Question

A gas has the van der Waals constants, \(\mathrm{a}=1.49 \mathrm{~L}^{2}\) \(\mathrm{atm} \mathrm{mol}^{-2}\) and \(\mathrm{b}=0.04 \mathrm{~L} \mathrm{~mol}^{-1} .\) Its Boyle's temperature is nearly: (a) \(50^{\circ} \mathrm{C}\) (b) \(354^{\circ} \mathrm{C}\) (c) \(454 \mathrm{~K}\) (d) \(408 \mathrm{~K}\)

Short answer

The Boyle's temperature is approximately 454 K.

Step-by-step solution

Understanding Boyle's Temperature

Boyle's temperature is the temperature at which a real gas behaves like an ideal gas over a range of pressures. Mathematically, it can be expressed as \( T_B = \frac{a}{Rb} \), where \( a \) and \( b \) are the van der Waals constants, and \( R \) is the ideal gas constant with a value of approximately \( 0.0821 \text{ L atm } \text{K}^{-1} \text{mol}^{-1} \).

Substituting Van der Waals Constants

Substitute the given values of \( a = 1.49 \text{ L}^2 \text{ atm mol}^{-2} \) and \( b = 0.04 \text{ L mol}^{-1} \) into the formula \( T_B = \frac{a}{Rb} \).

Calculating the Temperature in Kelvin

Calculate \( T_B \) using the equation: \( T_B = \frac{1.49}{0.0821 \times 0.04} \). This calculation gives \( T_B = \frac{1.49}{0.003284} \approx 454 \text{ K} \).

Converting Temperature if Needed

The calculated Boyle's temperature is already in Kelvin (\( 454 \text{ K} \)). If converting to Celsius is needed, use the formula \( T(\text{°C}) = T(\text{K}) - 273.15 \). However, since 454 K exactly matches one of the answer choices, no conversion is necessary.

Key concepts

Understanding Van der Waals Constants

Van der Waals constants are essential for characterizing real gases. They account for the non-ideal behavior of gases by introducing two parameters:

• a: This constant reflects the magnitude of intermolecular forces present. Larger values of 'a' signify stronger attractions between the molecules.
• b: Often termed as the volume correction, it accounts for the finite size of gas particles. Unlike ideal gas particles, which are point masses with no volume, real gas particles occupy space, and 'b' corrects for this.

Real gases deviate from ideal behavior, notably under high pressures and low temperatures. The van der Waals equation is a modification of the ideal gas law, integrating these constants:\[(P + \frac{an^2}{V^2})(V - nb) = nRT\]Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, and \(R\) is the ideal gas constant. The van der Waals equation allows for a more accurate prediction of gas behavior under conditions where the ideal gas law may fall short.

Ideal Gas Behavior

Ideal gases are a theoretical construct in which gases are assumed to behave perfectly under all conditions. The main assumptions for ideal gases include:

• Gas particles move randomly and collide elastically with container walls and each other.
• The volume of individual gas molecules is negligible compared to the volume of their container.
• No intermolecular forces are acting between the gas particles except for collisions.

Under certain conditions, such as high temperatures and low pressures, real gases tend to behave similarly to ideal gases. This is because:

• High temperatures energize molecules, making intermolecular attractions less impactful.
• Low pressure means molecules are far apart, reducing the significance of their volume and attractions.

Boyle's Temperature is critical as it marks a point where real gases approximate ideal behavior over a range of pressures, due to balanced forces and negligible volume effects.

Temperature Conversion Made Simple

Understanding temperature scales is vital in scientific calculations. The two primary scales commonly used are Kelvin (K) and Celsius (°C):

• Kelvin: The absolute temperature scale, starting from absolute zero, with no negative values. It's crucial in scientific work.
• Celsius: Based on the freezing and boiling points of water, 0°C and 100°C respectively.

Converting between these two scales is straightforward. For a Kelvin temperature \(T(K)\), the Celsius equivalent \(T(°C)\) is calculated using the formula:\[T(\text{°C}) = T(\text{K}) - 273.15\]Conversely,\[T(\text{K}) = T(\text{°C}) + 273.15\]Having a solid grasp of these conversion formulas ensures accuracy when solving problems, like determining Boyle's Temperature, where such conversions might be necessary.