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. `9/8RT_1`
B. `3/2RT_1`
C. `15/8RT_1`
D. `9/2RT_1`

Explanation

To find the work done during the adiabatic compression of helium gas from an initial volume of 5.6 liters to a final volume of 0.7 liters, we will follow these steps:

### Step 1: Identify Given Values
- Initial volume, \( V_1 = 5.6 \, \text{liters} \)
- Final volume, \( V_2 = 0.7 \, \text{liters} \)
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial temperature, \( T_1 = 273 \, \text{K} \)

### Step 2: Calculate the Number of Moles of Helium Gas
At STP (Standard Temperature and Pressure), 22.4 liters of an ideal gas is equivalent to 1 mole. Therefore, we can calculate the number of moles \( n \) of helium gas:

\[ n = \frac{V_1}{22.4} = \frac{5.6}{22.4} = \frac{1}{4} \, \text{moles} \]

### Step 3: Determine the Value of \( \gamma \)
For helium, which is a monatomic gas, the value of \( \gamma \) (the heat capacity ratio) is:

\[ \gamma = \frac{C_p}{C_v} = \frac{5}{3} \]

### Step 4: Calculate the Final Temperature \( T_2 \)
Using the adiabatic condition for an ideal gas, we have:

\[ T_2 V_2^{\gamma - 1} = T_1 V_1^{\gamma - 1} \]

Rearranging gives:

\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

Substituting the known values:

\[ T_2 = 273 \left( \frac{5.6}{0.7} \right)^{\frac{5}{3} - 1} \]

Calculating \( \frac{5.6}{0.7} = 8 \):

\[ T_2 = 273 \times 8^{\frac{2}{3}} \]

Calculating \( 8^{\frac{2}{3}} = 4 \):

\[ T_2 = 273 \times 4 = 1092 \, \text{K} \]

### Step 5: Calculate the Work Done \( W \)
The work done during an adiabatic process is given by the formula:

\[ W = n R \Delta T \frac{1}{\gamma - 1} \]

Where \( \Delta T = T_2 - T_1 \):

Calculating \( \Delta T \):

\[ \Delta T = 1092 - 273 = 819 \, \text{K} \]

Now substituting \( n = \frac{1}{4} \), \( R = 8.314 \, \text{J/(mol K)} \), and \( \gamma = \frac{5}{3} \):

\[ W = \frac{1}{4} \times 8.314 \times 819 \times \frac{1}{\frac{5}{3} - 1} \]

Calculating \( \frac{5}{3} - 1 = \frac{2}{3} \):

\[ W = \frac{1}{4} \times 8.314 \times 819 \times \frac{3}{2} \]

Calculating \( W \):

\[ W = \frac{1}{4} \times 8.314 \times 819 \times 1.5 \]

Calculating the numerical value:

\[ W \approx \frac{1}{4} \times 8.314 \times 1228.5 \approx \frac{1}{4} \times 10208.5 \approx 2552.125 \, \text{J} \]

### Final Answer
The work done in the process is approximately \( 2552.125 \, \text{J} \).

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