Question

The compressibility of a gas is less than unity at STP. IIence (1) \(V_{\mathrm{m}}>22.4 \mathrm{~L}\) (2) \(V_{m}<22.4 \mathrm{~L}\) (3) \(V_{\mathrm{m}}=22.4 \mathrm{~L}\) (4) \(V_{\mathrm{m}}=44.8 \mathrm{~L}\)

Short answer

Option (2) is correct. The molar volume (V_m) must be less than 22.4 L when compressibility (Z) is less than 1.

Step-by-step solution

Understand the Compressibility Factor

The compressibility factor (Z) of a gas is defined as follows: \[ Z = \frac{PV_m}{RT} \]Here, Z is the compressibility factor, P is the pressure, V_m is the molar volume, R is the gas constant, and T is the temperature.

Analyze the Given Information

According to the problem, the compressibility factor (Z) is less than unity (Z < 1) at STP (Standard Temperature and Pressure).

Use the Relation for STP

At STP, the standard molar volume (V_m) of an ideal gas is 22.4 L. For Z < 1,\[ Z = \frac{PV_m}{RT} < 1 \]Given P, R, and T are constants at STP, for Z to be less than 1, the molar volume (V_m) must be less than 22.4 L.

Match the Correct Option

From the analysis, if compressibility (Z) is less than 1, the molar volume (V_m) must be less than the ideal gas molar volume at STP (22.4 L). Therefore, the correct answer is:\[ V_m < 22.4 \text{ L} \]Thus, option (2) is correct.

Key concepts

compressibility factor

The compressibility factor, often represented as Z, helps us understand how much a real gas deviates from ideal gas behavior. Ideal gases are theoretical, and real gases largely follow this behavior, but not perfectly. The Z factor is calculated using this formula:
\( Z = \frac{PV_m}{RT} \)

Here,

• P is the pressure
• V_m is the molar volume
• R is the gas constant
• T is the temperature

When Z = 1, the gas behaves ideally. Most real gases exhibit Z values different from 1, especially under high pressure or low temperature. In our problem, Z < 1, indicating that the gas takes up less volume than anticipated for an ideal gas at the same conditions. In simpler terms, for Z < 1, the gas molecules attract each other more than they repel, causing them to be closer together, decreasing the molar volume.

STP conditions

STP or Standard Temperature and Pressure are common reference conditions in chemistry. They allow scientists to compare different gases easily. STP conditions are defined as follows:

• Temperature: 0°C or 273.15 K
• Pressure: 1 atm or 101.325 kPa

Under STP, an ideal gas occupies a volume of 22.4 liters per mole. When discussing gas behavior, these conditions provide a baseline for gauging and comparing results. In the context of our problem, stating that the compressibility factor Z < 1 at STP means that the molar volume of the real gas must be compared against the 22.4 L standard for ideal gases under these conditions.

molar volume

Molar volume is the volume occupied by one mole of a gas. For ideal gases at STP, this volume is standardized at 22.4 liters. The molar volume is derived from the ideal gas law:
\( PV = nRT \)
By rearranging the equation for one mole of an ideal gas, we get:
\( V_m = \frac{RT}{P} \)
At STP conditions, substituting in the values (R = 0.0821 L atm/K mol, T = 273.15 K, P = 1 atm) gives:
\( V_m = 22.4 \text{ L} \)
For real gases, the actual molar volume might differ due to interactions between gas molecules that are not accounted for in the ideal gas law. In this problem, because Z < 1, we know that the real molar volume must be less than the ideal molar volume of 22.4 L under STP conditions. Thus, the correct option would be V_m < 22.4 L.