Question

Soft drinks are canned under 4 atm of \(\mathrm{CO}_{2}\) and release \(\mathrm{CO}_{2}\) when the can is opened. (a) How many moles of \(\mathrm{CO}_{2}\) are dissolved in \(355 \mathrm{~mL}\) of soda in a can before it is opened? (b) After the soda has gone flat? (c) What volume (in L) would the released \(\mathrm{CO}_{2}\) occupy at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\left(k_{\mathrm{H}}\right.\) for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.3 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{atm} ; P_{\mathrm{CO}_{2}}\) in air is \(4 \times 10^{-4}\) atm )?

Short answer

0.04686 moles of CO2 are dissolved in the soda before opening. The soda retains about 4.686 x 10^{-6} moles of CO2, and 1.1468 L of CO2 is released at 1 atm and 25°C.

Step-by-step solution

Determine the moles of \(\text{CO}_2\) dissolved in the soda before opening

Use Henry's law which states that \[ C = k_H \times P \] \[ C = (3.3 \times 10^{-2} \ \frac{mol}{L \ atm}) \times 4 \ atm = 0.132 \ \frac{mol}{L} \] The volume of the soda is 355 mL, convert it to liters: \[ 355 \ mL = 0.355 \ L \] Therefore, the moles of \(\text{CO}_2\) are: \[ n = C \times V = 0.132 \frac{mol}{L} \times 0.355 \ L = 0.04686 \ mol \]

Find moles of \(\text{CO}_2\) in soda after it goes flat

After the soda is flat, the partial pressure of \(\text{CO}_2\) is very low - almost equivalent to its partial pressure in air. Using Henry's law again: \[ C = (3.3 \times 10^{-2} \frac{mol}{L \ atm}) \times (4 \times 10^{-4} \ atm) = 1.32 \times 10^{-5} \frac{mol}{L} \] The moles of \(\text{CO}_2\) in 0.355 L is: \[ n = 1.32 \times 10^{-5} \frac{mol}{L} \times 0.355 \ L = 4.686 \times 10^{-6} \ mol \]

Calculate the volume of released \(\text{CO}_2\) at 1 atm and 25°C

Using the ideal gas law, \[ PV = nRT \] Constants are: \[ P = 1 \ atm, \ T = 25^{\circ} \text{C} = 298 \text{K}, \ R = 0.0821 \frac{L \ atm}{K \ mol} \] The amount of \(\text{CO}_2\) released is: \[n_{\text{released}} = n_{\text{initial}} - n_{\text{flat}} = 0.04686 \ mol - 4.686 \times 10^{-6} \ mol \approx 0.04686 \ mol \] Applying the values to the ideal gas law: \[ V = \frac{nRT}{P} = \frac{0.04686 \ mol \times 0.0821 \frac{L \ atm}{K \ mol} \times 298 \text{K}}{1 \ atm} = 1.1468 \ L \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, expressed as \( PV = nRT \). This equation relates the pressure \( P \), volume \( V \), the amount in moles \( n \), and temperature \( T \) of a gas. The constant \( R \), known as the Gas Constant, is generally \( 0.0821 \frac{L \times atm}{K \times mol} \). To use this law:

• Ensure the units match: Pressure in atm, Volume in liters, Temperature in Kelvin
• Convert temperatures to Kelvin by adding 273 to the Celsius temperature.

In our problem, the Ideal Gas Law helped us find the volume of \( \text{CO}_2 \) released when the soda is opened. We calculated the amount of \( \text{CO}_2 \) released, found it to be about 0.04686 mol, and then used \( PV = nRT \) to find the volume at standard pressure and temperature.

Partial Pressure

Partial Pressure refers to the pressure exerted by a single type of gas in a mixture of gases. According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of individual gases:

\[ P_{total} = P_1 + P_2 + \text{...} \]

Using this principle, if a soda can contains \( \text{CO}_2 \) under 4 atm, the partial pressure of \( \text{CO}_2 \) is 4 atm. When the soda goes flat, \( \text{CO}_2 \) equilibrates with its low partial pressure in the air, around \( 4 \times 10^{-4} \) atm.

This drastic drop in partial pressure explains why much less \( \text{CO}_2 \) remains dissolved in flat soda. Henry's Law helps quantify this relationship, showing how gas solubility in a liquid is proportional to its partial pressure above the liquid.

Gas Solubility

Gas solubility is governed by Henry's Law, which states the concentration of a gas in a liquid \( C \) is directly proportional to the partial pressure \( P \) of the gas above the liquid:

\[ C = k_H \times P \]

Here, \( k_H \) is Henry’s constant specific to the gas at a given temperature. For \( \text{CO}_2 \) at 25°C, \( k_H \) is given as \( 3.3 \times 10^{-2} \frac{mol}{L \times atm} \).

Using this law, the solubility of \( \text{CO}_2 \) in soda before opening the can (under 4 atm) was calculated to be 0.132 mol/L. After the soda goes flat, we used a partial pressure of \( 4 \times 10^{-4} \) atm to find the solubility as \( 1.32 \times 10^{-5} \frac{mol}{L} \).

Thus, understanding gas solubility is crucial to grasp how gases like \( \text{CO}_2 \) behave in solutions and how they escape when pressure drops, such as when you open a soda can.