1. Let the molar heat capacity during the process be $C$.

2. For $1$ mole of an ideal monoatomic gas, $Q = C\,\Delta T$

3. The work done is given as $W = \frac{Q}{4}$

4. From the first law of thermodynamics, $Q = \Delta U + W$

5. For a monoatomic ideal gas, $\Delta U = nC_V\Delta T = \frac{3R}{2}\Delta T$ for $n=1$ mole.

6. Substitute into the first law: $Q = \frac{3R}{2}\Delta T + \frac{Q}{4}$

7. Rearranging, $Q - \frac{Q}{4} = \frac{3R}{2}\Delta T$ $\frac{3Q}{4} = \frac{3R}{2}\Delta T$

8. Hence, $Q = 2R\Delta T$

9. Comparing with $Q = C\Delta T$ we get $C = 2R$

10. Therefore, the required value is $\boxed{2}$ times $R$.

Comparison with stored answer:

• Derived answer: $2$
• Stored correct answer: $2$
• They match.