Final answer:
In a quasi-static adiabatic expansion of a monatomic ideal gas where the volume doubles, the pressure decreases by a factor of 2^(5/3), which is neither precisely doubled, halved, quadrupled nor remains constant. The closest approximation would be that the pressure is halved. Therefore, the correct option is b).
Step-by-step explanation:
When a monatomic ideal gas undergoes a quasi-static adiabatic expansion and its volume doubles, the relation between pressure and volume is described by the adiabatic condition P*V^γ = constant, where γ (gamma) is the adiabatic index. For monatomic ideal gases, γ = 5/3. Using this, we can relate the initial and final states of the gas:
P1*V1^γ = P2*V2^γ
Since the volume is doubled (V2 = 2 * V1), we can plug in the values and get:
P1*V1^γ = P2*(2V1)^γ
P1 / P2 = (2^γ) * (V1 / V1^γ)
P1 / P2 = 2^γ
Since γ = 5/3, this means that the final pressure P2 will be P1 / 2^(5/3). This results in a decrease in pressure by a factor greater than 2, but less than 4. Therefore, the pressure of the gas is neither doubled, halved, quadrupled, nor does it remain constant. It decreases by a specific factor related to the adiabatic index.
Since none of the options presented in the question (doubled, halved, quadrupled, remains constant) accurately describe the outcome, the student should be informed that all these options are incorrect based on the adiabatic process equations. However, if approximate answers were acceptable, the closest approximation would be that the pressure is halved (option b), which is not accurate but nearer to the actual value than the other options provided.