Solution:

For an ideal gas undergoing an isothermal process, the temperature remains constant (\( Delta T = 0 \)). Since the internal energy of an ideal gas depends only on temperature, the change in internal energy \( Delta U = 0 \). Thus, Reason (R) is true. From the first law of thermodynamics, \( Delta U = Q - W_{by} \), where \( W_{by} \) is work done by the gas. Since \( Delta U = 0 \), we have \( Q = W_{by} \). Given that the gas absorbs \( 25 text{ J} \) heat, \( Q = +25 text{ J} \). Therefore, work done by the gas is \( W_{by} = +25 text{ J} \). Work done on the gas is \( W_{on} = -W_{by} = -25 text{ J} \). This matches Assertion (A), so (A) is true. Reason (R) correctly explains (A) because the condition \( Delta U = 0 \) is central to deriving the relationship between heat and work in this process.