A processor has 40 distinct instructions and 24 general purpose registers. A 32-bit instruction word has an opcode, two register operands and an immediate operand. The number of bits available for the immediate operand field is __________.

Let us calculate the bits required for each field of the 32-bit instruction:

1. Opcode field: To support 40 distinct instructions:

   $$\text{Opcode bits} = \lceil \log_2 40 \rceil = 6\text{ bits} \quad (2^5 = 32 < 40 \le 2^6 = 64)$$

2. Register operand fields: To address 24 general purpose registers:

   $$\text{Bits per register} = \lceil \log_2 24 \rceil = 5\text{ bits} \quad (2^4 = 16 < 24 \le 2^5 = 32)$$

   Since there are two register operands, they require:

   $$2 \times 5 = 10\text{ bits}$$

3. Immediate operand field:

   $$\text{Total instruction length} = 32\text{ bits}$$

   $$\text{Bits for immediate operand} = 32 - (\text{Opcode bits} + 2 \times \text{Register bits})$$

   $$\text{Bits for immediate operand} = 32 - (6 + 10) = 32 - 16 = 16\text{ bits}$$

Thus, the number of bits available for the immediate operand field is 16.