Question

The air in a hot-air balloon at 744 torr is heated from \(17^{\circ} \mathrm{C}\) to \(60.0^{\circ} \mathrm{C}\). Assuming that the amount (mol) of air and the pressure remain constant, what is the density of the air at each temperature? (The average molar mass of air is \(28.8 \mathrm{~g} / \mathrm{mol} .)\)

Short answer

At 17°C, the density is approximately 1.18 g/L. At 60°C, the density is approximately 1.03 g/L.

Step-by-step solution

Convert Temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^{\text{C}}) + 273.15 \] At 17^{\text{C}}, the temperature in Kelvin is: \[ T_1 = 17 + 273.15 = 290.15 \text{ K} \] At 60^{\text{C}}, the temperature in Kelvin is: \[ T_2 = 60 + 273.15 = 333.15 \text{ K} \]

Apply the Ideal Gas Law (Initial Condition)

Use the ideal gas law to find the initial density of air at 17^{\text{C}}. The ideal gas law is: \[ PV = nRT \] Rearrange to find density (\( \rho \)): \[ \rho = \frac{MP}{RT} \] where \( M \) is the molar mass, \( P \) is the pressure, \( R \) is the gas constant (0.0821 L·atm/(mol·K)), and \( T \) is the temperature.Given: Pressure (P) = 744 torr (convert to atm: 744 torr / 760 torr/atm = 0.979 atm) Molar Mass (M) = 28.8 g/mol Temperature (T_1) = 290.15 K\[ \rho_1 = \frac{28.8 \times 0.979}{0.0821 \times 290.15} \approx 1.18 \ g/L \]

Apply the Ideal Gas Law (Final Condition)

Now find the density of air at 60^{\text{C}}. Using the same formula: \[ \rho = \frac{MP}{RT} \] Given: Pressure (P) = 0.979 atm Molar Mass (M) = 28.8 g/mol Temperature (T_2) = 333.15 K\[ \rho_2 = \frac{28.8 \times 0.979}{0.0821 \times 333.15} \approx 1.03 \ g/L \]

Final Answer

The densities of air are: At 17^{\text{C}}, the density is approximately \( 1.18 \ g/L \). At 60^{\text{C}}, the density is approximately \( 1.03 \ g/L \).

Key concepts

Density

Density is a measure of how much mass is contained within a given volume. It’s usually expressed in grams per liter (g/L) or kilograms per cubic meter (kg/m³).
For a gas, density can be affected by pressure, temperature, and the type of gas itself. In general, increased temperature decreases density - this is why hot air balloons rise when the air inside them is heated!

In calculation, we use the formula: \(\rho = \frac{M \times P}{R \times T}\),
where \(M\) is molar mass, \(P\) is pressure, \(R\) is the gas constant, and \(T\) is temperature.

Ideal Gas Law

The ideal gas law is a fundamental equation that describes the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. The formula is: \[ PV = nRT \]
Here:
* \(P\) is the pressure
* \(V\) is the volume
* \(n\) is the number of moles
* \(R\) is the universal gas constant (\(0.0821 \textrm{L} \textrm{​atm} /\textrm{mol} \textrm{​ K}\)
* \(T\) is the temperature in Kelvin

We can rearrange this equation to find density by using the fact that \(n\) (moles) is equal to the mass (m) divided by the molar mass (M): \[ \rho = \frac{MP}{RT} \]
This is exactly what we did in the exercise to find the density at two different temperatures.

Temperature Conversion

In many scientific calculations, the temperature needs to be in Kelvin (K), not Celsius (°C). The conversion formula from Celsius to Kelvin is simple:
\(T(\textrm{K}) = T(\textrm{°C}) + 273.15\)

For example:
* 17°C = 17 + 273.15 = 290.15 K
* 60°C = 60 + 273.15 = 333.15 K
This conversion is crucial because the Kelvin scale is based on absolute zero, the theoretical point where particles have minimum thermal motion.

Pressure Units

Pressure is the force exerted per unit area. It can be expressed in different units, including atmospheres (atm), torr, and pascals (Pa).

In the problem, the pressure is given in torr, but we need it in atmospheres for the ideal gas law calculation. The conversion factor is:
\textrm{1 atm} = \textrm{760 torr} \.

Therefore:
744 torr can be converted to atmospheres with:
\(\frac{744 \text{ torr}}{760 \text{ torr/atm}} \approx 0.979 \text{ atm} \)
This helps standardize measurements, making calculations easier.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

For air, the average molar mass is given as 28.8 g/mol.
This value represents the weighted average of the molar masses of the different gases that make up air, such as nitrogen, oxygen, and argon.

The molar mass is used in the ideal gas law to calculate other properties like density.
We can apply it as:
\( \rho = \frac{M \times P} {R \times T}\),
where Molar Mass (M) is essential for determining the mass in this context.