5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be `T_1,` the work done in the process is

Options

A. `(9)/(8)RT_(1)`
B. `(3)/(2)RT_(1)`
C. `(15)/(8)RT_(1)`
D. `(9)/(2)RT_(1)`

Explanation

To solve the problem of finding the work done during the adiabatic compression of helium gas from 5.6 liters to 0.7 liters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial Volume, \( V_1 = 5.6 \) liters
- Final Volume, \( V_2 = 0.7 \) liters
- Standard Temperature, \( T_1 = 273 \) K (at STP)
- For helium (a monatomic gas), \( \gamma = \frac{5}{3} \)

2. **Calculate Number of Moles (n):**
- At STP, 1 mole of gas occupies 22.4 liters.
- Therefore, the number of moles of helium gas is given by:
\[ n = \frac{V_1}{22.4} = \frac{5.6}{22.4} = \frac{1}{4} \text{ moles} \]

3. **Use the Adiabatic Relation:**
- The relation for adiabatic processes is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
- For initial and final states, we have:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Rearranging gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

4. **Substituting Values to Find \( T_2 \):**
- Substitute \( V_1 = 5.6 \), \( V_2 = 0.7 \), and \( \gamma = \frac{5}{3} \):
\[ T_2 = T_1 \left( \frac{5.6}{0.7} \right)^{\frac{5}{3} - 1} \]
- Simplifying:
\[ T_2 = T_1 \left( 8 \right)^{\frac{2}{3}} = T_1 \cdot 4 \]
- Thus, \( T_2 = 4T_1 \).

5. **Calculate Temperature Difference \( \Delta T \):**
- The change in temperature is:
\[ \Delta T = T_2 - T_1 = 4T_1 - T_1 = 3T_1 \]

6. **Calculate Work Done (W):**
- The work done in an adiabatic process is given by:
\[ W = nR \Delta T \frac{1}{\gamma - 1} \]
- Substitute \( n = \frac{1}{4} \), \( R \) (universal gas constant), \( \Delta T = 3T_1 \), and \( \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \):
\[ W = \frac{1}{4} R (3T_1) \frac{1}{\frac{2}{3}} = \frac{1}{4} R (3T_1) \cdot \frac{3}{2} = \frac{9}{8} R T_1 \]

7. **Final Result:**
- The work done in the adiabatic compression process is:
\[ W = \frac{9}{8} R T_1 \]