For ideal gas compressibility factor Z is:

Q: For ideal gas compressibility factor Z is:

For an ideal gas, the compressibility factor $ Z $ is defined as the ratio of the molar volume of the gas to the molar volume of an ideal gas at the same temperature and pressure. Mathematically, it is expressed as: \[ Z = \frac{PV}{nRT} \] where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the universal gas constant, and $ T $ is the temperature. 1. **Explanation for Option 2 (Correct Answer):** - For an ideal gas, the behavior is perfectly described by the ideal gas law, $ PV = nRT $. When we substitute this into the expression for $ Z $, we get: \[ Z = \frac{PV}{nRT} = \frac{nRT}{nRT} = 1 \] - Therefore, for an ideal gas, the compressibility factor $ Z $ is exactly 1. This indicates that the gas behaves exactly as predicted by the ideal gas law without any deviations. 2. **Explanation for why other options are incorrect:** - **Option 1 (0):** A compressibility factor of 0 would imply that the gas occupies no volume or exerts no pressure, which is physically impossible for any real or ideal gas. - **Option 3 (greater than 1):** A compressibility factor greater than 1 indicates that the gas is less compressible than predicted by the ideal gas law, which is typical for real gases at high pressures. However, this does not apply to an ideal gas, which by definition does not deviate from the ideal gas law. - **Option 4 (less than 0):** A negative compressibility factor is not physically meaningful, as it would imply negative pressure or volume, which is not possible for any gas. In conclusion, for an ideal gas, the compressibility factor $ Z $ is exactly 1, making Option 2 the correct answer. This reflects the fact that ideal gases perfectly follow the ideal gas law without any deviations.

Question

Accepted Answer

For an ideal gas, the compressibility factor $ Z $ is defined as the ratio of the molar volume of the gas to the molar volume of an ideal gas at the same temperature and pressure. Mathematically, it is expressed as:

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

where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the universal gas constant, and $ T $ is the temperature.

1. **Explanation for Option 2 (Correct Answer):**
   - For an ideal gas, the behavior is perfectly described by the ideal gas law, $ PV = nRT $. When we substitute this into the expression for $ Z $, we get:
     $$ Z = \frac{PV}{nRT} = \frac{nRT}{nRT} = 1 $$
   - Therefore, for an ideal gas, the compressibility factor $ Z $ is exactly 1. This indicates that the gas behaves exactly as predicted by the ideal gas law without any deviations.

2. **Explanation for why other options are incorrect:**
   - **Option 1 (0):** A compressibility factor of 0 would imply that the gas occupies no volume or exerts no pressure, which is physically impossible for any real or ideal gas.
   - **Option 3 (greater than 1):** A compressibility factor greater than 1 indicates that the gas is less compressible than predicted by the ideal gas law, which is typical for real gases at high pressures. However, this does not apply to an ideal gas, which by definition does not deviate from the ideal gas law.
   - **Option 4 (less than 0):** A negative compressibility factor is not physically meaningful, as it would imply negative pressure or volume, which is not possible for any gas.

In conclusion, for an ideal gas, the compressibility factor $ Z $ is exactly 1, making Option 2 the correct answer. This reflects the fact that ideal gases perfectly follow the ideal gas law without any deviations.