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 –2.4) K
B. (T + 4)K
C. (T – 4) K
D. (T + 2.4)K

Explanation

To solve the problem, we need to find the final temperature of the gas after it has done work adiabatically. We will use the relationship between work done, initial temperature, final temperature, and the specific heat ratio.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature, \( T_1 = T \) (in Kelvin)
- Work done, \( W = 6R \) (in Joules)
- Ratio of specific heats, \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \)

2. **Use the Formula for Work Done in an Adiabatic Process:**
The work done by an ideal gas in an adiabatic process can be expressed as:
\[ W = \frac{R}{1 - \gamma} (T_2 - T_1) \]
Here, \( T_2 \) is the final temperature we want to find.

3. **Substitute the Known Values into the Equation:**
Plugging in the values we have:
\[ 6R = \frac{R}{1 - \frac{5}{3}} (T_2 - T) \]

4. **Simplify the Denominator:**
Calculate \( 1 - \frac{5}{3} \):
\[ 1 - \frac{5}{3} = \frac{3}{3} - \frac{5}{3} = -\frac{2}{3} \]
Thus, the equation becomes:
\[ 6R = \frac{R}{-\frac{2}{3}} (T_2 - T) \]

5. **Multiply Both Sides by \(-\frac{2}{3}\):**
\[ 6R \cdot -\frac{2}{3} = R(T_2 - T) \]
This simplifies to:
\[ -4R = R(T_2 - T) \]

6. **Divide Both Sides by \( R \):**
\[ -4 = T_2 - T \]

7. **Solve for \( T_2 \):**
Rearranging gives:
\[ T_2 = T - 4 \]

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