{2} (T_2 - T_K) \]

### Step 6: Solve for \( T_2 - T_K \)
Now, multiply both sides by \(-\frac{2}{3}\):
\[ T_2 - T_K = -\frac{2}{3} \times 6 \]
\[ T_2 - T_K = -4 \]

### Step 7: Find the final temperature \( T_2 \)
Now, rearranging gives:
\[ T_2 = T_K - 4 \]

### Final Answer
The final temperature of the gas is:
\[ T_2 = T_K - 4 \text{ Kelvin} \]

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