Work Done in Adiabatic Compression of Helium Gas

To find the work done in the adiabatic compression of the helium gas, we can follow these steps:

Step 1: Calculate the number of moles of helium gas ($n$)
At Standard Temperature and Pressure (STP), $1\text{ mole}$ of an ideal gas occupies a volume of $22.4\text{ liters}$.
Given the initial volume $V_1 = 5.6\text{ liters}$, the number of moles $n$ is:
$n = \frac{V_1}{22.4} = \frac{5.6}{22.4} = \frac{1}{4}\text{ mole}$

Step 2: Determine the adiabatic exponent ($\gamma$)
Helium ($\text{He}$) is a monoatomic gas, so its adiabatic index is:
$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$

Step 3: Find the final temperature ($T_2$)
For an adiabatic process, the relation between temperature and volume is given by:
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

Substituting $V_1 = 5.6\text{ L}$, $V_2 = 0.7\text{ L}$, and $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$:
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} = T_1 \left( \frac{5.6}{0.7} \right)^{2/3} = T_1 (8)^{2/3}$

Since $8^{2/3} = \left(2^3\right)^{2/3} = 2^2 = 4$:
$T_2 = 4 T_1$

Step 4: Calculate the work done in the process
The work done in an adiabatic process is given by:
$W = \frac{n R (T_1 - T_2)}{\gamma - 1}$

Substituting the values into the formula:
$W = \frac{\frac{1}{4} R (T_1 - 4 T_1)}{\frac{5}{3} - 1} = \frac{\frac{1}{4} R (-3 T_1)}{\frac{2}{3}} = -\frac{9}{8} R T_1$

The magnitude of work done in compressing the gas (work done on the gas) is:
$|W| = \frac{9}{8} R T_1$

Therefore, the correct option is (A).