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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- Number of moles (n) = 1 mole
- Initial temperature (T_i) = T K
- Work done (W) = 6R joules
- Ratio of specific heats (γ) = 5/3

Since the process is adiabatic, we know that the heat transfer (ΔQ) is zero.

### Step 2: Use the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + W \]
Since ΔQ = 0 for an adiabatic process, we have:
\[ 0 = \Delta U + W \]
This implies:
\[ \Delta U = -W \]
Substituting the value of W:
\[ \Delta U = -6R \]

### Step 3: Relate Change in Internal Energy to Temperature Change
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where \(C_v\) is the molar specific heat at constant volume. For a gas with γ = C_p / C_v, we can express \(C_v\) as:
\[ C_v = \frac{R}{\gamma - 1} \]
Substituting γ = 5/3:
\[ C_v = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2} \]

### Step 4: Substitute ΔU into the Equation
Now substituting \(C_v\) into the equation for ΔU:
\[ \Delta U = n C_v \Delta T \Rightarrow -6R = 1 \cdot \frac{3R}{2} \Delta T \]
This simplifies to:
\[ -6R = \frac{3R}{2} \Delta T \]

### Step 5: Solve for ΔT
Now, we can solve for ΔT:
\[ \Delta T = \frac{-6R \cdot 2}{3R} = -4 \]

### Step 6: Find the Final Temperature
The change in temperature (ΔT) is related to the initial and final temperatures:
\[ \Delta T = T_f - T_i \]
Substituting ΔT = -4:
\[ -4 = T_f - T \]
Thus, we can rearrange to find the final temperature:
\[ T_f = T - 4 \]

### Conclusion
The final temperature of the gas is:
\[ \boxed{T - 4} \text{ K} \]