For an ideal gas in an isothermal process, temperature is constant, so internal energy \(Delta U = 0\). Thus (A) is false. In adiabatic expansion, work \(W > 0\) is done by the gas and heat \(Q = 0\), so \(Delta U = -W < 0\), meaning internal energy decreases. Thus (R) is false. Both (A) and (R) are false.

For free adiabatic expansion, \( Q = 0 \) and \( W = 0 \). According to the first law of thermodynamics, \( \Delta U = Q - W = 0 \). For an ideal gas, \( \Delta U = nC_v\Delta T \), so \( \Delta T = 0 \), meaning the final temperature is the same as the initial temperature. However, the final state (P,V,T) is not entirely the same as the initial state, only T is.

Thus, Assertion (A) is false. If temperature increases, work done by the gas can be positive, but it is not a direct consequence, and expansion (positive work) typically cools the gas. So, Reason (R) is also false. Therefore, both (A) and (R) are false.

In an adiabatic process, the system is perfectly insulated so that no heat exchange occurs \(Q = 0\). According to the First Law of Thermodynamics \(\Delta U = Q - W_{by}\), this means any change in internal energy is due entirely to work: \(\Delta U = -W_{by}\)

For a cyclic process, the system returns to its initial state. Internal energy \( U \) is a state function, meaning its change depends only on the initial and final states. Since the initial and final states are the same in a cyclic process, the change in internal energy \( \Delta U \) is zero. Thus, Assertion (A) is true. According to the first law of thermodynamics, \( \Delta U = Q - W \). Since \( \Delta U = 0 \), it implies \( Q = W \). For a cyclic process, net work done \( W \) is generally not zero (e.g., heat engines operate in cycles to produce net work). Thus, Reason (R) is false. Therefore, (A) is true but (R) is false.

The ideal gas equation \( PV = nRT \) describes the behavior of an ideal gas. Real gases approximate ideal gas behavior at low densities (and high temperatures) because intermolecular forces become negligible and the volume of gas molecules is tiny compared to the total volume.

Thus, Assertion (A) is true, and Reason (R) is true. Reason (R) directly explains why real gases obey the ideal gas equation at low densities. Therefore, (R) is the correct explanation of (A).

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.

For an ideal gas, internal energy \(U\) depends solely on temperature \(T\). Therefore, if \(T\) is constant, \(Delta U = 0\). Reason (R) correctly explains Assertion (A).

From the ideal gas equation \(PV = nRT\), if \(T = 0K\), then \(PV = 0\). Since volume \(V\) cannot be zero, pressure \(P\) must be zero. Reason (R) directly explains Assertion (A).

For an ideal monoatomic gas, \(C_v = \frac{3}{2}R\) as it only has 3 translational degrees of freedom. This number \(f=3\) remains constant with temperature. Thus, \(C_v\) is constant. Reason (R) correctly explains Assertion (A).

Assertion (A) is true by Boyle's Law (\( PV = \text{constant} \) at constant \( T \)). Reason (R) explains (A) microscopically: increasing volume at constant temperature reduces the density of molecules and thus the frequency of collisions with the piston, leading to decreased pressure.