For ideal gas compressibility factor Z is:

Q: For ideal gas compressibility factor Z is:

The compressibility factor $ Z $ is a measure used to describe how much a real gas deviates from ideal gas behavior. For an ideal gas, the compressibility factor is defined as: \[ Z = \frac{PV}{nRT} \] where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the ideal gas constant, and $ T $ is the temperature. For an ideal gas, the equation of state is perfectly described by the ideal gas law: \[ PV = nRT \] Substituting this into the expression for $ Z $, we get: \[ Z = \frac{nRT}{nRT} = 1 \] Therefore, for an ideal gas, the compressibility factor $ Z $ is exactly 1. This is why Option 2 is the correct answer. Now, let's explain why the other options are incorrect: 1. **Option 1 (0):** A compressibility factor of 0 would imply that the product of pressure and volume is zero, which is not possible for a gas under normal conditions. This would mean either the pressure or the volume is zero, which is not feasible for a gas that occupies space and exerts pressure. 3. **Option 3 (greater than 1):** A compressibility factor greater than 1 indicates that the gas is less compressible than an ideal gas, often due to repulsive forces dominating at high pressures. This behavior is typical of real gases but not of an ideal gas, which by definition does not experience intermolecular forces. 4. **Option 4 (less than 0):** A compressibility factor less than 0 is not physically meaningful. It would imply negative pressure or volume, which is not possible for a gas. This option is not applicable to any real or ideal gas scenario. In conclusion, the compressibility factor $ Z $ for an ideal gas is exactly 1, making Option 2 the correct choice.

Question

Accepted Answer

The compressibility factor $ Z $ is a measure used to describe how much a real gas deviates from ideal gas behavior. For an ideal gas, the compressibility factor is defined as:

$$ Z = \frac{PV}{nRT} $$

where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the ideal gas constant, and $ T $ is the temperature. For an ideal gas, the equation of state is perfectly described by the ideal gas law:

$$ PV = nRT $$

Substituting this into the expression for $ Z $, we get:

$$ Z = \frac{nRT}{nRT} = 1 $$

Therefore, for an ideal gas, the compressibility factor $ Z $ is exactly 1. This is why Option 2 is the correct answer.

Now, let's explain why the other options are incorrect:

1. **Option 1 (0):** A compressibility factor of 0 would imply that the product of pressure and volume is zero, which is not possible for a gas under normal conditions. This would mean either the pressure or the volume is zero, which is not feasible for a gas that occupies space and exerts pressure.

3. **Option 3 (greater than 1):** A compressibility factor greater than 1 indicates that the gas is less compressible than an ideal gas, often due to repulsive forces dominating at high pressures. This behavior is typical of real gases but not of an ideal gas, which by definition does not experience intermolecular forces.

4. **Option 4 (less than 0):** A compressibility factor less than 0 is not physically meaningful. It would imply negative pressure or volume, which is not possible for a gas. This option is not applicable to any real or ideal gas scenario.

In conclusion, the compressibility factor $ Z $ for an ideal gas is exactly 1, making Option 2 the correct choice.